NCERT Exemplar for Class 12 Maths Chapter-7 (Book Solutions)

Solved Examples

Short Answer (S.A.)

1. Integrate $\left( \frac{2a}{\sqrt{x}}-\frac{b}{{{x}^{2}}}+3c\sqrt[3]{{{x}^{2}}} \right)$ w.r.t. $x.$

Ans: let $I=~\int \left( \frac{2a}{\sqrt{x}}-\frac{b}{{{x}^{2}}}+3c\sqrt[3]{{{x}^{2}}} \right)dx$

⇒ $I=2a\int {{x}^{-\frac{1}{2}}}dx-b\int {{x}^{-2}}dx+3c\int {{x}^{\frac{2}{3}}}~dx$

⇒ $I=2a\times 2\times \sqrt{x}+\frac{b}{x}+\frac{9}{5}~c~{{x}^{\frac{5}{3}}}+C$

⇒ $I=4a\sqrt{x}+\frac{b}{x}+\frac{9}{5}~c~{{x}^{\frac{5}{3}}}+C$

Hence $\int \left( \frac{2a}{\sqrt{x}}-\frac{b}{{{x}^{2}}}+3c\sqrt[3]{{{x}^{2}}} \right)dx=4a\sqrt{x}+\frac{b}{x}+\frac{9}{5}~c~{{x}^{\frac{5}{3}}}+C.$

2. Evaluate $\int \frac{3ax}{{{b}^{2}}+{{c}^{2}}{{x}^{2}}}~dx$

Ans: let $I=\int \frac{3ax}{{{b}^{2}}+{{c}^{2}}{{x}^{2}}}~dx$

Now, put ${{b}^{2}}+{{c}^{2}}{{x}^{2}}=t$

⇒ $2{{c}^{2}}x~dx=dt$

⇒ $x~dx=\frac{dt}{2{{c}^{2}}}$

⇒ $3ax~dx=\frac{3a~dt}{2{{c}^{2}}}$

Therefore, 

⇒ $I=\frac{3a~}{2{{c}^{2}}}\int \frac{dt}{t}~$

⇒ $I=\frac{3a~}{2{{c}^{2}}}~\log \left| t \right|+C$

⇒ $I=\frac{3a~}{2{{c}^{2}}}~\log \left| {{b}^{2}}+{{c}^{2}}{{x}^{2}} \right|+C$

Hence, $\int \frac{3ax}{{{b}^{2}}+{{c}^{2}}{{x}^{2}}}~dx=\frac{3a~}{2{{c}^{2}}}~\log \left| {{b}^{2}}+{{c}^{2}}{{x}^{2}} \right|+C.$

3. Verify the following using the concept of integration as an antiderivative.

$\int \frac{{{x}^{3}}}{x+1}dx=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\log \left| x+1 \right|+C$ 

Ans: here, L.H.S$~=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\log \left| x+1 \right|+C~$

Now, differentiate w.r.t $x$ , we get

⇒ L.H.S$~=\frac{d}{dx}\left( x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\log \left| x+1 \right|+C \right)$

$=1-\frac{2x}{2}+\frac{3{{x}^{2}}}{3}-\frac{1}{\left( x+1 \right)}+0$

$=1-x+{{x}^{2}}-\frac{1}{\left( x+1 \right)}$

$~=\frac{\left( x+1 \right)-x\left( x+1 \right)+{{x}^{2}}\left( x+1 \right)-1}{\left( x+1 \right)}$

$=\frac{x+1-{{x}^{2}}-x+{{x}^{3}}+{{x}^{2}}-1}{\left( x+1 \right)}$

$=\frac{{{x}^{3}}}{\left( x+1 \right)}$

Therefore, $\left( x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\log \left| x+1 \right|+C \right)=\int \frac{{{x}^{3}}}{\left( x+1 \right)}~dx$

Hence verified.

4. Evaluate $\int \sqrt{\frac{1+x}{1-x}}~dx,$ $x\ne 1$.

Ans: Let $I=\int \sqrt{\frac{1+x}{1-x}}~dx$

⇒ $I=\int \sqrt{\frac{\left( 1+x \right)\left( 1+x \right)}{\left( 1-x \right)\left( 1+x \right)}}~dx$

⇒ $I=\int \sqrt{\frac{{{\left( 1+x \right)}^{2}}}{1-{{x}^{2}}}}~dx$

⇒ $I=\int \frac{1+x}{\sqrt{1-{{x}^{2}}}}~dx$

⇒ $I=\int \frac{1}{\sqrt{1-{{x}^{2}}}}~dx+\int \frac{x}{\sqrt{1-{{x}^{2}}}}~dx$

⇒ $I=si{{n}^{-1}}x+\int \frac{x}{\sqrt{1-{{x}^{2}}}}~dx$

Now, put $1-{{x}^{2}}={{t}^{2}}$

⇒ $-2x~dx=2t~dt$

⇒ $x~dx=-t~dt$

Therefore, 

⇒ $I=si{{n}^{-1}}x-\int \frac{t~dt}{t}~$

⇒ $I=si{{n}^{-1}}x-\int ~dt$

⇒ $I=si{{n}^{-1}}x-t+C$

⇒ $I=si{{n}^{-1}}x-\sqrt{1-{{x}^{2}}}+C$

Hence, $\int \sqrt{\frac{1+x}{1-x}}~dx=si{{n}^{-1}}x-\sqrt{1-{{x}^{2}}}+C$

5. Evaluate $\int \frac{dx}{\sqrt{\left( x-\alpha  \right)\left( \beta -x \right)}}$, $\beta >\alpha $

Ans: Let $I=\int \frac{dx}{\sqrt{\left( x-\alpha  \right)\left( \beta -x \right)}}$

Now, put $x-\alpha ={{t}^{2}}$

⇒ $dx=2t~dt$

And $\beta -x=\beta -\left( {{t}^{2}}+\alpha  \right)=\beta -\alpha -{{t}^{2}}$

Therefore, 

⇒ $I=\int \frac{2t~dt}{\sqrt{{{t}^{2}}\left( \beta -\alpha -{{t}^{2}} \right)}}$

⇒ $I=2\int \frac{~dt}{\sqrt{\left( \beta -\alpha -{{t}^{2}} \right)}}$

⇒ $I=2\int \frac{~dt}{\sqrt{\left[ {{\left( \sqrt{\beta -\alpha } \right)}^{2}}-{{t}^{2}} \right]}}$

⇒ $I=2~si{{n}^{-1}}\frac{t}{\sqrt{\beta -\alpha }}+C$

⇒ $I=2~si{{n}^{-1}}\frac{\sqrt{x-\alpha }}{\sqrt{\beta -\alpha }}+C$

Hence, $\int \frac{dx}{\sqrt{\left( x-\alpha  \right)\left( \beta -x \right)}}=2~si{{n}^{-1}}\frac{\sqrt{x-\alpha }}{\sqrt{\beta -\alpha }}+C.$

6. Evaluate $\int ta{{n}^{8}}x~se{{c}^{4}}x~dx$

Ans: let $I=\int ta{{n}^{8}}x~se{{c}^{4}}x~dx$

⇒ $I=\int ta{{n}^{8}}x~(se{{c}^{2}}x)~se{{c}^{2}}x~dx$

⇒ $I=\int ta{{n}^{8}}x~\left( 1+ta{{n}^{2}}x \right)~se{{c}^{2}}x~dx$

Now, put $\tan x=t$

⇒ $se{{c}^{2}}x~dx=dt$

Therefore, 

⇒ $I=\int {{t}^{8}}~\left( 1+{{t}^{2}} \right)~dt$

⇒ $I=\int \left( {{t}^{8}}+{{t}^{10}} \right)~~dt$

⇒ $I=\frac{{{t}^{9}}}{9}+\frac{{{t}^{11}}}{11}+C$

⇒ $I=\frac{ta{{n}^{9}}x}{9}+\frac{ta{{n}^{11}}x}{11}+C$

Hence, $\int ta{{n}^{8}}x~se{{c}^{4}}x~dx=\frac{ta{{n}^{9}}x}{9}+\frac{ta{{n}^{11}}x}{11}+C$

7. Find $\int \frac{{{x}^{3}}}{{{x}^{4}}+3{{x}^{2}}+2}dx$

Ans: Let $I=\int \frac{{{x}^{3}}}{{{x}^{4}}+3{{x}^{2}}+2}dx$

⇒ $I=\int \frac{{{x}^{2}}.x~dx}{{{x}^{4}}+3{{x}^{2}}+2}$

Now, put ${{x}^{2}}=t$

⇒ $2x~dx=dt$

⇒ $x~dx=\frac{dt}{2}~$

Therefore, 

⇒ $I=\frac{1}{2}\int \frac{t~dt}{{{t}^{2}}+3t+2}$

⇒ $I=\frac{1}{2}\int \frac{t~dt}{\left( t+1 \right)\left( t+2 \right)}$

Consider $\frac{t~}{\left( t+1 \right)\left( t+2 \right)}=\frac{A}{t+1}+\frac{B}{t+2}$

⇒ $\frac{t~}{\left( t+1 \right)\left( t+2 \right)}=\frac{At+2A+Bt+B}{\left( t+1 \right)\left( t+2 \right)}$

⇒ $\frac{t~}{\left( t+1 \right)\left( t+2 \right)}=\frac{\left( A+B \right)t+2A+B}{\left( t+1 \right)\left( t+2 \right)}$

⇒ $t=\left( A+B \right)t+2A+B$

On comparing both sides, we get

⇒ $A+B=1$ and $2A+B=0$

On solving, we get

⇒ $A=-1,~B=2$

Therefore, 

⇒ $I=\frac{1}{2}\int \frac{t~dt}{\left( t+1 \right)\left( t+2 \right)}$

⇒ $I=\frac{1}{2}\int \left[ \frac{-1}{t+1}+\frac{2}{t+2} \right]dt$

⇒ $I=\frac{1}{2}~\left( -\log \left| t+1 \right|+2\log \left| t+2 \right| \right)+C$

⇒ $I=-\frac{1}{2}\log \left| t+1 \right|+\log \left| t+2 \right|+C$

⇒ $I=-\log \left| \sqrt{t+1} \right|+\log \left| t+2 \right|+C$

⇒ $I=\log \left| \frac{t+2}{\sqrt{t+1}} \right|+C$

⇒ $I=\log \left| \frac{{{x}^{2}}+2}{\sqrt{{{x}^{2}}+1}} \right|+C$

Hence, $I=\int \frac{{{x}^{3}}}{{{x}^{4}}+3{{x}^{2}}+2}dx=\log \left| \frac{{{x}^{2}}+2}{\sqrt{{{x}^{2}}+1}} \right|+C$

8. Find $\int \frac{dx}{2~si{{n}^{2}}x+~5co{{s}^{2}}x}$

Ans: Let $I=\int \frac{dx}{2~si{{n}^{2}}x+~5co{{s}^{2}}x}$

Dividing numerator and denominator by $co{{s}^{2}}x,$ we get

⇒ $I=\int \frac{se{{c}^{2}}xdx}{2~ta{{n}^{2}}x+~5}$

Put $\tan x=t$

⇒ $se{{c}^{2}}x~dx=dt$

Therefore,

⇒ $I=\int \frac{dt}{2~{{t}^{2}}+~5}$

⇒ $I=\frac{1}{2}\int \frac{dt}{~{{t}^{2}}+\frac{5}{2}}$

⇒ $I=\frac{1}{2}\int \frac{dt}{~{{t}^{2}}+{{\left( \sqrt{\frac{5}{2}} \right)}^{2}}}$

⇒ $I=\frac{1}{2}\times \sqrt{\frac{2}{5}~}~ta{{n}^{-1}}\frac{\sqrt{2}t}{\sqrt{5}}+C$

⇒ $I=\frac{1}{\sqrt{10}}~ta{{n}^{-1}}\left( \frac{\sqrt{2}~tan~x}{\sqrt{5}} \right)+C$

Hence, $\int \frac{dx}{2~si{{n}^{2}}x+~5co{{s}^{2}}x}=\frac{1}{\sqrt{10}}~ta{{n}^{-1}}\left( \frac{\sqrt{2}~tan~x}{\sqrt{5}} \right)+C.$

9. Evaluate $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx$ as a limit of sums.

Ans: here, we have $a=-1$ , $b=2$ and $h=\frac{2+1}{n}\Rightarrow nh=3$

And $f\left( x \right)=7x-5$

Now, we have 

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( -1 \right)+f\left( -1+h \right)+f\left( -1+2h \right)\ldots +f(-1+\left( n-1 \right)h \right]$

Also we have, 

⇒ $f\left( -1 \right)=-7-5=-12$

⇒ $f\left( -1+h \right)=-7+7h-5=-12+7h$

⇒ $f\left( -1+\left( n-1 \right)h \right)=7\left( n-1 \right)h-12=$

Therefore, 

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\underset{h\to 0}{\mathop{\lim }}\,h\left[ -12+\left( -7h-12 \right)+\left( -14h-12 \right)\ldots +(7\left( n-1 \right)h-12 \right]$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 7h\left[ 1+2+\ldots +\left( n-10 \right) \right]-12n \right]$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 7h\times \frac{n\left( n-1 \right)}{2}-12n \right]$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\underset{h\to 0}{\mathop{\lim }}\,h\left[ \frac{7}{2}\left( nh \right)\left( nh-h \right)-12nh \right]$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\frac{7}{2}\times 3\times 3-12\times 3$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\frac{63}{2}-36$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\frac{63-72}{2}$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\frac{-9}{2}$

Hence, $\underset{-1}{\overset{2}{\mathop \int }}\,\left( 7x-5 \right)dx=\frac{-9}{2}.$

10. Evaluate $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{ta{{n}^{7}}x}{co{{t}^{7}}x+~ta{{n}^{7}}x}dx$

Ans: Let $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{ta{{n}^{7}}x}{co{{t}^{7}}x+~ta{{n}^{7}}x}dx$ …………. (i)

⇒ $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{ta{{n}^{7}}\left( \frac{\pi }{2}-x \right)}{co{{t}^{7}}\left( \frac{\pi }{2}-x \right)+~ta{{n}^{7}}\left( \frac{\pi }{2}-x \right)}dx$

⇒ $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{co{{t}^{7}}x}{ta{{n}^{7}}x+~co{{t}^{7}}x}dx$ …………….. (ii)

Now, adding eq (i) and (ii), we get 

⇒ $2I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{ta{{n}^{7}}x+~co{{t}^{7}}x}{ta{{n}^{7}}x+~co{{t}^{7}}x}dx$

⇒ $2I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,dx$

⇒ $2I=\left[ x \right]_{0}^{\frac{\pi }{2}}$

⇒ $2I=\frac{\pi }{2}$

⇒ $I=\frac{\pi }{4}$

Hence, $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{ta{{n}^{7}}x}{co{{t}^{7}}x+~ta{{n}^{7}}x}dx=\frac{\pi }{4}.$

11. Find $\underset{2}{\overset{8}{\mathop \int }}\,\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}~dx$

Ans: Let $I=\underset{2}{\overset{8}{\mathop \int }}\,\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}~dx$ ………….. (i)

⇒ $I=\underset{2}{\overset{8}{\mathop \int }}\,\frac{\sqrt{8+2-\left( 10-x \right)}}{\sqrt{8+2-x}+\sqrt{8+2-~\left( 10-x \right)}}~dx$

⇒ $I=\underset{2}{\overset{8}{\mathop \int }}\,\frac{\sqrt{10-10+x}}{\sqrt{10-x}+\sqrt{10-~10+x}}~dx$

⇒ $I=\underset{2}{\overset{8}{\mathop \int }}\,\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}~dx$ ……………… (ii)

Now, adding eq (i) and (ii), we get

⇒ $2I=\underset{2}{\overset{8}{\mathop \int }}\,\frac{\sqrt{10-x}~+\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}~dx$

⇒ $2I=\underset{2}{\overset{8}{\mathop \int }}\,dx$

⇒ $2I=\left[ x \right]_{2}^{8}$

⇒ $2I=8-2$

⇒ $2I=6$

⇒ $I=3$

Hence, $\underset{2}{\overset{8}{\mathop \int }}\,\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}~dx=3.$

12. Find $\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\sqrt{1+\sin 2x~}~dx$

Ans: Let $I=\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\sqrt{1+\sin 2x~}~dx$

⇒ $I=\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\sqrt{si{{n}^{2}}x+co{{s}^{2}}x+2~sinx~.~cos~x~}~dx$

⇒ $I=\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\sqrt{{{\left( \sin x+\cos x \right)}^{2}}~}~dx$

⇒ $I=\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\left( \sin x+\cos x \right)~dx$

⇒ $I=~\left[ -\cos x+\sin x \right]_{0}^{\frac{\pi }{4}}$

⇒ $I=\left[ -\cos \frac{\pi }{4}+\sin \frac{\pi }{4}-\left( -\cos 0+\sin 0 \right) \right]~$

⇒ $I=\left[ -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\left( -1+0 \right) \right]$

⇒ $I=1$

Hence, $\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\sqrt{1+\sin 2x~}~dx=1.$

13. Find $\int {{x}^{2}}ta{{n}^{-1}}x~dx$

Ans: Let $I=\int {{x}^{2}}ta{{n}^{-1}}x~dx$

⇒ $I=ta{{n}^{-1}}x\int {{x}^{2}}~dx-\int \left( \frac{d}{dx}\left( ta{{n}^{-1}}x \right)\int {{x}^{2}}~dx \right)dx$

⇒ $I=\frac{{{x}^{3}}}{3}ta{{n}^{-1}}x-\int \frac{1}{1+~{{x}^{2}}}.\frac{{{x}^{3}}}{3}~dx$

⇒ $I=\frac{{{x}^{3}}}{3}ta{{n}^{-1}}x-\frac{1}{3}\int \frac{{{x}^{3}}}{1+~{{x}^{2}}}~dx$

⇒ $I=\frac{{{x}^{3}}}{3}ta{{n}^{-1}}x-\frac{1}{3}\int \left( x-\frac{x}{1+~{{x}^{2}}} \right)~dx$

⇒ $I=\frac{{{x}^{3}}}{3}ta{{n}^{-1}}x-\frac{1}{3}\int x~dx+\frac{1}{3}\int \frac{x}{1+~{{x}^{2}}}~dx~$

⇒ $I=\frac{{{x}^{3}}}{3}ta{{n}^{-1}}x-\frac{{{x}^{2}}}{6}+\frac{1}{3}\int \frac{x}{1+~{{x}^{2}}}~dx$

Now, put $1+~{{x}^{2}}=t$

⇒ $2x~dx=dt$

⇒ $x~dx=\frac{dt}{2}$

⇒ $I=\frac{{{x}^{3}}}{3}ta{{n}^{-1}}x-\frac{{{x}^{2}}}{6}+\frac{1}{6}\int \frac{dt}{t}~$

⇒ $I=\frac{{{x}^{3}}}{3}ta{{n}^{-1}}x-\frac{{{x}^{2}}}{6}+\frac{1}{6}\log \left| t \right|+C$

⇒ $I=\frac{{{x}^{3}}}{3}ta{{n}^{-1}}x-\frac{{{x}^{2}}}{6}+\frac{1}{6}\log \left| 1+~{{x}^{2}} \right|+C$

14. Find $\int \sqrt{10-4x+4{{x}^{2}}~}~dx$

Ans: Let $I=\int \sqrt{10-4x+4{{x}^{2}}~}~dx$

⇒ $I=\int \sqrt{4{{x}^{2}}-4x+10~}~dx$

⇒ $I=\int \sqrt{4\left( {{x}^{2}}-x+\frac{10}{4} \right)~}~dx$

⇒ $I=2\int \sqrt{\left[ {{x}^{2}}-x+{{\left( \frac{1}{2} \right)}^{2}}-{{\left( \frac{1}{2} \right)}^{2}}+\frac{10}{4} \right]~}~dx$

⇒ $I=2\int \sqrt{\left[ {{\left( x-\frac{1}{2} \right)}^{2}}-\frac{1}{4}+\frac{10}{4} \right]~}~dx$

⇒ $I=2\int \sqrt{\left[ {{\left( x-\frac{1}{2} \right)}^{2}}+\frac{9}{4} \right]~}~dx$

⇒ $I=2\int \sqrt{\left[ {{\left( x-\frac{1}{2} \right)}^{2}}+{{\left( \frac{3}{2} \right)}^{2}} \right]~}~dx$

⇒ $I=~\left[ \frac{\left( x-\frac{1}{2} \right)}{2}\sqrt{4{{x}^{2}}-4x+10~}+\frac{{{\left( \frac{3}{2} \right)}^{2}}}{2}\log \left| \left( x-\frac{1}{2} \right)+\sqrt{4{{x}^{2}}-4x+10~} \right| \right]$

⇒ $I=\left( \frac{2x-1}{4} \right)\sqrt{4{{x}^{2}}-4x+10~}+\frac{9}{4}\log \left| \left( x-\frac{1}{2} \right)+\sqrt{4{{x}^{2}}-4x+10~} \right|$

Long Answer (L.A.)

15. Evaluate $\int \frac{{{x}^{2}}~dx}{{{x}^{4}}+{{x}^{2}}-2}$

Ans: Let $I=\int \frac{{{x}^{2}}~dx}{{{x}^{4}}+{{x}^{2}}-2}$

Now, put ${{x}^{2}}=t$

Therefore, 

⇒ $\frac{{{x}^{2}}~}{{{x}^{4}}+{{x}^{2}}-2}=\frac{t~}{{{t}^{2}}+t-2}=\frac{t~}{\left( t+2 \right)\left( t-1 \right)}$

⇒ $\frac{t~}{\left( t+2 \right)\left( t-1 \right)}=\frac{A}{t+2~}+\frac{B}{t-1~}$

⇒ $\frac{t~}{\left( t+2 \right)\left( t-1 \right)}=\frac{At-A+Bt+2B}{\left( t+2 \right)\left( t-1 \right)~}$

⇒ $t=\left( A+B \right)t+2B-A$

On comparing both side, we get

⇒ $A+B=1$ and $2B-A=0$

On solving these equations, we get

⇒ $B=\frac{1}{3}~$ and $A=\frac{2}{3}$

So, $\frac{{{x}^{2}}~}{{{x}^{4}}+{{x}^{2}}-2}=\frac{2}{3}\frac{1}{{{x}^{2}}+2}+\frac{1}{3}\frac{1}{{{x}^{2}}-1}$

Therefore, 

⇒ $I=\int \left( \frac{2}{3}\frac{1}{{{x}^{2}}+2}+\frac{1}{3}\frac{1}{{{x}^{2}}-1} \right)dx$

⇒ $I=\frac{2}{3}\int \frac{1}{{{x}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}dx+~\frac{1}{3}\int \frac{1}{{{x}^{2}}-1}dx$

⇒ $I=\frac{2}{3}~\times \frac{1}{\sqrt{2}}~ta{{n}^{-1}}\frac{x}{\sqrt{2}}+\frac{1}{3}~\times \frac{1}{2}\log \left| \frac{x-1}{x+1} \right|+C$

⇒ $I=\frac{\sqrt{2}}{3}~~ta{{n}^{-1}}\frac{x}{\sqrt{2}}+\frac{1}{6}~\log \left| \frac{x-1}{x+1} \right|+C$

Hence, $\int \frac{{{x}^{2}}~dx}{{{x}^{4}}+{{x}^{2}}-2}=\frac{\sqrt{2}}{3}~~ta{{n}^{-1}}\frac{x}{\sqrt{2}}+\frac{1}{6}~\log \left| \frac{x-1}{x+1} \right|+C.$

16. Evaluate $\int \frac{{{x}^{3}}+x}{{{x}^{4}}-9~}~dx$

Ans: Let $I=\int \frac{{{x}^{3}}+x}{{{x}^{4}}-9~}~dx$

⇒ $I=\int \frac{{{x}^{3}}}{{{x}^{4}}-9~}~dx+\int \frac{x}{{{x}^{4}}-9~}~dx$

⇒ $I={{I}_{1}}+{{I}_{2}}$ ………….. (i)

where ${{I}_{1}}=\int \frac{{{x}^{3}}}{{{x}^{4}}-9~}~dx$ and ${{I}_{2}}=\int \frac{x}{{{x}^{4}}-9~}~dx$

Now, ${{I}_{1}}=\int \frac{{{x}^{3}}}{{{x}^{4}}-9~}~dx$

Put ${{x}^{4}}-9=t$

⇒ $4{{x}^{3}}~dx=dt$

⇒ ${{x}^{3}}~dx=\frac{dt}{4}$

Therefore, 

⇒ ${{I}_{1}}=\frac{1}{4}\int \frac{dt}{t~}~$

⇒ ${{I}_{1}}=\frac{1}{4}\log \left| t \right|+{{C}_{1}}$

⇒ ${{I}_{1}}=\frac{1}{4}\log \left| {{x}^{4}}-9 \right|+{{C}_{1}}$ ……….. (ii)

Now, ${{I}_{2}}=\int \frac{x}{{{x}^{4}}-9~}~dx$

Put ${{x}^{2}}=t$

⇒ $2x~dx=dt$

⇒ $x~dx=\frac{dt}{2}$

Therefore, 

⇒ ${{I}_{2}}=\frac{1}{2}\int \frac{dt}{{{t}^{2}}-{{3}^{2}}~}~$

⇒ ${{I}_{2}}=\frac{1}{2}\times \frac{1}{2\times 3}\log \left| \frac{t-3}{t+3} \right|+{{C}_{2}}$

⇒ ${{I}_{2}}=\frac{1}{12}\log \left| \frac{{{x}^{2}}-3}{{{x}^{2}}+3} \right|+{{C}_{2}}$

Now, put the value of ${{I}_{1}}$ and ${{I}_{2}}$ in eq (i), we get 

⇒  $I=\frac{1}{4}\log \left| {{x}^{4}}-9 \right|+{{C}_{1}}+\frac{1}{12}\log \left| \frac{{{x}^{2}}-3}{{{x}^{2}}+3} \right|+{{C}_{2}}$

⇒  $I=\frac{1}{4}\log \left| {{x}^{4}}-9 \right|+\frac{1}{12}\log \left| \frac{{{x}^{2}}-3}{{{x}^{2}}+3} \right|+C$

17. Show that $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{si{{n}^{2}}x~dx}{\sin x+\cos x}=\frac{1}{\sqrt{2}}\log \left( \sqrt{2}+1 \right)$

Ans: let $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{si{{n}^{2}}x~dx}{\sin x+\cos x}~$ ………. (i)

⇒ $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{si{{n}^{2}}\left( \frac{\pi }{2}-x \right)~dx}{\sin \left( \frac{\pi }{2}-x \right)+\cos \left( \frac{\pi }{2}-x \right)}$

⇒ $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{co{{s}^{2}}x~dx}{\cos x+\sin x}~$…………. (ii)

Adding eq (i) and (ii), we get 

⇒ $2I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\left( si{{n}^{2}}x+co{{s}^{2}}x \right)~dx}{\cos x+\sin x}$

⇒ $2I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{~dx}{\cos x+\sin x}$

⇒ $2I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{~dx}{\cos x~\frac{1}{\sqrt{2}}+\sin x.~\frac{1}{\sqrt{2}}}$

⇒ $2I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{~dx}{\cos x~\cos ~\frac{\pi }{4}~+\sin x.~\text{sin }\!\!~\!\!\text{ }\frac{\pi }{4}~}$

⇒ $2I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{~dx}{\cos \left( x-~\frac{\pi }{4} \right)~}$

⇒ $2I=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\sec \left( x-~\frac{\pi }{4} \right)~dx~$

⇒ $2I=\frac{1}{\sqrt{2}}\left[ \log \left\{ \sec \left( x-~\frac{\pi }{4} \right)+\tan \left( x-~\frac{\pi }{4} \right)~~~ \right\} \right]_{0}^{\frac{\pi }{2}}$

⇒ $2I=\frac{1}{\sqrt{2}}\left[ \log \left\{ \sec \left( \frac{\pi }{2}-~\frac{\pi }{4} \right)+\tan \left( \frac{\pi }{2}-~\frac{\pi }{4} \right)~~~ \right\}-\log \left\{ \sec \left( -~\frac{\pi }{4} \right)+\tan \left( -~\frac{\pi }{4} \right)~~~ \right\} \right]$

⇒ $2I=\frac{1}{\sqrt{2}}\left[ \log \left\{ \sec \left( ~\frac{\pi }{4} \right)+\tan \left( ~\frac{\pi }{4} \right)~~~ \right\}-\log \left\{ \sec \left( ~\frac{\pi }{4} \right)-\tan \left( ~\frac{\pi }{4} \right)~~~ \right\} \right]$

⇒ $2I=\frac{1}{\sqrt{2}}\left[ \log \left( \sqrt{2}+1 \right)-\log \left( \sqrt{2}-1 \right) \right]$

⇒ $2I=\frac{1}{\sqrt{2}}~\times \log \left| \frac{\sqrt{2}+1}{\sqrt{2}-1} \right|$

⇒ $2I=\frac{1}{\sqrt{2}}~\times \log \left| \frac{{{\left( \sqrt{2}+1 \right)}^{2}}}{2-1} \right|$

⇒ $2I=\frac{1}{\sqrt{2}}~\times 2\log \left( \sqrt{2}+1 \right)$

⇒ $I=\frac{1}{\sqrt{2}}~\log \left( \sqrt{2}+1 \right)$

Hence proved 

18. Find $\underset{0}{\overset{1}{\mathop \int }}\,x{{\left( ta{{n}^{-1}}x \right)}^{2}}~dx$

Ans: Let $I=\underset{0}{\overset{1}{\mathop \int }}\,x{{\left( ta{{n}^{-1}}x \right)}^{2}}~dx$

Applying integrating by parts, we get

⇒ $I=\left[ \frac{{{x}^{2}}}{2}{{\left( ta{{n}^{-1}}x \right)}^{2}} \right]_{0}^{1}~-\underset{0}{\overset{1}{\mathop \int }}\,\frac{2~ta{{n}^{-1}}x}{1~+~{{x}^{2}}}.~\frac{{{x}^{2}}}{2}dx$ 

⇒ $I=\left[ \frac{1}{2}{{\left( ta{{n}^{-1}}1 \right)}^{2}}-0 \right]~-\underset{0}{\overset{1}{\mathop \int }}\,\frac{~ta{{n}^{-1}}x}{1~+~{{x}^{2}}}{{x}^{2}}dx$

⇒ $I=\left[ \frac{1}{2}{{\left( \frac{\pi }{4} \right)}^{2}} \right]~-\underset{0}{\overset{1}{\mathop \int }}\,\frac{~\left( {{x}^{2}}+1-1 \right)ta{{n}^{-1}}x}{1~+~{{x}^{2}}}dx$

⇒ $I=\frac{{{\pi }^{2}}}{32}~-\underset{0}{\overset{1}{\mathop \int }}\,\frac{~\left( {{x}^{2}}+1-1 \right)ta{{n}^{-1}}x}{1~+~{{x}^{2}}}dx$

⇒ $I=\frac{{{\pi }^{2}}}{32}~-\underset{0}{\overset{1}{\mathop \int }}\,\frac{~\left( {{x}^{2}}+1 \right)ta{{n}^{-1}}x}{1~+~{{x}^{2}}}dx+\underset{0}{\overset{1}{\mathop \int }}\,\frac{~ta{{n}^{-1}}x}{1~+~{{x}^{2}}}dx~$

⇒ $I=\frac{{{\pi }^{2}}}{32}~-\underset{0}{\overset{1}{\mathop \int }}\,ta{{n}^{-1}}x~dx+\underset{0}{\overset{1}{\mathop \int }}\,\frac{~ta{{n}^{-1}}x}{1~+~{{x}^{2}}}dx~$

⇒ $I=\frac{{{\pi }^{2}}}{32}~-{{I}_{1}}+{{I}_{2}}$ ……….. (i)

Where, ${{I}_{1}}=\underset{0}{\overset{1}{\mathop \int }}\,ta{{n}^{-1}}x~dx$ and ${{I}_{2}}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{~ta{{n}^{-1}}x}{1~+~{{x}^{2}}}dx$

Now, ${{I}_{1}}=\underset{0}{\overset{1}{\mathop \int }}\,ta{{n}^{-1}}x~dx$

⇒ ${{I}_{1}}=\left[ x~ta{{n}^{-1}}x \right]_{0}^{1}-~\underset{0}{\overset{1}{\mathop \int }}\,\frac{x}{1+{{x}^{2}}}~dx$

⇒ ${{I}_{1}}=\left[ 1.~ta{{n}^{-1}}1-0 \right]-~\underset{0}{\overset{1}{\mathop \int }}\,\frac{x}{1+{{x}^{2}}}~dx$

⇒ ${{I}_{1}}=\frac{\pi }{4}-~\underset{0}{\overset{1}{\mathop \int }}\,\frac{x}{1+{{x}^{2}}}~dx$

Now, put $1+{{x}^{2}}=t$

⇒ $2x~dx=dt$

⇒ $x~dx=\frac{dt}{2}$

And at $x=0~\to t=1$ 

⇒ at $x=1~\to t=2$

Therefore, 

⇒ ${{I}_{1}}=\frac{\pi }{4}-\frac{1}{2}\underset{1}{\overset{2}{\mathop \int }}\,\frac{dt}{t}~$

⇒ ${{I}_{1}}=\frac{\pi }{4}-\frac{1}{2}\left[ \log t \right]_{1}^{2}$

⇒ ${{I}_{1}}=\frac{\pi }{4}-\frac{1}{2}\left[ \log 2-\log 1 \right]$

⇒ ${{I}_{1}}=\frac{\pi }{4}-\frac{1}{2}\log 2$ ………….. (ii)

Now, ${{I}_{2}}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{~ta{{n}^{-1}}x}{1~+~{{x}^{2}}}dx$

Put $ta{{n}^{-1}}x=t$

⇒ $\frac{1}{1+{{x}^{2}}}~dx=dt$

And at $x=0\to t=0$

⇒ at $x=1\to t=\frac{\pi }{4}$ 

Therefore, 

⇒ ${{I}_{2}}=\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,t~dt$

⇒ ${{I}_{2}}=\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{\frac{\pi }{4}}$

⇒ ${{I}_{2}}=\frac{1}{2}\left[ {{t}^{2}} \right]_{0}^{\frac{\pi }{4}}$

⇒ ${{I}_{2}}=\frac{1}{2}\left[ \frac{{{\pi }^{2}}}{16}-0 \right]$

⇒ ${{I}_{2}}=\frac{{{\pi }^{2}}}{32}$

Now, put the value of ${{I}_{2}}$ and ${{I}_{2}}$ in eq(i), we get

⇒ $I=\frac{{{\pi }^{2}}}{32}~-\frac{\pi }{4}+\frac{1}{2}\log 2+\frac{{{\pi }^{2}}}{32}$

⇒ $I=\frac{{{\pi }^{2}}}{16}~-\frac{\pi }{4}+\frac{1}{2}\log 2$

Hence, $\underset{0}{\overset{1}{\mathop \int }}\,x{{\left( ta{{n}^{-1}}x \right)}^{2}}~dx=\frac{{{\pi }^{2}}}{16}~-\frac{\pi }{4}+\frac{1}{2}\log 2.$

19. Evaluate $\underset{-1}{\overset{2}{\mathop \int }}\,f\left( x \right)dx$, where $f\left( x \right)=~\left| x+1 \right|+\left| x \right|+\left| x-1 \right|$.

Ans: here, we can redefine $f\left( x \right)$ as 

f\left( x \right)=\left\{ \begin{matrix} 2-x,~~~~if-1<x\le 0  \\ x+2,~~~if~~0<x\le 1  \\   3x,~~~~~~~~~~~if~1<x\le 2  \\ \end{matrix} \right.$

Therefore, $\underset{-1}{\overset{2}{\mathop \int }}\,f\left( x \right)dx=~\underset{-1}{\overset{0}{\mathop \int }}\,\left( 2-x \right)~dx+\underset{0}{\overset{1}{\mathop \int }}\,\left( x+2 \right)dx+\underset{1}{\overset{2}{\mathop \int }}\,3x~dx$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,f\left( x \right)dx=~\left[ 2x-\frac{{{x}^{2}}}{2} \right]_{-1}^{0}+\left[ \frac{{{x}^{2}}}{2}+2x \right]_{0}^{1}+3\left[ \frac{{{x}^{2}}}{2} \right]_{1}^{2}$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,f\left( x \right)dx=~\left[ 0-\left( -2-\frac{1}{2} \right) \right]+\left[ \left( \frac{1}{2}+2 \right)-0 \right]+3\left[ 2-\frac{1}{2} \right]$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,f\left( x \right)dx=\frac{5}{2}+\frac{5}{2}+\frac{9}{2}$

⇒ $\underset{-1}{\overset{2}{\mathop \int }}\,f\left( x \right)dx=\frac{19}{2}$.

Objective Type Questions 

Choose the correct answer from the given four option in each of the Examples from 20 to 30.

20. $\int {{e}^{x}}\left( \cos x-\sin x \right)~dx$ is equal to

(A) ${{e}^{x}}\cos x+C$                               

(B) ${{e}^{x}}\sin x+C$         

(C) $-{{e}^{x}}\cos x+C$                           

(D) $-{{e}^{x}}\sin x+C$       

Ans: here, as we know that $\int {{e}^{x}}\left[ f\left( x \right)+{f}'\left( x \right) \right]~dx={{e}^{x}}~f\left( x \right)+C~$

Now, if $f\left( x \right)=\cos x$, then ${f}'\left( x \right)=-\sin x$

Therefore, $\int {{e}^{x}}\left( \cos x-\sin x \right)~dx={{e}^{x}}\cos x+C$                               

Hence, option (A) is the correct answer.

21. $\int \frac{dx}{si{{n}^{2}}x~~co{{s}^{2}}x}$ is equal to

(A)  $\tan x+\cot x+C$                       

(B) ${{\left( \tan x+\cot x \right)}^{2}}+C$       

(C)   $\tan x-\cot x+C$                      

(D)  ${{\left( \tan x-\cot x \right)}^{2}}+C$       

Ans: Let $I=\int \frac{dx}{si{{n}^{2}}x~~co{{s}^{2}}x}$

⇒ $I=\int \frac{\left( si{{n}^{2}}x~+~co{{s}^{2}}x \right)~dx}{si{{n}^{2}}x~~co{{s}^{2}}x}$

⇒ $I=\int \frac{si{{n}^{2}}x~dx}{si{{n}^{2}}x~~co{{s}^{2}}x}+\int \frac{co{{s}^{2}}x~dx}{si{{n}^{2}}x~~co{{s}^{2}}x}$

⇒ $I=\int \frac{~dx}{~~co{{s}^{2}}x}+\int \frac{~dx}{si{{n}^{2}}x~~}$

⇒ $I=\int se{{c}^{2}}x~dx+\int cose{{c}^{2}}x~dx$

⇒ $I=\tan x-\cot x+C$

Hence, option (C) is the correct answer.

22. If $\int \frac{3{{e}^{x}}-5{{e}^{-x}}}{4{{e}^{x}}+5{{e}^{-x}}}~dx=ax+b\log \left| 4{{e}^{x}}+5{{e}^{-x}} \right|+C$, then 

(A)  $a=-\frac{1}{8}~,~b=\frac{7}{8}$                    

(B) $a=\frac{1}{8}~,~b=\frac{7}{8}$

(C) $a=-\frac{1}{8}~,~b=-\frac{7}{8}$                  

(D) $a=\frac{1}{8}~,~b=-\frac{7}{8}$

Ans: here, we have $\int \frac{3{{e}^{x}}-5{{e}^{-x}}}{4{{e}^{x}}+5{{e}^{-x}}}~dx=ax+b\log \left| 4{{e}^{x}}+5{{e}^{-x}} \right|+C$

Now, differentiating both sides, we have 

⇒ $\frac{d}{dx}\left( \int \frac{3{{e}^{x}}-5{{e}^{-x}}}{4{{e}^{x}}+5{{e}^{-x}}}~dx \right)=\frac{d}{dx}\left( ax+b\log \left| 4{{e}^{x}}+5{{e}^{-x}} \right|+C \right)$

⇒ $\frac{3{{e}^{x}}-5{{e}^{-x}}}{4{{e}^{x}}+5{{e}^{-x}}}=a+b\times \frac{1}{4{{e}^{x}}+5{{e}^{-x}}}\times \left( 4{{e}^{x}}-5{{e}^{-x}} \right)$

⇒ $\frac{3{{e}^{x}}-5{{e}^{-x}}}{4{{e}^{x}}+5{{e}^{-x}}}=a+b\left( \frac{4{{e}^{x}}-5{{e}^{-x}}}{4{{e}^{x}}+5{{e}^{-x}}} \right)$

⇒ $\left( 3{{e}^{x}}-5{{e}^{-x}} \right)=a\left( 4{{e}^{x}}+5{{e}^{-x}} \right)+b\left( 4{{e}^{x}}-5{{e}^{-x}} \right)$

⇒ $\left( 3{{e}^{x}}-5{{e}^{-x}} \right)=4a{{e}^{x}}+5a{{e}^{-x}}+4b{{e}^{x}}-5b{{e}^{-x}}$

⇒ $3{{e}^{x}}-5{{e}^{-x}}=4{{e}^{x}}\left( a+b \right)-5{{e}^{-x}}\left( -a+b \right)$

Now, comparing both sides, we get

⇒ $4\left( a+b \right)=3$ and $\left( -a+b \right)=1$

⇒ $\left( a+b \right)=\frac{3}{4}$ …… (i)  and $\left( -a+b \right)=1$ ………(ii)

Adding eq (i) and (ii), we get 

⇒ $2b=\frac{7}{4}$

⇒ $b=\frac{7}{8}$. 

Now, put the value of $b$ in eq (ii), 

⇒$~-a+\frac{7}{8}=1$

⇒$~-a=1-\frac{7}{8}$

⇒$~a=-\frac{1}{8}$

Here, $a=-\frac{1}{8}$ and $b=\frac{7}{8}$

Hence, option (A) is the correct answer.

23: $\underset{a+c}{\overset{b+c}{\mathop \int }}\,f\left( x \right)dx~$is equal to 

(A)  $\underset{a}{\overset{b}{\mathop \int }}\,f\left( x-c \right)dx$                           

(B) $\underset{a}{\overset{b}{\mathop \int }}\,f\left( x+c \right)dx$

(C) $\underset{a}{\overset{b}{\mathop \int }}\,f\left( x \right)dx$                                    

(D) $\underset{a-c}{\overset{b-c}{\mathop \int }}\,f\left( x \right)dx$

Ans: Let $I=\underset{a+c}{\overset{b+c}{\mathop \int }}\,f\left( x \right)dx$

Now, put $x=t+c$

⇒ $dx=dt$

And at $x=a+c$ ⇒ $t=a$

⇒ at $x=b+c$ ⇒ $t=b$

Therefore, 

⇒ $I=\underset{a}{\overset{b}{\mathop \int }}\,f\left( t+c \right)dt$

⇒ $I=\underset{a}{\overset{b}{\mathop \int }}\,f\left( x+c \right)dx$

Hence, option (B) is the correct answer.

24. If $f$ and $g$ are continuous function in $\left[ 0,~1 \right]$ satisfying $f\left( x \right)=f\left( a-x \right)$ and $g\left( x \right)+g\left( a-x \right)=a,~$then $\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right).g\left( x \right)dx$ is equal to

(A) $\frac{a}{2}~$                                                          

(B) $\frac{a}{2}~\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right)~dx$

(C) $\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right)~dx$                                         

(D) $a\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right)~dx$

Ans: Let $I=\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right).g\left( x \right)dx$

⇒ $I=\underset{0}{\overset{a}{\mathop \int }}\,f\left( x-a \right).g\left( x-a \right)dx$

⇒ $I=\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right).\left( a-g\left( x \right) \right)dx$

⇒ $I=\underset{0}{\overset{a}{\mathop \int }}\,\left[ a~f\left( x \right)-f\left( x \right).g\left( x \right) \right]dx$

⇒ $I=a\underset{0}{\overset{a}{\mathop \int }}\,~f\left( x \right)dx-\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right).g\left( x \right)dx$

⇒ $I=a\underset{0}{\overset{a}{\mathop \int }}\,~f\left( x \right)dx-I$

⇒ $2I=a\underset{0}{\overset{a}{\mathop \int }}\,~f\left( x \right)dx$

⇒ $I=\frac{a}{2}\underset{0}{\overset{a}{\mathop \int }}\,~f\left( x \right)dx$

Hence, option (B) is the correct answer.

25: If $x=\underset{0}{\overset{y}{\mathop \int }}\,\frac{dt}{\sqrt{1+9{{t}^{2}}}}$ and $\frac{{{d}^{2}}y}{d{{x}^{2}}}=ay$, then $a$ is equal to 

(A)  3                                                

(B) 6

(C) 9  

(D) 1

Ans: here, we have $x=\underset{0}{\overset{y}{\mathop \int }}\,\frac{dt}{\sqrt{1+9{{t}^{2}}}}$

Differentiate w. r.t $y$, we get 

⇒ $\frac{dx}{dy}=\frac{1}{\sqrt{1+9{{y}^{2}}}}$

⇒ $\frac{dy}{dx}=\sqrt{1+9{{y}^{2}}}$

Now, differentiate w. r.t $x$ , we get 

⇒ $\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{18~y}{2\sqrt{1+9{{y}^{2}}}}.\frac{dy}{dx}$

⇒ $\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{9~y}{\sqrt{1+9{{y}^{2}}}}.\sqrt{1+9{{y}^{2}}}$

⇒ $\frac{{{d}^{2}}y}{d{{x}^{2}}}=9y$

Thus, value of $x$ is 9.

Hence, option (C) is the correct answer.

26. $\underset{-1}{\overset{1}{\mathop \int }}\,\frac{{{x}^{3}}+\left| x \right|+1}{{{x}^{2}}+2\left| x \right|+1}dx~$is equal to 

(A) $\log 2$                                             

(B) $2~\mathbf{log}2$

(C) $\frac{1}{2}~\mathbf{log}2$                                          

(D) $4~\mathbf{log}2$

Ans: Let $I=\underset{-1}{\overset{1}{\mathop \int }}\,\frac{{{x}^{3}}+\left| x \right|+1}{{{x}^{2}}+2\left| x \right|+1}dx$

⇒ $I=\underset{-1}{\overset{1}{\mathop \int }}\,\frac{{{x}^{3}}}{{{x}^{2}}+2\left| x \right|+1}dx+\underset{-1}{\overset{1}{\mathop \int }}\,\frac{\left| x \right|+1}{{{x}^{2}}+2\left| x \right|+1}dx$

Here, $\underset{-1}{\overset{1}{\mathop \int }}\,\frac{{{x}^{3}}}{{{x}^{2}}+2\left| x \right|+1}dx$ is odd function and $\underset{-1}{\overset{1}{\mathop \int }}\,\frac{\left| x \right|+1}{{{x}^{2}}+2\left| x \right|+1}dx$ is even function.

⇒ $I=0+2\underset{0}{\overset{1}{\mathop \int }}\,\frac{\left| x \right|+1}{{{\left( \left| x \right|+1 \right)}^{2}}}dx$

⇒ $I=2\underset{0}{\overset{1}{\mathop \int }}\,\frac{x~+1}{{{\left( x~+1 \right)}^{2}}}dx$

⇒ $I=2\underset{0}{\overset{1}{\mathop \int }}\,\frac{1}{\left( x~+1 \right)}dx$

⇒ $I=2\left[ \log \left| x+1 \right| \right]_{0}^{1}$

⇒ $I=2\left[ \log 2-\log 1 \right]$

⇒ $I=2\log 2$

Hence, option (B) is the correct answer.

27. If $\underset{0}{\overset{1}{\mathop \int }}\,\frac{{{e}^{t}}}{1+~t}~dt=a$, then $\underset{0}{\overset{1}{\mathop \int }}\,\frac{{{e}^{t}}}{{{\left( 1+~t \right)}^{2}}}~dt$ is equal to 

(A) $a-1+\frac{e}{2}~$                            

(B) $a+1-\frac{e}{2}$

(C) $a-1-\frac{e}{2}$                             

(D) $a+1+\frac{e}{2}$

Ans: here, we have $\underset{0}{\overset{1}{\mathop \int }}\,\frac{{{e}^{t}}}{1+~t}~dt=a$, 

⇒ $a=\underset{0}{\overset{1}{\mathop \int }}\,\frac{{{e}^{t}}}{1+~t}~dt$

Applying integrating by parts, we get

⇒ $a=\left[ {{e}^{t}}\times \frac{1}{\left( 1+~t \right)} \right]_{0}^{1}+\underset{0}{\overset{1}{\mathop \int }}\,\frac{{{e}^{t}}}{{{\left( 1+~t \right)}^{2}}}~dt$

⇒ $a=\left[ \frac{e}{2}-1 \right]+\underset{0}{\overset{1}{\mathop \int }}\,\frac{{{e}^{t}}}{{{\left( 1+~t \right)}^{2}}}~dt$

⇒ $\underset{0}{\overset{1}{\mathop \int }}\,\frac{{{e}^{t}}}{{{\left( 1+~t \right)}^{2}}}~dt=a+1-\frac{e}{2}$

Hence, option (B) is the correct answer.

28. $\underset{-2}{\overset{2}{\mathop \int }}\,\left| x\cos \pi x \right|dx$ is equal to 

(A) $\frac{8}{\pi }$ 

(B) $\frac{4}{\pi }$.

(C) $\frac{2}{\pi }$                

(D)   $\frac{1}{\pi }$                

Ans: Let $I=\underset{-2}{\overset{2}{\mathop \int }}\,\left| x\cos \pi x \right|dx$

Here, it is a even function, therefore 

⇒ $I=2\underset{0}{\overset{2}{\mathop \int }}\,\left| x\cos \pi x \right|dx$

⇒ $\frac{I}{2}=\underset{0}{\overset{\frac{1}{2}}{\mathop \int }}\,x\cos \pi xdx+\underset{\frac{1}{2}}{\overset{\frac{3}{2}}{\mathop \int }}\,-x\cos \pi xdx+\underset{\frac{3}{2}}{\overset{2}{\mathop \int }}\,x\cos \pi xdx$

⇒ $\frac{I}{2}=\left[ \frac{x}{\pi }\sin \pi x+\frac{1}{{{\pi }^{2}}}\cos \pi x \right]_{0}^{\frac{1}{2}}-\left[ \frac{x}{\pi }\sin \pi x+\frac{1}{{{\pi }^{2}}}\cos \pi x \right]_{\frac{1}{2}}^{\frac{3}{2}}+\left[ \frac{x}{\pi }\sin \pi x+\frac{1}{{{\pi }^{2}}}\cos \pi x \right]_{\frac{3}{2}}^{0}$

⇒ $\frac{I}{2}=\left[ \frac{1}{2\pi }-\frac{1}{{{\pi }^{2}}} \right]-\left[ -\frac{3}{2\pi }-\frac{1}{2\pi } \right]+\left[ \frac{1}{{{\pi }^{2}}}+\frac{3}{2\pi } \right]$

⇒ $\frac{I}{2}=\frac{1}{2\pi }-\frac{1}{{{\pi }^{2}}}+\frac{3}{2\pi }+\frac{1}{2\pi }+\frac{1}{{{\pi }^{2}}}+\frac{3}{2\pi }$

⇒ $\frac{I}{2}=\frac{8}{2\pi }$

⇒ $I=\frac{8}{\pi }$

Hence, option (A) is the correct answer.

Fill in the blanks in each of the  29 to 32.

29. $\int \frac{si{{n}^{6}}x}{co{{s}^{8}}x}~dx=$ …………..

Ans: Let $I=\int \frac{si{{n}^{6}}x}{co{{s}^{8}}x}~dx$

⇒ $I=\int \frac{si{{n}^{6}}x}{co{{s}^{6}}x.~~co{{s}^{2}}x}~dx$

⇒ $I=\int ta{{n}^{6}}x~.~se{{c}^{2}}x~dx$

Put $\tan x=t$

⇒ $se{{c}^{2}}x~dx=dt$

Therefore, 

⇒ $I=\int {{t}^{6}}~~dt$

⇒ $I=\frac{{{t}^{7}}}{7}+C$

⇒ $I=\frac{ta{{n}^{7}}x}{7}+C$

Hence, $\int \frac{si{{n}^{6}}x}{co{{s}^{8}}x}~dx=\frac{ta{{n}^{7}}x}{7}+C$.

30. $\underset{-a}{\overset{a}{\mathop \int }}\,f\left( x \right)dx=0$ if $f$ is an ……………. function.

Ans: As we know that by the property of integration, $\underset{-a}{\overset{a}{\mathop \int }}\,f\left( x \right)dx=0$ if $f$ is an odd function.

31. $\underset{0}{\overset{2a}{\mathop \int }}\,f\left( x \right)dx=2~\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right)dx~,~$if $f\left( 2a-x \right)=$ ………..

Ans: As, we know that by the property of integration, we have  

⇒ $\underset{0}{\overset{2a}{\mathop \int }}\,f\left( x \right)dx=2~\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right)dx~,~$if $f\left( 2a-x \right)=f\left( x \right)$ 

$=0,$ if $f\left( 2a-x \right)=-f\left( x \right)$

Hence, $\underset{0}{\overset{2a}{\mathop \int }}\,f\left( x \right)dx=2~\underset{0}{\overset{a}{\mathop \int }}\,f\left( x \right)dx~,~$if $f\left( 2a-x \right)=f\left( x \right).$

32. $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{si{{n}^{n}}x}{si{{n}^{n}}x~+~co{{s}^{n}}x}dx=$ …………

Ans: Let $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{si{{n}^{n}}x}{si{{n}^{n}}x~+~co{{s}^{n}}x}dx$ …………….. (i)

⇒ $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{si{{n}^{n}}\left( \frac{\pi }{2}-x \right)}{si{{n}^{n}}\left( \frac{\pi }{2}-x \right)~+~co{{s}^{n}}\left( \frac{\pi }{2}-x \right)}dx$

⇒ $I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{co{{s}^{n}}x}{co{{s}^{n}}x~+~si{{n}^{n}}x}dx$ ……………… (ii)

Adding eq (i) and (ii), we get 

⇒ $2I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,dx$

⇒ $2I=\left[ x \right]_{0}^{\frac{\pi }{2}}$

⇒ $2I=\frac{\pi }{2}$

⇒ $I=\frac{\pi }{4}$

Hence, $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{si{{n}^{n}}x}{si{{n}^{n}}x~+~co{{s}^{n}}x}dx=\frac{\pi }{4}.$

Exercise 7.3

Short Answer (S.A)

Verify the following: 

1. $\int \frac{2x-1}{2x+3}~dx=x-\mathbf{log}\left| {{\left( 2x-3 \right)}^{2}} \right|+C$

Ans: Let $I=~\int \frac{2x-1}{2x+3}~dx$

$\Rightarrow I=~\int \frac{2x+3-3-1}{2x+3}~dx$

$\Rightarrow I=~\int \frac{2x+3-4}{2x+3}~dx$

$\Rightarrow I=~\int \left( 1-\frac{4}{2x+3} \right)~dx$

$\Rightarrow I=~\int ~dx-\int \frac{4}{2x+3}~dx$

$\Rightarrow I=~x-\int \frac{4}{2~\left( x+\frac{3}{2} \right)}~dx$

$~\Rightarrow I=~x-2\log \left( x+\frac{3}{2} \right)+C$

$~\Rightarrow I=~x-2\text{ }\!\!~\!\!\text{ log}\left( \frac{2x+3}{2} \right)+C$

$~~\Rightarrow I=~x-2\text{ }\!\!~\!\!\text{ }[\log \left( 2x+3 \right)-\log 2]+C$

$~~\Rightarrow I=~x-2\text{ }\!\!~\!\!\text{ log}\left( 2x+3 \right)+2\log 2+C$

$\Rightarrow I=~x-\text{ }\!\!~\!\!\text{ log}~|{{\left( 2x+3 \right)}^{2}}|+C$

Thus, $\int \frac{2x-1}{2x+3}~dx=x-\mathbf{log}\left| {{\left( 2x-3 \right)}^{2}} \right|+C$

Hence proved.

2. $\int \frac{2x+3}{{{x}^{2}}+3x}~dx=\mathbf{log}\left| {{x}^{2}}+3x \right|+C$

Ans: let $I=~\int \frac{2x~+~3}{{{x}^{2}}+~3x}~dx$

Put ${{x}^{2}}+3x=t$

⇒ $\left( 2x+3 \right)dx=dt$

$\therefore I=\int \frac{1}{t}~dt$ 

⇒ $I=\log \left| t \right|+C$

⇒ $I=\log \left| \left( {{x}^{2}}+3x \right) \right|+C$

Thus, $\int \frac{2x+3}{{{x}^{2}}+3x}~dx=\text{log}\left| {{x}^{2}}+3x \right|+C$

Hence proved. 

Evaluate the Following: 

3. $\int \frac{\left( {{x}^{2}}+2 \right)dx}{x+1}$

Ans: Let $\text{I}=\int \frac{\left( {{x}^{2}}+2 \right)dx}{x+1}$ 

⇒ $\text{I}=\int \left( x-1+\frac{3}{x+1} \right)dx$

⇒ $\text{I}=\int xdx-\int dx+\int \frac{3}{x+1}dx$

⇒ $\text{I}=\frac{{{x}^{2}}}{2}-x+3~\text{log}\left| x+1 \right|+C$

Hence, $\int \frac{\left( {{x}^{2}}+2 \right)dx}{x+1}=\frac{{{x}^{2}}}{2}-x+3~\text{log}\left| x+1 \right|+C.$

4. $\int \frac{{{e}^{6\log x}}-{{e}^{5\log x}}}{{{e}^{4\log x}}-{{e}^{3\log x}}}~dx$

Ans: Let $\text{I}=\int \frac{{{e}^{6\log x}}-{{e}^{5\log x}}}{{{e}^{4\log x}}-{{e}^{3\log x}}}~dx$

⇒ $\text{I}=\int \frac{{{e}^{\log {{x}^{6}}}}-{{e}^{\log {{x}^{5}}}}}{{{e}^{\log {{x}^{4}}}}-{{e}^{\log {{x}^{3}}}}}~dx$

⇒ $\text{I}=\int \frac{{{x}^{6}}-{{x}^{5}}}{{{x}^{4}}-{{x}^{3}}}~dx$                                [${{e}^{\log x}}=x]$

⇒ $\text{I}=\int \frac{{{x}^{5}}\left( x-1 \right)}{{{x}^{3}}\left( x-1 \right)}~dx$

⇒ $\text{I}=\int {{x}^{2}}~dx$

⇒ $\text{I}=\frac{{{x}^{3}}}{3}+C$

Hence, $\int \frac{{{e}^{6\log x}}-{{e}^{5\log x}}}{{{e}^{4\log x}}-{{e}^{3\log x}}}~dx=\frac{{{x}^{3}}}{3}+C.$

5. $\int \frac{\left( 1+\cos x \right)}{x+\sin x}dx$

Ans: Let $\text{I}=\int \frac{\left( 1+\cos x \right)}{x+\sin x}dx~$

Put $x+\sin x=t$

⇒ $\left( 1+\cos x \right)dx=dt$

⇒ $\text{I}=\int \frac{1}{t}dx$

⇒ $\text{I}=\log \left| t \right|+C$

⇒ $\text{I}=\log \left| \left( x+\cos x \right) \right|+C$

6. $\int \frac{dx}{1+\cos x}$

Ans: Let $I=\int \frac{dx}{1+\cos x}$

⇒ $I=\int \frac{dx}{2co{{s}^{2}}\frac{x}{2}~}$

⇒ $I=\frac{1}{2}\int \frac{dx}{co{{s}^{2}}\frac{x}{2}~}$

⇒ $I=\frac{1}{2}\int se{{c}^{2}}\frac{x}{2}~dx$

⇒ $I=\frac{1}{2}\times 2\tan \frac{x}{2}~+C$                [$\int se{{c}^{2}}x~dx=\tan x+C]$

⇒ $I=\tan \frac{x}{2}~+C$

Hence, $\int \frac{dx}{1+\cos x}=\tan \frac{x}{2}~+C$.

7. $\int ta{{n}^{2}}x.~se{{c}^{4}}x~dx$

Ans: Let $\text{I}=\int ta{{n}^{2}}x~.~se{{c}^{4}}x~dx$

⇒ $\text{I}=\int ta{{n}^{2}}x~.~se{{c}^{2}}x.~se{{c}^{2}}x~dx$

⇒ $\text{I}=\int ta{{n}^{2}}x\left( 1+ta{{n}^{2}}x~ \right)se{{c}^{2}}x~dx$

Put $\text{tan }\!\!~\!\!\text{ x}=t$

⇒ $\text{se}{{\text{c}}^{2}}\text{x }\!\!~\!\!\text{ dx}=\text{dt}$

Therefore, 

⇒ $\text{I}=\int {{t}^{2}}\left( 1+{{t}^{2}}~ \right)dt$

⇒ $\text{I}=\int ({{t}^{2}}+{{t}^{4}})dt$

⇒ $\text{I}=\frac{{{t}^{3}}}{3}+\frac{{{t}^{5}}}{5}+C$

⇒ $\text{I}=\frac{ta{{n}^{3}}x}{3}+\frac{ta{{n}^{5}}x}{5}+C$

8. $\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2x}}dx$

Ans: Let $\mathbf{I}=~\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2x}}dx$

 ⇒ $\mathbf{I}=~\int \frac{\sin x+\cos x}{\sqrt{si{{n}^{2}}x+co{{s}^{2}}x+2.\sin x\cos x}}dx$

⇒ $\mathbf{I}=~\int \frac{\sin x+\cos x}{\sqrt{{{\left( \sin x+\cos x \right)}^{2}}}}dx$

⇒ $\mathbf{I}=~\int \frac{(\sin x+\cos x)}{\left( \sin x+\cos x \right)}dx$

⇒ $\mathbf{I}=~\int dx$

⇒ $\mathbf{I}=~x+C$

Hence, $\int \frac{\sin x+\cos x}{\sqrt{1+\sin 2x}}dx=x+C$.

9. $\int \sqrt{1+\sin x}~dx$

Ans: Let $\mathbf{I}=\int \sqrt{1+\sin x}~dx~$

⇒ $\mathbf{I}=\int \sqrt{si{{n}^{2}}\frac{x}{2}+co{{s}^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}~dx$

⇒ $\mathbf{I}=\int \sqrt{{{\left( \sin \frac{x}{2}+\cos \frac{x}{2} \right)}^{2}}}~dx$

⇒ $\text{I}=\int \left( \sin \frac{x}{2}+\cos \frac{x}{2} \right)~dx$

⇒ $\mathbf{I}=\int \sin \frac{x}{2}~dx+~\int \cos \frac{x}{2}~dx$

⇒ $\text{I}=-2\text{cos }\!\!~\!\!\text{ }\frac{\text{x}}{2}+\text{ }\!\!~\!\!\text{ }2\text{ }\!\!~\!\!\text{ sin }\!\!~\!\!\text{ }\frac{\text{x}}{2}+\text{C}$

Hence, $\int \sqrt{1+\sin x}~dx=-2\text{cos }\!\!~\!\!\text{ }\frac{\text{x}}{2}+\text{ }\!\!~\!\!\text{ }2\text{ }\!\!~\!\!\text{ sin }\!\!~\!\!\text{ }\frac{\text{x}}{2}+\text{C}.$

10. $\int \frac{x}{\sqrt{x}+1}dx$

Ans: Let $\text{I}=\text{ }\!\!~\!\!\text{ }\int \frac{\text{x}}{\sqrt{\text{x}}+1}\text{dx}$

Put $\sqrt{x}=z$

⇒ $\frac{1}{2\sqrt{x}}dx=dz$

⇒ $dx=2\sqrt{x}~dz$

⇒ $dx=2z~dz$

Therefore,

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }\int \frac{{{z}^{2}}}{\text{z }\!\!~\!\!\text{ }+1}\times 2\text{z }\!\!~\!\!\text{ dz}$

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }2\int \frac{{{z}^{3}}}{\text{z }\!\!~\!\!\text{ }+1}\text{ }\!\!~\!\!\text{ dz}$

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }2\int \frac{{{z}^{3}}+1-1}{\text{z }\!\!~\!\!\text{ }+1}\text{ }\!\!~\!\!\text{ dz}$

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }2\int \frac{{{z}^{3}}+1}{\text{z }\!\!~\!\!\text{ }+1}\text{ }\!\!~\!\!\text{ dz}-\text{ }\!\!~\!\!\text{ }2\int \frac{1}{\text{z }\!\!~\!\!\text{ }+1}\text{ }\!\!~\!\!\text{ dz}$

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }2\int \frac{\left( \text{z }\!\!~\!\!\text{ }+1 \right)\left( {{z}^{2}}-z+1 \right)}{\text{z }\!\!~\!\!\text{ }+1}\text{ }\!\!~\!\!\text{ dz}-\text{ }\!\!~\!\!\text{ }2\int \frac{1}{\text{z }\!\!~\!\!\text{ }+1}\text{ }\!\!~\!\!\text{ dz}$

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }2\int \left( {{z}^{2}}-z+1 \right)\text{ }\!\!~\!\!\text{ dz}-\text{ }\!\!~\!\!\text{ }2\int \frac{1}{\text{z }\!\!~\!\!\text{ }+1}\text{ }\!\!~\!\!\text{ dz}$

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }2\left( \frac{{{z}^{3}}}{3}-\frac{{{z}^{2}}}{2}+z \right)-\text{ }\!\!~\!\!\text{ }2\text{ }\!\!~\!\!\text{ log }\!\!~\!\!\text{ }\left| z+1 \right|+C$

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }2\left( \frac{x\sqrt{x}}{3}-\frac{x}{2}+\sqrt{x} \right)-\text{ }\!\!~\!\!\text{ }2\text{ }\!\!~\!\!\text{ log }\!\!~\!\!\text{ }\left| \sqrt{x}+1 \right|+C$

⇒ $\text{I}=\text{ }\!\!~\!\!\text{ }2\left( \frac{x\sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\text{log }\!\!~\!\!\text{ }\left| \sqrt{x}+1 \right| \right)+C$

Hence, $\int \frac{\text{x}}{\sqrt{\text{x}}+1}\text{dx}=~2\left( \frac{x\sqrt{x}}{3}-\frac{x}{2}+\sqrt{x}-\text{log }\!\!~\!\!\text{ }\left| \sqrt{x}+1 \right| \right)+C~.$

11. $\int \sqrt{\frac{a+x}{a-x}}~dx$

Ans: Let $\text{I}=\int \sqrt{\frac{a~+~x~}{a~-~x~}}~dx$

⇒ $\text{I}=\int \sqrt{\frac{\left( a~+~x \right)~\left( a~+~x \right)~}{\left( a~-~x \right)~\left( a~+~x \right)~}}~dx$

⇒ $\text{I}=\int \sqrt{\frac{\left( a~+~x \right){{~}^{2}}~}{\left( {{a}^{2}}-{{x}^{2}} \right)~}}~dx$

⇒ $\text{I}=\int \frac{a~+~x~~}{\sqrt{\left( {{a}^{2}}-{{x}^{2}} \right)~}}~dx$

⇒ $\text{I}=\int \frac{a~~~}{\sqrt{\left( {{a}^{2}}-{{x}^{2}} \right)~}}~dx+\int \frac{~x~~}{\sqrt{\left( {{a}^{2}}-{{x}^{2}} \right)~}}~dx$

⇒ $\text{I}=a\text{ }\!\!~\!\!\text{ }si{{n}^{-1}}\left( \frac{x}{a} \right)+\int \frac{~x~~}{\sqrt{\left( {{a}^{2}}-{{x}^{2}} \right)~}}~dx$

Put  ${{a}^{2}}-{{x}^{2}}={{t}^{2}}$

⇒ $-2x~dx=2t~dt$

⇒ $x~dx=-t~dt$

Therefore, 

⇒ $\text{I}=a\text{ }\!\!~\!\!\text{ }si{{n}^{-1}}\left( \frac{x}{a} \right)+\int \frac{-~t~dt~}{\sqrt{{{t}^{2}}~}}~$

⇒ $\text{I}=a\text{ }\!\!~\!\!\text{ }si{{n}^{-1}}\left( \frac{x}{a} \right)-\int \frac{t~dt~}{t}~$

⇒ $\text{I}=a\text{ }\!\!~\!\!\text{ }si{{n}^{-1}}\left( \frac{x}{a} \right)-\int ~dt$

⇒ $\text{I}=a\text{ }\!\!~\!\!\text{ }si{{n}^{-1}}\left( \frac{x}{a} \right)-t+~~C$

⇒ $\text{I}=a\text{ }\!\!~\!\!\text{ }si{{n}^{-1}}\left( \frac{x}{a} \right)-\sqrt{\left( {{a}^{2}}-{{x}^{2}} \right)~}+~~C$

12. $\int \frac{{{x}^{\frac{1}{2}}}}{1+~{{x}^{\frac{3}{4}}}}~dx$

Ans: Let $\text{I}=\int \frac{{{x}^{\frac{1}{2}}}}{1+~{{x}^{\frac{3}{4}}}}~dx$

Put  $x=~{{z}^{4}}$

⇒ $dx=4{{z}^{3}}dz$

Therefore, 

⇒ $\text{I}=\int \frac{{{\left( {{z}^{4}} \right)}^{\frac{1}{2}}}}{1+~{{\left( {{z}^{4}} \right)}^{\frac{3}{4}}}}~\times 4{{z}^{3}}dz$

⇒ $\text{I}=4\int \frac{{{z}^{2}}.{{z}^{3}}}{1+~{{z}^{3}}}~dz$

Now, put $1+~{{z}^{3}}=t$

⇒ $3{{z}^{2}}dz=dt$

⇒ ${{z}^{2}}dz=\frac{dt}{3}$

Therefore, we get

⇒ $\text{I}=4\int \frac{\left( t-1 \right)}{t}.\frac{dt}{3}$

⇒ $\text{I}=\frac{4}{3}\int \left( 1-\frac{1}{t}~ \right)dt$

⇒ $\text{I}=\frac{4}{3}\text{t}-\frac{4}{3}\text{ }\!\!~\!\!\text{ log}\left| t \right|+{{C}_{1}}$

⇒ $\text{I}=\frac{4}{3}\left( 1+{{z}^{3}} \right)-\frac{4}{3}\text{ }\!\!~\!\!\text{ log}\left| 1+{{z}^{3}} \right|+{{C}_{1}}$

⇒ $\text{I}=\frac{4}{3}\left( 1+{{x}^{\frac{3}{4}}} \right)-\frac{4}{3}\text{ }\!\!~\!\!\text{ log}\left| 1+{{x}^{\frac{3}{4}}} \right|+{{C}_{1}}$

⇒ $\text{I}=\frac{4}{3}+\frac{4}{3}{{x}^{\frac{3}{4}}}-\frac{4}{3}\text{ }\!\!~\!\!\text{ log}\left| 1+{{x}^{\frac{3}{4}}} \right|+{{C}_{1}}$

⇒ $\text{I}=\frac{4}{3}{{x}^{\frac{3}{4}}}-\frac{4}{3}\text{ }\!\!~\!\!\text{ log}\left| 1+{{x}^{\frac{3}{4}}} \right|+C$

13. $\int \frac{\sqrt{1+~{{x}^{2}}}}{{{x}^{4}}}~dx$ 

Ans: Let $\text{I}=\int \frac{\sqrt{1+~{{x}^{2}}}}{{{x}^{4}}}~dx$

⇒ $\text{I}=\int \frac{\sqrt{1+~{{x}^{2}}}}{x.~~{{x}^{3}}}~dx$

⇒ $\text{I}=\int \sqrt{\frac{1+~{{x}^{2}}}{{{x}^{2}}}}~\frac{dx}{{{x}^{3}}}$

⇒ $\text{I}=\int \sqrt{\left( \frac{1}{{{x}^{2}}}+1 \right)}~\frac{dx}{{{x}^{3}}}$

Put $\left( \frac{1}{{{x}^{2}}}+1 \right)={{t}^{2}}$

⇒ $-\frac{2}{{{x}^{3}}}dx=2t~dt$

⇒ $\frac{dx}{{{x}^{3}}}=-~t~dt$

Therefore, 

⇒ $\text{I}=\int \sqrt{{{t}^{2}}}~\left( -t \right)~dt$

⇒ $\text{I}=-\int t.~t~dt$

⇒ $\text{I}=-\int {{t}^{2}}~~dt$

⇒ $\text{I}=-\frac{{{t}^{3}}}{3}+C$

⇒ $\text{I}=-\frac{1}{3}{{\left( \frac{1}{{{x}^{2}}}+1 \right)}^{\frac{3}{2}}}+C$

14. $\int \frac{dx}{\sqrt{16-9{{x}^{2}}}}$

Ans: Let $\text{I}=\int \frac{dx}{\sqrt{16-9{{x}^{2}}}}$

⇒ $\text{I}=\int \frac{dx}{\sqrt{9~\left( \frac{16}{9}-{{x}^{2}} \right)}}$

⇒ $\text{I}=\frac{1}{3}\int \frac{dx}{\sqrt{{{\left( ~\frac{4}{3} \right)}^{2}}-{{x}^{2}}}}$

⇒ $\text{I}=\frac{1}{3}~si{{n}^{-1}}\left( \frac{x}{\frac{4}{3}} \right)+C$

⇒ $\text{I}=\frac{1}{3}~si{{n}^{-1}}\left( \frac{3x}{4} \right)+C$

15. $\int \frac{dt}{\sqrt{3t-~2{{t}^{2}}}}$

Ans: Let $\text{I}=\int \frac{dt}{\sqrt{3t-~2{{t}^{2}}}}$

⇒ $\text{I}=\int \frac{dt}{\sqrt{-2\left( {{t}^{2}}-\frac{3t}{2} \right)}}$

⇒ $\text{I}=\frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{-\left( {{t}^{2}}-\frac{3t}{2} \right)}}$

⇒ $\text{I}=\frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{-\left[ {{t}^{2}}-\frac{3t}{2}+{{\left( \frac{3}{4} \right)}^{2}}-{{\left( \frac{3}{4} \right)}^{2}} \right]}}$

⇒ $\text{I}=\frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{-\left[ {{\left( t-\frac{3}{4} \right)}^{2}}-{{\left( \frac{3}{4} \right)}^{2}} \right]}}$

⇒ $\text{I}=\frac{1}{\sqrt{2}}\int \frac{dt}{\sqrt{{{\left( \frac{3}{4} \right)}^{2}}-{{\left( t-\frac{3}{4} \right)}^{2}}}}$

⇒ $\text{I}=\frac{1}{\sqrt{2}}~si{{n}^{-1}}\left( \frac{t-\frac{3}{4}}{\frac{3}{4}} \right)+C$

⇒ $\text{I}=\frac{1}{\sqrt{2}}~si{{n}^{-1}}\left( \frac{4t-3}{3} \right)+C$

16. $\int \frac{3\mathbf{x}-1}{\sqrt{{{\mathbf{x}}^{2}}+9}}~\mathbf{dx}$

Ans: Let $\text{I}=\int \frac{3\text{x}-1}{\sqrt{{{\text{x}}^{2}}+9}}\text{ }\!\!~\!\!\text{ dx}$

⇒ $\text{I}=\int \frac{3\text{x}}{\sqrt{{{\text{x}}^{2}}+9}}\text{ }\!\!~\!\!\text{ dx}-\int \frac{\text{dx}}{\sqrt{{{\text{x}}^{2}}+9}}\text{ }\!\!~\!\!\text{ }$

⇒ $\text{I}={{\text{I}}_{1}}-{{\text{I}}_{2}}$  ………………. (i),

Where,  ${{\text{I}}_{1}}=\int \frac{3\text{x}}{\sqrt{{{\text{x}}^{2}}+9}}\text{ }\!\!~\!\!\text{ dx}~$ and $\text{ }\!\!~\!\!\text{ }{{\text{I}}_{2}}=\int \frac{\text{dx}}{\sqrt{{{\text{x}}^{2}}+9}}\text{ }\!\!~\!\!\text{ }$

Therefore, 

⇒${{\text{I}}_{1}}=\int \frac{3\text{x}}{\sqrt{{{\text{x}}^{2}}+9}}\text{ }\!\!~\!\!\text{ dx}$

Put ${{\text{x}}^{2}}+9={{t}^{2}}$

⇒ $2x~dx=2t~dt$

⇒ $x~dx=t~dt$

⇒$~{{\text{I}}_{1}}=\int \frac{3\text{t }\!\!~\!\!\text{ dt}}{\sqrt{{{t}^{2}}}}$

⇒$~{{\text{I}}_{1}}=\int \frac{3\text{t }\!\!~\!\!\text{ dt}}{\text{t}}$

⇒$~{{\text{I}}_{1}}=\int 3~dt$

⇒$\text{ }\!\!~\!\!\text{ }{{\text{I}}_{1}}=3t+~{{C}_{1}}$  

⇒$\text{ }\!\!~\!\!\text{ }{{\text{I}}_{1}}=3\sqrt{{{x}^{2}}+9}+~{{C}_{1}}$ ……….. (ii)

Also we have, 

$\Rightarrow \text{ }\!\!~\!\!\text{ }{{\text{I}}_{2}}=\int \frac{\text{dx}}{\sqrt{{{\text{x}}^{2}}+9}}\text{ }\!\!~\!\!\text{ }$ 

$\Rightarrow \text{ }\!\!~\!\!\text{ }{{\text{I}}_{2}}=\int \frac{\text{dx}}{\sqrt{{{\text{x}}^{2}}+{{3}^{2}}}}\text{ }\!\!~\!\!\text{ }$ 

$\Rightarrow \text{ }\!\!~\!\!\text{ }{{\text{I}}_{2}}=\text{log}\left| x+\sqrt{{{\text{x}}^{2}}+9} \right|+{{C}_{2}}$  ………….. (iii)

Put the value of $\text{ }\!\!~\!\!\text{ }{{\text{I}}_{1}}$ and $\text{ }\!\!~\!\!\text{ }{{\text{I}}_{2}}$ in eq (i), we get 

$\Rightarrow \text{I}=3\sqrt{{{x}^{2}}+9}+~{{C}_{1}}-\text{ }\!\!~\!\!\text{ log}\left| x+\sqrt{{{\text{x}}^{2}}+9} \right|-{{C}_{2}}$  

$\Rightarrow \text{I}=3\sqrt{{{x}^{2}}+9}-\text{ }\!\!~\!\!\text{ log}\left| x+\sqrt{{{\text{x}}^{2}}+9} \right|+C$

17. $\int \sqrt{5-2x+{{x}^{2}}}~dx$

Ans: Let $\text{I}=~\int \sqrt{5-2x+{{x}^{2}}}~dx$

⇒ $\text{I}=~\int \sqrt{{{x}^{2}}-2x+5}~dx$. 

⇒ $\text{I}=~\int \sqrt{{{x}^{2}}-2x+1+4}~dx$

⇒ $\text{I}=~\int \sqrt{({{x}^{2}}-2x+1)+{{2}^{2}}}~dx$

⇒ $\text{I}=~\int \sqrt{{{\left( x-1 \right)}^{2}}+{{2}^{2}}}~dx$

⇒ $\text{I}=\frac{x-1}{2}\sqrt{{{\left( x-1 \right)}^{2}}+{{2}^{2}}}+\frac{{{2}^{2}}}{2}~\text{log}\left| x-1+\sqrt{{{\left( x-1 \right)}^{2}}+{{2}^{2}}} \right|+C~~$

⇒ $\text{I}=\frac{\left( x-1 \right)}{2}\sqrt{5-2x+{{x}^{2}}}+2~\text{log}\left| x-1+\sqrt{5-2x+{{x}^{2}}} \right|+C$. 

18. $\int \frac{x}{{{x}^{4}}-1}dx$. 

Ans: Let $\text{I}=~\int \frac{x}{{{x}^{4}}-1}dx$ 

⇒ $\text{I}=~\int \frac{x}{{{\left( {{x}^{2}} \right)}^{2}}-1}dx$

Put ${{x}^{2}}=t$.  

⇒ $2x~dx=dt$. 

⇒ $x~dx=\frac{dt}{2}$

Therefore, 

⇒ $\text{I}=\frac{1}{2}~\int \frac{dt}{{{\left( t \right)}^{2}}-1}$

⇒ $\text{I}=\frac{1}{2}.\frac{1}{2\times 1}\text{log}\left| \frac{t-1}{t+1} \right|+C$

⇒ $\text{I}=\frac{1}{4}\text{ }\!\!~\!\!\text{ log}\left| \frac{{{x}^{2}}-1}{{{x}^{2}}+1} \right|+C$

19. $\int \frac{{{x}^{2}}}{1-~{{x}^{4}}}~dx$

Ans: Let $\text{I}=~\int \frac{{{x}^{2}}}{1-~{{x}^{4}}}~dx$

⇒ $\text{I}=~\int \frac{\frac{{{x}^{2}}}{2}+\frac{{{x}^{2}}}{2}}{{{1}^{2}}-~{{\left( {{x}^{2}} \right)}^{2}}}~dx$

⇒ $\text{I}=~\int \frac{\frac{1}{2}~+~~\frac{{{x}^{2}}}{2}-~\frac{1}{2}+\frac{{{x}^{2}}}{2}}{\left( 1-{{x}^{2}} \right)\left( 1~+~{{x}^{2}} \right)}~dx$

⇒ $\text{I}=~\int \frac{\frac{1}{2}\left( 1~+~~{{x}^{2}} \right)-~\frac{1}{2}\left( 1-{{x}^{2}} \right)}{\left( 1-{{x}^{2}} \right)\left( 1~+~{{x}^{2}} \right)}~dx$

⇒ $\text{I}=\frac{1}{2}\int \frac{\left( 1~+~~{{x}^{2}} \right)}{\left( 1-{{x}^{2}} \right)\left( 1~+~{{x}^{2}} \right)}~dx-\frac{1}{2}\int \frac{\left( 1-~~{{x}^{2}} \right)}{\left( 1-{{x}^{2}} \right)\left( 1~+~{{x}^{2}} \right)}~dx$

⇒ $\text{I}=\frac{1}{2}\int \frac{dx}{\left( 1-{{x}^{2}} \right)}~-\frac{1}{2}\int \frac{dx}{\left( 1~+~{{x}^{2}} \right)}~$

⇒ $\text{I}=\frac{1}{2}.\frac{1}{2\times 1}\text{log}\left| \frac{1+x}{1-x~} \right|-\frac{1}{2}~ta{{n}^{-1}}x+C$

⇒ $\text{I}=\frac{1}{4}\text{log}\left| \frac{1+x}{1-x~} \right|-\frac{1}{2}~ta{{n}^{-1}}x+C$. 

20. $\int \sqrt{2ax-~{{x}^{2}}}~dx$

Ans: Let $\text{I}=\int \sqrt{2ax-~{{x}^{2}}}~dx$

⇒ $\text{I}=\int \sqrt{-\left( ~{{x}^{2}}-2ax \right)}~dx$

⇒ $\text{I}=\int \sqrt{-\left( ~{{x}^{2}}-2ax+{{a}^{2}}-{{a}^{2}} \right)}~dx$. 

⇒ $\text{I}=\int \sqrt{-\left[ ({{x}^{2}}-2ax+{{a}^{2}})-{{a}^{2}} \right]}~dx$. 

⇒ $\text{I}=\int \sqrt{-\left[ {{\left( x-a \right)}^{2}}-{{a}^{2}} \right]}~dx$. 

⇒ $\text{I}=\int \sqrt{{{a}^{2}}-{{\left( x-a \right)}^{2}}}~dx$. 

⇒ $\text{I}=\frac{\left( x-a \right)}{2}\sqrt{{{a}^{2}}-{{\left( x-a \right)}^{2}}}+\frac{{{a}^{2}}}{2}~si{{n}^{-1}}\left( \frac{x-a}{a} \right)+C$. 

⇒ $\text{I}=\frac{\left( x-a \right)}{2}\sqrt{2ax-~{{x}^{2}}}~+\frac{{{a}^{2}}}{2}~si{{n}^{-1}}\left( \frac{x-a}{a} \right)+C$

21. $\int \frac{si{{n}^{-1}}x}{{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{2}}}}dx$. 

Let $\text{I}=\int \frac{si{{n}^{-1}}x}{{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{2}}}}~dx$. 

⇒ $\text{I}=\int \frac{si{{n}^{-1}}x}{\left( 1~-~{{x}^{2}} \right)\sqrt{1~-~{{x}^{2}}}}~dx$. 

Put  $si{{n}^{-1}}x=t$.   

⇒ $\frac{1}{\sqrt{1-{{x}^{2}}}}dx=dt$

And $si{{n}^{-1}}x=t$  ⇒ $x=sint$

$\text{cost}=\sqrt{1-{{x}^{2}}}$

Therefore, 

$\text{I}=\int \frac{t.dt}{\left( 1~-~si{{n}^{2}}t \right)}~$

⇒ $\text{I}=\int \frac{t.~~dt}{co{{s}^{2}}t}~$

⇒ $\text{I}=\int t.~se{{c}^{2}}t~dt$

⇒ $\text{I}=t\int se{{c}^{2}}t~dt-\int \left( \frac{d}{dt}t.\int se{{c}^{2}}t~dt \right)dt$

⇒ $\text{I}=\text{t }\!\!~\!\!\text{ tant}-\int \text{tantdt}$

⇒ $\text{I}=\text{t }\!\!~\!\!\text{ tant}+\text{log}\left| \text{cost} \right|+\text{C}$

⇒ $\text{I}=si{{n}^{-1}}x\text{ }\!\!~\!\!\text{ }\times \frac{\sin t}{\cos t}+\text{log}\left| \sqrt{1-{{x}^{2}}} \right|+\text{C}$

⇒ $\text{I}=\frac{x~si{{n}^{-1}}x}{\sqrt{1-{{x}^{2}}}}\text{ }\!\!~\!\!\text{ }+\text{log}\left| \sqrt{1-{{x}^{2}}} \right|+\text{C}$

Hence, $\int \frac{si{{n}^{-1}}x}{{{\left( 1-{{x}^{2}} \right)}^{\frac{3}{2}}}}~dx=\frac{x~si{{n}^{-1}}x}{\sqrt{1-{{x}^{2}}}}\text{ }\!\!~\!\!\text{ }+\text{log}\left| \sqrt{1-{{x}^{2}}} \right|+\text{C}.$

22. $\int \frac{\left( \cos 5x+\cos 4x \right)}{1-2\cos 3x}dx$ 

Ans: Let $\text{I}=~\int \frac{\left( \cos 5x+\cos 4x \right)}{1-2\cos 3x}dx$. 

Applying $[\cos C+\cos D=2\cos \frac{C+D}{2}.~\cos \frac{C-D}{2}$ d $\cos 2x=2co{{s}^{2}}x-1]$

⇒ $\text{I}=~\int \frac{2\cos \frac{9x}{2}~.\cos \frac{x}{2}}{1-2~\left( 2~\text{co}{{\text{s}}^{2}}\frac{3x}{2}-1 \right)}dx$

⇒ $\text{I}=~\int \frac{2\cos \frac{9x}{2}~.\cos \frac{x}{2}}{1-4~\text{co}{{\text{s}}^{2}}\frac{~3x}{2}+2}dx$

⇒ $\text{I}=~\int \frac{2\cos \frac{9x}{2}~.\cos \frac{x}{2}}{3-4~\text{co}{{\text{s}}^{2}}\frac{3x}{2}}dx$

Now, multiply and divide by $\cos \frac{3x}{2}$, we get

⇒ $\text{I}=~\int \frac{2\cos \frac{9x}{2}~.\cos \frac{x}{2}.~\cos \frac{3x}{2}~}{3\cos \frac{3x}{2}-4~\text{co}{{\text{s}}^{3}}\frac{3x}{2}}dx$

⇒ $\text{I}=~-\int \frac{2\cos \frac{9x}{2}~.\cos \frac{x}{2}.~\cos \frac{3x}{2}~}{4~\text{co}{{\text{s}}^{3}}\frac{3x}{2}-3\cos \frac{3x}{2}}dx$

⇒ $\text{I}=~-\int \frac{2\cos \frac{9x}{2}~.\cos \frac{x}{2}.~\cos \frac{3x}{2}~}{\cos 3.\frac{3x}{2}}dx$

⇒ $\text{I}=~-\int \frac{2\cos \frac{9x}{2}~.\cos \frac{x}{2}.~\cos \frac{3x}{2}~}{\cos \frac{9x}{2}}dx$

⇒ $\text{I}=~-\int 2\cos \frac{x}{2}.~\cos \frac{3x}{2}dx$

⇒ $\text{I}=~-\int \left[ \cos \left( \frac{3x}{2}+\frac{x}{2} \right)+\cos \left( \frac{3x}{2}-\frac{x}{2} \right) \right]dx$

⇒ $\text{I}=~-\int (\cos 2x+\cos x)dx$

⇒ $\text{I}=~-\int \cos 2xdx-\int \cos xdx$

⇒ $\text{I}=~-\left( \frac{\sin 2x}{2} \right)-\sin x~+C$

Hence, $\int \frac{\left( \cos 5x+\cos 4x \right)}{1-2\cos 3x}dx=-\left( \frac{\sin 2x}{2} \right)-\sin x~+C.$

23. $\int \frac{si{{n}^{6}}x~+~co{{s}^{6}}x~}{si{{n}^{2}}x~.co{{s}^{2}}x}dx$

Ans: Let $\text{I}=~\int \frac{si{{n}^{6}}x~+~co{{s}^{6}}x~}{si{{n}^{2}}x~.co{{s}^{2}}x}dx$

⇒ $\text{I}=~\int \frac{{{\left( si{{n}^{2}}x \right)}^{3}}~+~{{\left( co{{s}^{2}}x \right)}^{3}}~}{si{{n}^{2}}x~.co{{s}^{2}}x}dx$

⇒ $\text{I}=~\int \frac{\left( si{{n}^{2}}x+~co{{s}^{2}}x~~ \right)~\left( si{{n}^{4}}x-si{{n}^{2}}x~.co{{s}^{2}}x+co{{s}^{4}}x \right)~}{si{{n}^{2}}x~.co{{s}^{2}}x}dx$

⇒ $\text{I}=~\int \frac{~\left( si{{n}^{4}}x-si{{n}^{2}}x~.co{{s}^{2}}x+co{{s}^{4}}x \right)~}{si{{n}^{2}}x~.co{{s}^{2}}x}dx$

⇒ $\text{I}=~\int \frac{~si{{n}^{4}}x}{si{{n}^{2}}x~.co{{s}^{2}}x}dx-\int \frac{~si{{n}^{2}}x~.co{{s}^{2}}x}{si{{n}^{2}}x~.co{{s}^{2}}x}dx+\int \frac{~co{{s}^{4}}x}{si{{n}^{2}}x~.co{{s}^{2}}x}~dx$

⇒ $\text{I}=~\int \frac{~si{{n}^{2}}x}{co{{s}^{2}}x}dx-\int dx+\int \frac{~co{{s}^{2}}x}{si{{n}^{2}}x~}~dx$

⇒ $\text{I}=~\int ta{{n}^{2}}x~dx-\int dx+\int co{{t}^{2}}x~dx$

⇒ $\text{I}=~\int \left( se{{c}^{2}}x-1~ \right)~dx-\int dx+\int (cose{{c}^{2}}x-1)~~dx$

⇒ $\text{I}=~\int (se{{c}^{2}}x~dx+\int cose{{c}^{2}}x~dx-3\int dx$

⇒ $\text{I}=~tanx-cotx-3x+C$

Hence, $\int \frac{si{{n}^{6}}x~+~co{{s}^{6}}x~}{si{{n}^{2}}x~.co{{s}^{2}}x}dx=~tanx-cotx-3x+C.$

24. $\int \frac{\sqrt{x}}{\sqrt{{{a}^{3}}-{{x}^{3}}}}dx$

Ans: Let $\text{I}=\int \frac{\sqrt{x}}{\sqrt{{{a}^{3}}-{{x}^{3}}}}dx$

⇒ $\text{I}=\int \frac{{{x}^{\frac{1}{2}}}}{\sqrt{{{\left( {{a}^{\frac{3}{2}}} \right)}^{2}}-{{\left( {{x}^{\frac{3}{2}}} \right)}^{2}}}}dx$

Here, Put $~~{{x}^{\frac{3}{2}}}=t$

⇒ $\frac{3}{2}~{{x}^{\frac{1}{2}}}~dx=dt$

⇒ $~{{x}^{\frac{1}{2}}}~dx=\frac{2}{3}dt$

Therefore, 

⇒ $\text{I}=\frac{2}{3}\int \frac{dt}{\sqrt{{{\left( {{a}^{\frac{3}{2}}} \right)}^{2}}-{{\left( t \right)}^{2}}}}$

⇒ $\text{I}=\frac{2}{3}~si{{n}^{-1}}\left( \frac{t}{{{a}^{\frac{3}{2}}}} \right)+C$

⇒ $\text{I}=\frac{2}{3}~si{{n}^{-1}}\left( \frac{~~{{x}^{\frac{3}{2}}}}{{{a}^{\frac{3}{2}}}} \right)+C$

⇒ $\text{I}=\frac{2}{3}~si{{n}^{-1}}{{\left( \frac{x}{a} \right)}^{\frac{3}{2}}}+C$

Hence, $\int \frac{\sqrt{x}}{\sqrt{{{a}^{3}}-{{x}^{3}}}}dx=\frac{2}{3}~si{{n}^{-1}}{{\left( \frac{x}{a} \right)}^{\frac{3}{2}}}+C.$

25. $\int \frac{cos~x-cos~2x}{1-cosx}~dx$

Ans: Let $\text{I}=\int \frac{cos~x-cos~2x}{1-cosx}~dx$

⇒ $\text{I}=\int \frac{2~sin~\frac{3x}{2}.~~sin~\frac{x}{2}}{2~si{{n}^{2}}\frac{x}{2}}~dx$

⇒ $\text{I}=\int \frac{~sin~\frac{3x}{2}}{~~sin~\frac{x}{2}}~dx$

⇒ $\text{I}=\int \frac{~3~sin~\frac{x}{2}-4~si{{n}^{3}}~\frac{x}{2}}{~~sin~\frac{x}{2}}~dx$

⇒ $\text{I}=3\int dx-4\int ~si{{n}^{2}}~\frac{x}{2}~dx$

⇒ $\text{I}=3x-4\int ~\frac{1+cos~x}{2}~dx$

⇒ $\text{I}=3x-2\int ~\left( 1-cos~x \right)~dx$

⇒ $\text{I}=3x-2\left( x-sin~x \right)+C$

⇒ $\text{I}=3x-2x+2~sin~x+C$

⇒ $\text{I}=x+2~sin~x+C$

Hence, $\int \frac{cos~x-cos~2x}{1-cosx}~dx=x+2~sin~x+C.$

26. $\int \frac{dx}{x\sqrt{{{x}^{4}}-1}}$

Ans: Let $\text{I}=\int \frac{dx}{x\sqrt{{{x}^{4}}-1}}$

Multiply and divide by ${{x}^{3}}$, we get

⇒ $\text{I}=\int \frac{{{x}^{3}}dx}{{{x}^{4}}\sqrt{{{x}^{4}}-1}}$

Now, put $~{{x}^{4}}-1=~{{t}^{2}}$

⇒ $4{{x}^{3}}dx=2t~dt$

⇒ ${{x}^{3}}dx=\frac{t}{2}~dt$

Therefore, 

⇒ $\text{I}=\frac{1}{2}\int \frac{t~dt}{\left( 1+{{t}^{2}} \right)t}$

⇒ $\text{I}=\frac{1}{2}\int \frac{~dt}{\left( 1+{{t}^{2}} \right)}$

⇒ $\text{I}=\frac{1}{2}~ta{{n}^{-1}}t+C$

⇒ $\text{I}=\frac{1}{2}~ta{{n}^{-1}}\left( \sqrt{{{x}^{4}}-1} \right)+C$

Hence, $\int \frac{dx}{x\sqrt{{{x}^{4}}-1}}=\frac{1}{2}~ta{{n}^{-1}}\left( \sqrt{{{x}^{4}}-1} \right)+C.$

Evaluate the Following As Limit of Sums: 

27. $\underset{0}{\overset{2}{\mathop \int }}\,\left( {{x}^{2}}+3 \right)~dx$

Ans: Let $\text{I}=\underset{0}{\overset{2}{\mathop \int }}\,\left( {{x}^{2}}+3 \right)~dx~$

Here, we have $a=0,~b=2$ and $h=\frac{b-a}{n}=\frac{2}{n}$

⇒ $nh=2$ and $f\left( x \right)=\left( {{x}^{2}}+3 \right)$

Now, $\underset{0}{\overset{2}{\mathop \int }}\,\left( {{x}^{2}}+3 \right)~dx=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( 0 \right)+f\left( 0+h \right)+\ldots +f\left\{ 0+\left( n-1 \right)h \right\} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3+{{h}^{2}}+3+{{2}^{2}}{{h}^{2}}+3+\ldots +{{\left( n-1 \right)}^{2}}{{h}^{2}}+3 \right]$ 

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3n+{{h}^{2}}\left\{ {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+\ldots +{{\left( n-1 \right)}^{2}} \right\} \right]$ 

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 3n+{{h}^{2}}\times ~\frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$ 

$=\underset{h\to 0}{\mathop{\lim }}\,\left[ 3nh+{{h}^{3}}\times ~\frac{n\left( n-1 \right)\left( 2n-1 \right)}{6} \right]$ 

$=\underset{h\to 0}{\mathop{\lim }}\,\left[ 3nh+~\frac{nh\left( nh-h \right)\left( 2nh-h \right)}{6} \right]$ 

$=\underset{h\to 0}{\mathop{\lim }}\,\left[ 3.2+~\frac{2\left( 2-h \right)\left( 2.2-h \right)}{6} \right]$ 

$=6+\frac{4\times 4}{6}$ 

$=6+\frac{16}{6}$ 

$=6+\frac{8}{3}$ 

$=\frac{26}{3}$ 

Hence, $\text{I}=\underset{0}{\overset{2}{\mathop \int }}\,\left( {{x}^{2}}+3 \right)~dx=\frac{26}{3}$

28. $\underset{0}{\overset{2}{\mathop \int }}\,{{e}^{x}}~dx$

Ans: let $\text{I}=\underset{0}{\overset{2}{\mathop \int }}\,{{e}^{x}}~dx$

Here, we have $a=0,~b=2$ and $h=\frac{b-a}{n}=\frac{2}{n}$

⇒ $nh=2$ and $f\left( x \right)={{e}^{x}}$

Now, $\underset{0}{\overset{2}{\mathop \int }}\,{{e}^{x}}~dx=\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( 0 \right)+f\left( 0+h \right)+\ldots +f\left\{ 0+\left( n-1 \right)h \right\} \right]$

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ 1+{{e}^{h}}+{{e}^{2h}}+{{e}^{3h}}+\ldots +{{e}^{\left( n-1 \right)h}} \right]$ 

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ \frac{1.~~{{\left( {{e}^{h}} \right)}^{n}}-1}{{{e}^{h}}-1} \right]$ 

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ \frac{{{e}^{nh}}-1}{{{e}^{h}}-1} \right]$ 

$=\underset{h\to 0}{\mathop{\lim }}\,h\left[ \frac{{{e}^{2}}-1}{{{e}^{h}}-1} \right]$ 

$={{e}^{2}}~\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{{{e}^{h}}-1}-~\underset{h\to 0}{\mathop{\lim }}\,\frac{h}{{{e}^{h}}-1}$ 

$={{e}^{2}}-1$ 

Hence, $\text{I}=\underset{0}{\overset{2}{\mathop \int }}\,{{e}^{x}}~dx={{e}^{2}}-1$ .

29. $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\tan x~dx}{1+~{{m}^{2}}~tan{{~}^{2}}x}$

Ans: Let $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\tan x~dx}{1+~{{m}^{2}}~tan{{~}^{2}}x}$

⇒ \[I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\frac{sinx}{cos~x}~dx}{1+~{{m}^{2}}\frac{sin{{~}^{2}}x}{co{{s}^{2}}x}}\]

⇒ $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\frac{sinx}{cos~x}~dx}{\frac{co{{s}^{2}}x+{{m}^{2}}sin{{~}^{2}}x}{co{{s}^{2}}x}}$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{sin~x.~cosx~dx}{co{{s}^{2}}x+{{m}^{2}}sin{{~}^{2}}x}$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{sin~x.~cosx~dx}{1-sin{{~}^{2}}x\left( 1-{{m}^{2}} \right)}$

Now, put $si{{n}^{2}}x=t$

⇒ $2.\sin x.\cos x~dx=dt$

Here, at $x=0\Rightarrow t=0$ and at $x=\frac{\pi }{2}\Rightarrow t=1$

Therefore, 

⇒ $I=\frac{1}{2}\underset{0}{\overset{1}{\mathop \int }}\,\frac{dt}{1-t\left( 1-{{m}^{2}} \right)}$

⇒ $I=\frac{1}{2}~\left[ -\log \left| 1-t\left( 1-{{m}^{2}} \right) \right|\times \frac{1}{1-{{m}^{2}}} \right]_{0}^{1}$ 

⇒ $I=-\frac{1}{2}~\times \frac{1}{1-{{m}^{2}}}\left[ \log \left| 1-t\left( 1-{{m}^{2}} \right) \right| \right]_{0}^{1}$

⇒ $I=-\frac{1}{2\left( 1-{{m}^{2}} \right)}\left[ \log \left| 1-1\left( 1-{{m}^{2}} \right) \right|-\log \left| 1-0\left( 1-{{m}^{2}} \right) \right| \right]$ 

⇒ $I=-\frac{1}{2\left( 1-{{m}^{2}} \right)}\left[ \log \left| 1-1+{{m}^{2}} \right|-\log \left| 1 \right| \right]$ 

⇒ $I=-\frac{1}{2\left( 1-{{m}^{2}} \right)}\left[ \log \left| {{m}^{2}} \right| \right]$

⇒ $I=-\frac{1}{2\left( 1-{{m}^{2}} \right)}2\times \log \left| m \right|$

⇒ $I=-\frac{1}{\left( 1-{{m}^{2}} \right)}\log \left| m \right|$

Hence, $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\tan x~dx}{1+~{{m}^{2}}~tan{{~}^{2}}x}=-\frac{1}{\left( 1-{{m}^{2}} \right)}\log \left| m \right|$ .

30. $\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{\left( x-1 \right)\left( 2-x \right)}}$

Ans: Let $I=\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{\left( x-1 \right)\left( 2-x \right)}}$

⇒ $I=\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{2x-{{x}^{2}}-2+x}}$

⇒ $I=\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{-{{x}^{2}}+3x-2}}$

⇒ $I=\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{-\left( {{x}^{2}}-3x+2 \right)}}$

⇒ $I=\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{-\left[ {{x}^{2}}-3x+{{\left( \frac{3}{2} \right)}^{2}}-{{\left( \frac{3}{2} \right)}^{2}}+2 \right]}}$

⇒ $I=\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{-\left[ \left\{ {{x}^{2}}-3x+{{\left( \frac{3}{2} \right)}^{2}} \right\}-\left\{ \frac{9}{4}-2 \right\} \right]}}$

⇒ $I=\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{-\left[ {{\left( x-\frac{3}{2} \right)}^{2}}-{{\left( \frac{1}{2} \right)}^{2}} \right]}}$

⇒ $I=\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{\left[ {{\left( \frac{1}{2} \right)}^{2}}-{{\left( x-\frac{3}{2} \right)}^{2}} \right]}}$

⇒ $I=\left[ si{{n}^{-1}}\left( \frac{x-\frac{3}{2}}{\frac{1}{2}} \right) \right]_{1}^{2}$

⇒ $I=\left[ si{{n}^{-1}}\left( 2x-3 \right) \right]_{1}^{2}$

⇒ $I=\left[ si{{n}^{-1}}\left( 4-3 \right)-si{{n}^{-1}}\left( 2-3 \right) \right]$

⇒ $I=\left[ si{{n}^{-1}}\left( 1 \right)-si{{n}^{-1}}\left( -1 \right) \right]$

⇒ $I=\left[ \frac{\pi }{2}-\left( -\frac{\pi }{2} \right) \right]$

⇒ $I=\left[ \frac{\pi }{2}+\frac{\pi }{2} \right]$

⇒ $I=\pi $

Hence, $\underset{1}{\overset{2}{\mathop \int }}\,\frac{dx}{\sqrt{\left( x-1 \right)\left( 2-x \right)}}=\pi $

Evaluate the Following: 

31. $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{{{e}^{x}}+{{e}^{-x}}}$

Ans: Let $\text{I}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{{{e}^{x}}+{{e}^{-x}}}$

⇒ $\text{I}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{{{e}^{x}}+\frac{1}{{{e}^{x}}}}$

⇒ $\text{I}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{{{e}^{x}}.dx}{{{\left( {{e}^{x}} \right)}^{2}}+1}$

Now, put ${{e}^{x}}=t$

⇒ ${{e}^{x}}dx=dt$

Now, at $x=0~\Rightarrow t=1$ and at $x=1~\Rightarrow t=e$

Therefore, 

⇒ $\text{I}=\underset{1}{\overset{e}{\mathop \int }}\,\frac{dt}{{{t}^{2}}+1}$

⇒ $\text{I}=\left[ ta{{n}^{-1}}t \right]_{1}^{e}$

⇒ $\text{I}=\left( ta{{n}^{-1}}e-ta{{n}^{-1}}1 \right)$

⇒ $\text{I}=ta{{n}^{-1}}e-~\frac{\pi }{4}$

Hence, $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{{{e}^{x}}+{{e}^{-x}}}=ta{{n}^{-1}}e-~\frac{\pi }{4}$

32. $\underset{0}{\overset{1}{\mathop \int }}\,\frac{x~dx}{\sqrt{1+{{x}^{2}}}}$

Ans: Let $\text{I}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{x~dx}{\sqrt{1+{{x}^{2}}}}$

Now, put $1+{{x}^{2}}={{t}^{2}}$

⇒ $2x~dx=2t~dt$

⇒ $x~dx=t~dt$

here, at $x=0~\Rightarrow t=1$ and at $x=1~\Rightarrow t=\sqrt{2}$

Therefore, 

⇒ $\text{I}=\underset{1}{\overset{\sqrt{2}}{\mathop \int }}\,\frac{t~dt}{\sqrt{{{t}^{2}}}}$

⇒ $\text{I}=\underset{1}{\overset{\sqrt{2}}{\mathop \int }}\,\frac{t~dt}{t}$

⇒ $\text{I}=\underset{1}{\overset{\sqrt{2}}{\mathop \int }}\,dt$

⇒ $\text{I}=\left[ t \right]_{1}^{\sqrt{2}}$

⇒ $\text{I}=\left( \sqrt{2}-1 \right)$

Hence, $\underset{0}{\overset{1}{\mathop \int }}\,\frac{x~dx}{\sqrt{1+{{x}^{2}}}}=\left( \sqrt{2}-1 \right)$

33. $\underset{0}{\overset{\pi }{\mathop \int }}\,x\sin x~co{{s}^{2}}x~dx$

Ans: Let $\text{I}=\underset{0}{\overset{\pi }{\mathop \int }}\,x\sin x~co{{s}^{2}}x~dx$ ………. (i)

⇒ $\text{I}=\underset{0}{\overset{\pi }{\mathop \int }}\,\left( \pi -x \right)\sin \left( \pi -x \right)~co{{s}^{2}}\left( \pi -x \right)~dx$

⇒ $\text{I}=\underset{0}{\overset{\pi }{\mathop \int }}\,\left( \pi -x \right)\sin x~co{{s}^{2}}x~dx$ …………(ii)

Adding eq(i) and (ii), we get 

⇒ $2\text{I}=\pi \underset{0}{\overset{\pi }{\mathop \int }}\,\sin x~co{{s}^{2}}x~dx$

Now, put $\cos x=t$

⇒ $-\sin x~dx=dt$

⇒ $\sin x~dx=-~dt$

here, at $x=0~\Rightarrow t=1$ and at $x=\pi ~\Rightarrow t=-1$

Therefore, 

⇒ $2\text{I}=-\pi \underset{1}{\overset{-1}{\mathop \int }}\,{{t}^{2}}dt$

⇒ $2\text{I}=-\pi \left[ \frac{{{t}^{3}}}{3} \right]_{1}^{-1}$

⇒ $2\text{I}=-\frac{1}{3}\pi \left[ {{t}^{3}} \right]_{1}^{-1}$

⇒ $2\text{I}=-\frac{1}{3}\pi ~\left( -1-1 \right)$

⇒ 2$\text{I}=\frac{2\pi }{3}$

⇒ $\text{I}=\frac{\pi }{3}$

Hence, $\underset{0}{\overset{\pi }{\mathop \int }}\,x\sin x~co{{s}^{2}}x~dx=\frac{\pi }{3}$ .

34. $\underset{0}{\overset{\frac{1}{2}}{\mathop \int }}\,\frac{dx}{\left( 1+{{x}^{2}} \right)\sqrt{1-{{x}^{2}}}}$

Ans: Let $I=~\underset{0}{\overset{\frac{1}{2}}{\mathop \int }}\,\frac{dx}{\left( 1+{{x}^{2}} \right)\sqrt{1-{{x}^{2}}}}$

Put $x=\sin \theta $

⇒ $dx=\cos \theta ~d\theta $

Here, at $x=0\Rightarrow ~\theta =0$ and at $x=\frac{1}{2}\Rightarrow ~\theta =\frac{\pi }{6}$

Therefore, 

⇒ $I=~\underset{0}{\overset{\frac{\pi }{6}}{\mathop \int }}\,\frac{cos~\theta ~d\theta }{\left( 1+si{{n}^{2}}\theta  \right)\sqrt{1-si{{n}^{2}}\theta }}$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{6}}{\mathop \int }}\,\frac{cos~\theta ~d\theta }{\left( 1+si{{n}^{2}}\theta  \right)cos~\theta }$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{6}}{\mathop \int }}\,\frac{~d\theta }{\left( 1+si{{n}^{2}}\theta  \right)}$

Divide by $co{{s}^{2}}\theta $ in numerator and denominator 

⇒ $I=~\underset{0}{\overset{\frac{\pi }{6}}{\mathop \int }}\,\frac{\frac{1}{co{{s}^{2}}\theta }~d\theta }{\left( \frac{1}{co{{s}^{2}}\theta }+\frac{si{{n}^{2}}\theta }{co{{s}^{2}}\theta } \right)}$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{6}}{\mathop \int }}\,\frac{se{{c}^{2}}\theta ~d\theta }{\left( se{{c}^{2}}\theta +ta{{n}^{2}}\theta  \right)}$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{6}}{\mathop \int }}\,\frac{se{{c}^{2}}\theta ~d\theta }{\left( 1+ta{{n}^{2}}\theta +ta{{n}^{2}}\theta  \right)}$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{6}}{\mathop \int }}\,\frac{se{{c}^{2}}\theta ~d\theta }{\left( 1+2ta{{n}^{2}}\theta  \right)}$

Now, put $\tan \theta =t$

⇒ $se{{c}^{2}}\theta ~d\theta =dt$

Here, at $\theta =0\Rightarrow t=0$ and at $\theta =\frac{\pi }{6}\Rightarrow t=\frac{1}{\sqrt{3}}$

Therefore, 

⇒ $I=~\underset{0}{\overset{\frac{1}{\sqrt{3}}}{\mathop \int }}\,\frac{dt}{\left( 1+2{{t}^{2}} \right)}$

⇒ $I=\frac{1}{2}~\underset{0}{\overset{\frac{1}{\sqrt{3}}}{\mathop \int }}\,\frac{dt}{\left( \frac{1}{2}+{{t}^{2}} \right)}$

⇒ $I=\frac{1}{2}~\underset{0}{\overset{\frac{1}{\sqrt{3}}}{\mathop \int }}\,\frac{dt}{\left[ {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+{{t}^{2}} \right]}$

⇒ $I=\frac{1}{2}~\left[ \frac{1}{\frac{1}{\sqrt{2}}}~ta{{n}^{-1}}\left( \frac{t}{\frac{1}{\sqrt{2}}} \right) \right]_{0}^{\frac{1}{\sqrt{3}}}$

⇒ $I=\frac{1}{\sqrt{2}}~\left[ ~ta{{n}^{-1}}\left( \sqrt{2}~\times t \right) \right]_{0}^{\frac{1}{\sqrt{3}}}$

⇒ $I=~\frac{1}{\sqrt{2}}\left[ ta{{n}^{-1}}\left( \sqrt{2}~\times \frac{1}{\sqrt{3}} \right)-ta{{n}^{-1}}\left( 0 \right) \right]$

⇒ $I=~\frac{1}{\sqrt{2}}ta{{n}^{-1}}\left( ~\frac{\sqrt{2}}{\sqrt{3}} \right)$

Long Answer (L.A)

35. $\int \frac{{{x}^{2}}~dx}{{{x}^{4}}-{{x}^{2}}-12}$

Ans: Let $I=~\int \frac{{{x}^{2}}~dx}{{{x}^{4}}-{{x}^{2}}-12}$

⇒ $I=~\int \frac{{{x}^{2}}~dx}{{{x}^{4}}-4{{x}^{2}}+3{{x}^{2}}-12}$

⇒ $I=~\int \frac{{{x}^{2}}~dx}{{{x}^{2}}\left( {{x}^{2}}-4 \right)+3\left( {{x}^{2}}-4 \right)}$

⇒ $I=~\int \frac{{{x}^{2}}~dx}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+3 \right)}$

⇒ $I=~\int \frac{{{x}^{2}}-4+~4~dx}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+3 \right)}$

$\Rightarrow I=~\int \frac{\left( {{x}^{2}}-4 \right)~dx}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+3 \right)}+4\int \frac{dx}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+3 \right)}$ 

$\Rightarrow I=~\int \frac{~dx}{\left( {{x}^{2}}+3 \right)}+4\int \frac{\left[ \left( {{x}^{2}}-4 \right)-({{x}^{2}}+3 \right)\}dx}{\left( -7 \right)\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+3 \right)}$ 

$\Rightarrow I=~\int \frac{~dx}{\left( {{x}^{2}}+3 \right)}-\frac{4}{7}\int \frac{\left( {{x}^{2}}-4 \right)dx}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+3 \right)}+\frac{4}{7}\int \frac{({{x}^{2}}+3)dx}{\left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+3 \right)}$ 

$\Rightarrow I=~\int \frac{~dx}{\left( {{x}^{2}}+3 \right)}-\frac{4}{7}\int \frac{dx}{\left( {{x}^{2}}+3 \right)}+\frac{4}{7}\int \frac{dx}{\left( {{x}^{2}}-4 \right)}$ 

$\Rightarrow I=~\frac{3}{7}\int \frac{dx}{\left( {{x}^{2}}+3 \right)}+\frac{4}{7}\int \frac{dx}{\left( {{x}^{2}}-4 \right)}$ 

$\Rightarrow I=~\frac{3}{7}\int \frac{dx}{\left( {{x}^{2}}+{{\left( \sqrt{3} \right)}^{2}} \right)}+\frac{4}{7}\int \frac{dx}{\left( {{x}^{2}}-{{2}^{2}} \right)}$ 

$\Rightarrow I=~\frac{3}{7}\left[ \frac{1}{\sqrt{3}}~ta{{n}^{-1}}\frac{x}{\sqrt{3}} \right]+\frac{4}{7}.\frac{1}{2\times 2}\log \left| \frac{x-2}{x+2} \right|+C$ 

$\Rightarrow I=~\frac{\sqrt{3}}{7}~ta{{n}^{-1}}\frac{x}{\sqrt{3}}+\frac{1}{7}\log \left| \frac{x-2}{x+2} \right|+C$ 

Hence, $\int \frac{{{x}^{2}}~dx}{{{x}^{4}}-{{x}^{2}}-12}=\frac{\sqrt{3}}{7}~ta{{n}^{-1}}\frac{x}{\sqrt{3}}+\frac{1}{7}\log \left| \frac{x-2}{x+2} \right|+C.$

36. $\int \frac{{{x}^{2}}~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~\left( {{x}^{2}}+{{b}^{2}} \right)~}$

Ans: Let $I=~\int \frac{{{x}^{2}}+{{a}^{2}}-{{a}^{2}}~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~\left( {{x}^{2}}+{{b}^{2}} \right)~}$

⇒ $I=~\int \frac{({{x}^{2}}+{{a}^{2}})~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~\left( {{x}^{2}}+{{b}^{2}} \right)~}-~\int \frac{{{a}^{2}}~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~\left( {{x}^{2}}+{{b}^{2}} \right)~}$

⇒ $I=~\int \frac{~dx}{~\left( {{x}^{2}}+{{b}^{2}} \right)~}-~{{a}^{2}}\int \frac{\left[ \left( {{x}^{2}}+{{a}^{2}} \right)-\left( {{x}^{2}}+{{b}^{2}} \right) \right]~dx}{\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{x}^{2}}+{{a}^{2}} \right)~\left( {{x}^{2}}+{{b}^{2}} \right)~}$

⇒ $I=~\int \frac{~dx}{~\left( {{x}^{2}}+{{b}^{2}} \right)~}-\frac{{{a}^{2}}}{\left( {{a}^{2}}-{{b}^{2}} \right)}\int \frac{\left[ \left( {{x}^{2}}+{{a}^{2}} \right)-\left( {{x}^{2}}+{{b}^{2}} \right) \right]~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~\left( {{x}^{2}}+{{b}^{2}} \right)~}$

⇒ $I=~\int \frac{~dx}{~\left( {{x}^{2}}+{{b}^{2}} \right)~}-\frac{{{a}^{2}}}{\left( {{a}^{2}}-{{b}^{2}} \right)}\left[ \int \frac{\left( {{x}^{2}}+{{a}^{2}} \right)~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~\left( {{x}^{2}}+{{b}^{2}} \right)~}-\int \frac{\left( {{x}^{2}}+{{b}^{2}} \right)~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~\left( {{x}^{2}}+{{b}^{2}} \right)~} \right]$

⇒ $I=~\int \frac{~dx}{~\left( {{x}^{2}}+{{b}^{2}} \right)~}-\frac{{{a}^{2}}}{\left( {{a}^{2}}-{{b}^{2}} \right)}\left[ \int \frac{~dx}{~\left( {{x}^{2}}+{{b}^{2}} \right)~}-\int \frac{~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~~} \right]$

⇒ $I=\frac{-{{b}^{2}}}{\left( {{a}^{2}}-{{b}^{2}} \right)}~\int \frac{~dx}{~\left( {{x}^{2}}+{{b}^{2}} \right)~}+\frac{{{a}^{2}}}{\left( {{a}^{2}}-{{b}^{2}} \right)}\int \frac{~dx}{\left( {{x}^{2}}+{{a}^{2}} \right)~~}$

⇒ $I=\frac{{{b}^{2}}}{\left( {{b}^{2}}-{{a}^{2}} \right)}\frac{1}{b}ta{{n}^{-1}}\frac{x}{b}-\frac{{{a}^{2}}}{\left( {{b}^{2}}-{{a}^{2}} \right)}\frac{1}{a}~ta{{n}^{-1}}\frac{x}{a}+C$

⇒ $I=\frac{b}{\left( {{b}^{2}}-{{a}^{2}} \right)}ta{{n}^{-1}}\frac{x}{b}-\frac{a}{\left( {{b}^{2}}-{{a}^{2}} \right)}~ta{{n}^{-1}}\frac{x}{a}+C$

37. $\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{x}{1+\sin x}dx$

Ans: Let $I=\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{x}{1+\sin x}dx~$ ……….(i)

⇒ $I=\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{\left( \pi -x \right)}{1+\sin (\pi -x)}dx$

⇒ $I=\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{\left( \pi -x \right)}{1+\sin x}dx$ ………….(ii)

Now, adding eq(i) and (ii), we get

$\Rightarrow 2I=\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{\text{ }\!\!\pi\!\!\text{ }}{1+\sin x}dx$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{(1-\sin x)}{(1+\sin x)(1-\sin x)}dx$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{(1-\sin x)}{\left( 1-si{{n}^{2}}x \right)}dx$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{(1-\sin x)}{co{{s}^{2}}x}dx$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\underset{0}{\overset{\pi }{\mathop \int }}\,\left( \frac{1}{co{{s}^{2}}x}-\frac{\sin x}{co{{s}^{2}}x} \right)dx$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\underset{0}{\overset{\pi }{\mathop \int }}\,\left( se{{c}^{2}}x-\sec x.\tan x \right)dx$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\left[ \tan x-\sec x \right]_{0}^{\pi }$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\left[ \tan \pi -\sec \pi ~-(\tan 0-\sec 0~) \right]_{0}^{\pi }$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\left[ 0-\left( -1 \right)-\left( 0-1 \right) \right]$ 

$\Rightarrow 2I=\text{ }\!\!\pi\!\!\text{ }\left[ 1+1 \right]$ 

$\Rightarrow 2I=2\text{ }\!\!\pi\!\!\text{ }$ 

$\Rightarrow I=\text{ }\!\!\pi\!\!\text{ }$ 

Hence, $\underset{0}{\overset{\pi }{\mathop \int }}\,\frac{x}{1+\sin x}dx=\pi $

38. $\int \frac{2x-1}{\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)}dx$

Ans: Let $I=~\int \frac{2x-1}{\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)}dx$

Now, $\frac{2x-1}{\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)}=\frac{A}{\left( x-1 \right)}+\frac{B}{\left( x+2 \right)}+\frac{C}{\left( x-3 \right)}$

⇒ $2x-1=A\left( x+2 \right)\left( x-3 \right)+B\left( x-1 \right)\left( x-3 \right)+C\left( x-1 \right)\left( x+2 \right)$

Here, put $x=1$ in above equation, we get

⇒ $1=A\left( 1+2 \right)\left( 1-3 \right)+0+0$

⇒ $1=-6~A$

⇒ $A=-\frac{1}{6}~$

Now, put $x=-2$

⇒ $2\left( -2 \right)-1=0+B\left( -2-1 \right)\left( -2-3 \right)+0$

⇒ $-5=B\left( -3 \right)\left( -5 \right)$

⇒ $B=-\frac{1}{3}$

Now, put $x=3$, we get

⇒ $2\left( 3 \right)-1=0+0+C\left( 3-1 \right)\left( 3+2 \right)$

⇒ $5=C\left( 2 \right)\left( 5 \right)$

⇒ $C=\frac{1}{2}$

Therefore, 

$\Rightarrow \int \frac{2x-1}{\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)}dx=-\int \frac{dx}{6\left( x-1 \right)}-\int \frac{dx}{3\left( x+2 \right)}+\int \frac{dx}{2\left( x-3 \right)}$ 

$=-\frac{1}{6}\log \left| \left( x-1 \right) \right|-\frac{1}{3}\log \left| \left( x+2 \right) \right|+\frac{1}{2}\log \left| \left( x-3 \right) \right|+C$ 

$=-\log {{\left( x-1 \right)}^{\frac{1}{6}}}-\log {{\left( x+2 \right)}^{\frac{1}{3}}}+\log {{\left( x-3 \right)}^{\frac{1}{2}}}+C$ 

$=\log \left| \frac{\sqrt{\left( x-3 \right)}}{{{\left( x-1 \right)}^{\frac{1}{6}}}~{{\left( x+2 \right)}^{\frac{1}{3}}}} \right|+C$ 

Hence, $\int \frac{2x-1}{\left( x-1 \right)\left( x+2 \right)\left( x-3 \right)}dx=\log \left| \frac{\sqrt{\left( x-3 \right)}}{{{\left( x-1 \right)}^{\frac{1}{6}}}~{{\left( x+2 \right)}^{\frac{1}{3}}}} \right|+C$

39. $\int {{e}^{ta{{n}^{-1}}x}}\left( \frac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)dx$

Ans: Let $\text{I}=\int {{e}^{ta{{n}^{-1}}x}}\left( \frac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)dx$

⇒ $\text{I}=\int {{e}^{ta{{n}^{-1}}x}}\left( \frac{1+{{x}^{2}}}{1+{{x}^{2}}}+\frac{x}{1+{{x}^{2}}} \right)dx$

⇒ $\text{I}=\int {{e}^{ta{{n}^{-1}}x}}\left( 1+\frac{x}{1+{{x}^{2}}} \right)dx$

⇒ $\text{I}=\int {{e}^{ta{{n}^{-1}}x}}dx+~\int \frac{x{{e}^{ta{{n}^{-1}}x}}}{1+{{x}^{2}}}dx$

⇒ $\text{I}={{\text{I}}_{1}}+{{\text{I}}_{2}}$ …...(i), where,$~{{I}_{1}}=\int {{e}^{ta{{n}^{-1}}x}}dx~~\And ~~{{I}_{2}}=~\int \frac{x{{e}^{ta{{n}^{-1}}x}}}{1+{{x}^{2}}}dx$

Now, ${{\text{I}}_{2}}=~\int \frac{x{{e}^{ta{{n}^{-1}}x}}}{1+{{x}^{2}}}dx$

Put $ta{{n}^{-1}}x=t$

⇒ $\frac{1}{1+{{x}^{2}}}dx=dt$

Therefore, 

⇒ ${{\text{I}}_{2}}=~\int {{e}^{t}}.\tan t~dt$

⇒ ${{\text{I}}_{2}}=~\int {{e}^{t}}.\tan t~dt$

⇒ ${{\text{I}}_{2}}=\tan t\int {{e}^{t}}~dt-\int \left( \frac{d}{dx}\tan t.\int {{e}^{t}}~dt \right)dt$

⇒ ${{\text{I}}_{2}}=\tan t.~{{e}^{t}}-\int se{{c}^{2}}t.~~{{e}^{t}}dt+C$

⇒ ${{\text{I}}_{2}}=x.~{{e}^{ta{{n}^{-1}}x}}-\int (1+ta{{n}^{2}}t).~~{{e}^{ta{{n}^{-1}}x}}\frac{1}{1+{{x}^{2}}}dx+C$

⇒ ${{\text{I}}_{2}}=x.~{{e}^{ta{{n}^{-1}}x}}-\int (1+{{x}^{2}}).~~{{e}^{ta{{n}^{-1}}x}}\frac{1}{1+{{x}^{2}}}dx+C$

⇒ ${{\text{I}}_{2}}=x.~{{e}^{ta{{n}^{-1}}x}}-\int {{e}^{ta{{n}^{-1}}x}}dx+C~$

Put values of ${{\text{I}}_{1}}$ and ${{\text{I}}_{2}}$, in equation (i), 

⇒ $\text{I}=\int {{e}^{ta{{n}^{-1}}x}}dx+x.~{{e}^{ta{{n}^{-1}}x}}-\int {{e}^{ta{{n}^{-1}}x}}dx+C$

⇒ $\text{I}=x.~{{e}^{ta{{n}^{-1}}x}}+C$

Hence, value of $\int {{e}^{ta{{n}^{-1}}x}}\left( \frac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)dx=x.~{{e}^{ta{{n}^{-1}}x}}+C.$

40. $\int si{{n}^{-1}}\sqrt{\frac{x}{a+x}}dx$

Ans: Let $I=~\int si{{n}^{-1}}\sqrt{\frac{x}{a+x}~}~dx$

Put $x=a~ta{{n}^{2}}\theta $

⇒ $dx=a.~2\tan \theta .se{{c}^{2}}\theta ~d\theta $

Therefore, 

$\Rightarrow I=~\int si{{n}^{-1}}\sqrt{\frac{a~ta{{n}^{2}}\theta }{a+a~ta{{n}^{2}}\theta }~}~\times $ $~2a\tan \theta .se{{c}^{2}}\theta ~d\theta $

$\Rightarrow I=~\int si{{n}^{-1}}\sqrt{\frac{a~ta{{n}^{2}}\theta }{a\left( 1+~ta{{n}^{2}}\theta  \right)}~}~\times $ $~2a\tan \theta .se{{c}^{2}}\theta ~d\theta $

$\Rightarrow I=~\int si{{n}^{-1}}\sqrt{\frac{~ta{{n}^{2}}\theta }{\left( 1+~ta{{n}^{2}}\theta  \right)}~}~\times $ $~2a\tan \theta .se{{c}^{2}}\theta ~d\theta $

$\Rightarrow I=~\int si{{n}^{-1}}\sqrt{\frac{~ta{{n}^{2}}\theta }{se{{c}^{2}}\theta }~}~\times $ $~2a\tan \theta .se{{c}^{2}}\theta ~d\theta $

$\Rightarrow I=~\int si{{n}^{-1}}\sqrt{si{{n}^{2}}\theta ~}~\times $ $~2a\tan \theta .se{{c}^{2}}\theta ~d\theta $

$\Rightarrow I=~\int si{{n}^{-1}}\left( \sin \theta  \right)~\times $ $~2a\tan \theta .se{{c}^{2}}\theta ~d\theta $

$\Rightarrow I=~2a\int \theta ~\tan \theta .se{{c}^{2}}\theta ~d\theta $ 

$\Rightarrow I=~2a~\left[ \theta \int ~\tan \theta .se{{c}^{2}}\theta ~d\theta -\int \left( \frac{d}{d\theta }\theta .~~\int \tan \theta .se{{c}^{2}}\theta ~d\theta  \right)d\theta  \right]$ 

Put $\tan \theta =t$

$\Rightarrow $ $se{{c}^{2}}\theta ~d\theta =dt$

$\Rightarrow I=~2a~\left[ \theta \int ~t~dt-\int \left( ~~\int t~dt \right)d\theta  \right]$ 

$\Rightarrow I=~2a~\left[ \theta .~\frac{{{t}^{2}}}{2}-\int \frac{{{t}^{2}}}{2}d\theta  \right]$ 

$\Rightarrow I=~2a~\left[ \theta .~\frac{ta{{n}^{2}}\theta }{2}-\int \frac{ta{{n}^{2}}\theta }{2}d\theta  \right]$ 

$\Rightarrow I=~a~\left[ \theta \left( ta{{n}^{2}}\theta  \right)-\text{ }\!\!~\!\!\text{ }\int ta{{n}^{2}}\theta ~d\theta  \right]$ 

$\Rightarrow I=~a\theta \left( ta{{n}^{2}}\theta  \right)-$ $a\int (se{{c}^{2}}\theta -1)~d\theta $

$\Rightarrow I=~a\theta \left( ta{{n}^{2}}\theta  \right)-a\tan \theta +~a\theta +C$ 

$\Rightarrow I=~a\left[ \frac{x}{a}\left( ta{{n}^{-1}}\sqrt{\frac{x}{a}} \right)-\sqrt{\frac{x}{a}}+~ta{{n}^{-1}}\sqrt{\frac{x}{a}} \right]+C$ 

Hence, $\int si{{n}^{-1}}\sqrt{\frac{x}{a+x}~}~dx=a\left[ \frac{x}{a}\left( ta{{n}^{-1}}\sqrt{\frac{x}{a}} \right)-\sqrt{\frac{x}{a}}+~ta{{n}^{-1}}\sqrt{\frac{x}{a}} \right]+C$ .

41. $\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\sqrt{1+\cos x}}{{{\left( 1-\cos x \right)}^{\frac{5}{2}}}}~dx\ldots $

Ans: Let $\text{I}=\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\sqrt{1+\cos x}}{{{\left( 1-\cos x \right)}^{\frac{5}{2}}}}~dx$

⇒ $\text{I}=\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\sqrt{2\text{co}{{\text{s}}^{2}}\frac{x}{2}}}{{{\left( 2~si{{n}^{2}}\frac{x}{2} \right)}^{\frac{5}{2}}}}~dx$

⇒ $\text{I}=\frac{\sqrt{2}}{{{2}^{\frac{5}{2}}}}\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\text{cos }\!\!~\!\!\text{ }\frac{x}{2}}{si{{n}^{5}}\frac{x}{2}}~dx$

Put $\sin \frac{x}{2}=t$

⇒ $\frac{1}{2}\cos \frac{x}{2}~dx=dt$

⇒ $\cos \frac{x}{2}~dx=2dt$

Here, at $x=\frac{\pi }{3}~\Rightarrow t=\frac{1}{2}$ and at $x=\frac{\pi }{2}~\Rightarrow t=\frac{1}{\sqrt{2}}$

Therefore, 

⇒ $\text{I}=\frac{\sqrt{2}}{\sqrt{32}}\underset{\frac{1}{2}}{\overset{\frac{1}{\sqrt{2}}}{\mathop \int }}\,\frac{2~dt}{{{t}^{5}}}~$

⇒ $\text{I}=\frac{\sqrt{2}}{4\sqrt{2}}\times 2\underset{\frac{1}{2}}{\overset{\frac{1}{\sqrt{2}}}{\mathop \int }}\,~{{t}^{-5}}dt~$

⇒ $\text{I}=\frac{1}{2}~\left[ \frac{{{t}^{-4}}}{-4} \right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}$

⇒ $\text{I}=-\frac{1}{8}~\left[ {{t}^{-4}} \right]_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}}$

⇒ $\text{I}=-\frac{1}{8}~\left[ {{\left( \frac{1}{\sqrt{2}} \right)}^{-4}}-{{\left( \frac{1}{2} \right)}^{-4}} \right]$

⇒ $\text{I}=-\frac{1}{8}~\left[ {{\left( \sqrt{2} \right)}^{4}}-{{\left( 2 \right)}^{4}} \right]$

⇒ $\text{I}=-\frac{1}{8}~\left[ 4-16 \right]$

⇒ $\text{I}=\frac{1}{8}~\times 12$

⇒ $\text{I}=\frac{3}{2}~$

Hence, $\underset{\frac{\pi }{3}}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\sqrt{1+\cos x}}{{{\left( 1-\cos x \right)}^{\frac{5}{2}}}}~dx=\frac{3}{2}~.$

42. $\int {{e}^{-3x}}co{{s}^{3}}x~dx$

Ans: Let $\text{I}=~\int {{e}^{-3x}}co{{s}^{3}}x~dx$

⇒ $\text{I}=~\int {{e}^{-3x}}\left( \frac{\cos 3x+3\cos x}{4} \right)~dx$

⇒ $\text{I}=\frac{1}{4}\int \left( {{e}^{-3x}}\cos 3x+3{{e}^{-3x}}\cos x \right)~dx$

⇒ $\text{I}=\frac{1}{4}\int {{e}^{-3x}}\cos 3x~dx+\frac{3}{4}~\int {{e}^{-3x}}\cos x~dx$

⇒ $\text{I}=\frac{1}{4}\text{ }\!\!~\!\!\text{ }{{\text{I}}_{1}}+\frac{3}{4}\text{ }\!\!~\!\!\text{ }{{\text{I}}_{2}}$ ………… eq (i)

Where, ${{\text{I}}_{1}}=\int {{e}^{-3x}}\cos 3x~dx$ and ${{\text{I}}_{2}}=\int 3{{e}^{-3x}}\cos x~dx$

Now, ${{\text{I}}_{1}}=\int {{e}^{-3x}}\cos 3x~dx$

⇒ ${{\text{I}}_{1}}=\cos 3x\int {{e}^{-3x}}~dx-\int \left( \frac{d}{dx}\cos 3x~.\int {{e}^{-3x}}~dx \right)dx$

⇒ ${{\text{I}}_{1}}=\frac{{{e}^{-3x}}}{-3}\cos 3x-\int \left( -3\text{ }\!\!~\!\!\text{ sin}3x~.\frac{{{e}^{-3x}}}{-3} \right)dx$

⇒ ${{\text{I}}_{1}}=\frac{{{e}^{-3x}}}{-3}\cos 3x-\int \left( {{e}^{-3x}}\text{ }\!\!~\!\!\text{ sin}3x~. \right)dx$

⇒ ${{\text{I}}_{1}}=\frac{{{e}^{-3x}}}{-3}\cos 3x-\left[ \sin 3x\int {{e}^{-3x}}dx-\int \left( \frac{d}{dx}\sin 3x~.\int {{e}^{-3x}}~dx \right)dx \right]$

⇒ ${{\text{I}}_{1}}=\frac{{{e}^{-3x}}}{-3}\cos 3x-\left[ \frac{{{e}^{-3x}}}{-3}\sin 3x-\int \left( 3\cos 3x~.\frac{{{e}^{-3x}}}{-3} \right)dx \right]$

⇒ ${{\text{I}}_{1}}=\frac{{{e}^{-3x}}}{-3}\cos 3x-\left[ \frac{{{e}^{-3x}}}{-3}\sin 3x+\int \left( {{e}^{-3x}}\cos 3x~ \right)dx \right]$

⇒ ${{\text{I}}_{1}}=\frac{{{e}^{-3x}}}{-3}\cos 3x-\left[ \frac{{{e}^{-3x}}}{-3}\sin 3x+{{\text{I}}_{1}} \right]$

⇒ ${{\text{I}}_{1}}=\frac{{{e}^{-3x}}}{-3}\cos 3x+\frac{{{e}^{-3x}}}{3}\sin 3x-{{\text{I}}_{1}}$

⇒ $2{{\text{I}}_{1}}=\frac{{{e}^{-3x}}}{-3}\cos 3x+\frac{{{e}^{-3x}}}{3}\sin 3x$

⇒ $2{{\text{I}}_{1}}=\frac{1}{3}{{e}^{-3x}}(-\cos 3x+\sin 3x)+{{C}_{1}}$

⇒ ${{\text{I}}_{1}}=\frac{1}{6}{{e}^{-3x}}(\sin 3x-\cos 3x)+{{C}_{1}}$

Now, ${{\text{I}}_{2}}=\int {{e}^{-3x}}\cos x~dx$

⇒ ${{\text{I}}_{2}}=\int {{e}^{-3x}}\cos x~dx$

⇒ ${{\text{I}}_{2}}=\cos x\int {{e}^{-3x}}~dx-\int \left( \frac{d}{dx}\cos x.\int {{e}^{-3x}}~dx~ \right)dx$

⇒ ${{\text{I}}_{2}}=\frac{{{e}^{-3x}}}{-3}\cos x-\int \left( -\sin x.\frac{{{e}^{-3x}}}{-3}~ \right)dx$

⇒ ${{\text{I}}_{2}}=\frac{{{e}^{-3x}}}{-3}\cos x-\frac{1}{3}\int \left( {{e}^{-3x}}.\sin x~ \right)dx$

⇒ ${{\text{I}}_{2}}=\frac{{{e}^{-3x}}}{-3}\cos x-\frac{1}{3}\left[ \sin x\int {{e}^{-3x}}dx-\int \left( \frac{d}{dx}\sin x.\int {{e}^{-3x}}dx \right)dx \right]$

⇒ ${{\text{I}}_{2}}=\frac{{{e}^{-3x}}}{-3}\cos x-\frac{1}{3}\left[ \frac{{{e}^{-3x}}}{-3}\sin x-\int \left( \frac{{{e}^{-3x}}}{-3}\cos x \right)dx \right]$

⇒ ${{\text{I}}_{2}}=\frac{{{e}^{-3x}}}{-3}\cos x-\frac{1}{3}\left[ \frac{{{e}^{-3x}}}{-3}\sin x+\frac{1}{3}\int \left( {{e}^{-3x}}\cos x \right)dx \right]$

⇒ ${{\text{I}}_{2}}=\frac{{{e}^{-3x}}}{-3}\cos x+\frac{1}{9}{{e}^{-3x}}\sin x-\frac{1}{9}\int \left( {{e}^{-3x}}\cos x \right)dx$

⇒ ${{\text{I}}_{2}}=\frac{{{e}^{-3x}}}{-3}\cos x+\frac{1}{9}{{e}^{-3x}}\sin x-\frac{1}{9}{{\text{I}}_{2}}$

⇒ ${{\text{I}}_{2}}+\frac{{{\text{I}}_{2}}}{9}=\frac{{{e}^{-3x}}}{-3}\cos x+\frac{1}{9}{{e}^{-3x}}\sin x$

⇒ $\frac{10{{\text{I}}_{2}}}{9}=\frac{-3{{e}^{-3x}}.\cos x+{{e}^{-3x}}\sin x}{9}+{{C}_{2}}$

⇒ $10{{\text{I}}_{2}}={{e}^{-3x}}\left( \sin x-3\cos x \right)+{{C}_{2}}$

⇒ ${{\text{I}}_{2}}=\frac{1}{10}{{e}^{-3x}}\left( \sin x-3\cos x \right)+{{C}_{2}}$

Now, put the value of ${{\text{I}}_{1}}$ and ${{\text{I}}_{2}}$ in eq(i), we get

⇒ $\text{I}=\frac{1}{24}{{e}^{-3x}}(\sin 3x-\cos 3x)+\frac{3}{40}{{e}^{-3x}}\left( \sin x-3\cos x \right)+C\text{ }\!\!~\!\!\text{ }$

43. $\int \sqrt{\tan x}~dx$

Ans: Let $\text{I}=\int \sqrt{\tan x}~dx$

Put $\tan x=~{{t}^{2}}$

⇒ $se{{c}^{2}}x~dx=2t~dt$

⇒ $(1+ta{{n}^{2}}x)~dx=2t~dt$

⇒$~dx=\frac{2t}{(1+ta{{n}^{2}}x)~}~dt$

⇒$~dx=\frac{2t}{(1+{{t}^{4}})~}~dt$

Therefore, 

⇒ $\text{I}=\int \sqrt{{{t}^{2}}}~.~\frac{2t}{(1+{{t}^{4}})~}~dt$

⇒ $\text{I}=\int ~\frac{2{{t}^{2}}}{(1+{{t}^{4}})~}~dt$

⇒ $\text{I}=\int ~\frac{\left( 1+{{t}^{2}} \right)-\left( 1-{{t}^{2}} \right)}{(1+{{t}^{4}})~}~dt$

⇒ $\text{I}=\int ~\frac{\left( 1+{{t}^{2}} \right)}{(1+{{t}^{4}})~}~dt-\int ~\frac{\left( 1-{{t}^{2}} \right)}{(1+{{t}^{4}})~}~dt$

⇒ $\text{I}=\int ~\frac{\left( \frac{1}{{{t}^{2}}}~+1 \right)}{\left( \frac{1}{{{t}^{2}}}~+{{t}^{2}} \right)~}~dt-\int ~\frac{\left( \frac{1}{{{t}^{2}}}-1 \right)}{~\left( \frac{1}{{{t}^{2}}}~+{{t}^{2}} \right)}~dt$

⇒ $\text{I}=\int ~\frac{\left( \frac{1}{{{t}^{2}}}~+1 \right)}{{{\left( t~-\frac{1}{t} \right)}^{2}}+2~}~dt-\int ~\frac{\left( \frac{1}{{{t}^{2}}}-1 \right)}{~{{\left( t+~\frac{1}{t} \right)}^{2}}-2}~dt$

Put $u=\left( t~-\frac{1}{t} \right)~\Rightarrow du=~\left( \frac{1}{{{t}^{2}}}+1 \right)dt$

And $v=\left( t+\frac{1}{t} \right)~\Rightarrow du=~-\left( \frac{1}{{{t}^{2}}}-1 \right)dt$

Therefore,

⇒ $\text{I}=\int ~\frac{~du}{{{\left( u \right)}^{2}}+2~}~-\int ~\frac{-dv}{~{{\left( v \right)}^{2}}-2}~$

⇒ $\text{I}=\int ~\frac{~du}{{{\left( u \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}~}+\int ~\frac{dv}{~{{\left( v \right)}^{2}}-{{\left( \sqrt{2} \right)}^{2}}}$ 

⇒ $\text{I}=\frac{1}{\sqrt{2}}~ta{{n}^{-1}}\frac{u}{\sqrt{2}}+\frac{1}{2\sqrt{2}}\log \left| \frac{v-\sqrt{2}}{v+\sqrt{2}} \right|+C$ 

⇒ $\text{I}=\frac{1}{\sqrt{2}}~ta{{n}^{-1}}\frac{\left( t~-\frac{1}{t} \right)}{\sqrt{2}}+\frac{1}{2\sqrt{2}}\log \left| \frac{\left( t~+~\frac{1}{t} \right)-\sqrt{2}}{\left( t~+~\frac{1}{t} \right)+\sqrt{2}} \right|+C$

⇒ $\text{I}=\frac{1}{\sqrt{2}}~ta{{n}^{-1}}\frac{\left( \sqrt{\tan x}~-\frac{1}{\sqrt{\tan x}} \right)}{\sqrt{2}}+\frac{1}{2\sqrt{2}}\log \left| \frac{\left( \sqrt{\tan x}~+~\frac{1}{\sqrt{\tan x}} \right)-\sqrt{2}}{\left( \sqrt{\tan x}~+~\frac{1}{\sqrt{\tan x}} \right)+\sqrt{2}} \right|+C$

⇒ $\text{I}=\frac{1}{\sqrt{2}}~ta{{n}^{-1}}\frac{\left( \tan x~-1 \right)}{\sqrt{2\tan x}}+\frac{1}{2\sqrt{2}}\log \left| \frac{\tan x-\sqrt{2\tan x}~+1}{\tan x~+~\sqrt{2\tan x}~+1} \right|+C$

44. $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{dx}{{{\left( {{a}^{2}}co{{s}^{2}}x+~{{b}^{2}}si{{n}^{2}}x \right)}^{2}}}$

Ans: Let $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{dx}{{{\left( {{a}^{2}}co{{s}^{2}}x+~{{b}^{2}}si{{n}^{2}}x \right)}^{2}}}$

Divide numerator and denominator by $co{{s}^{4}}x$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\frac{dx}{co{{s}^{4}}x}}{\frac{{{\left( {{a}^{2}}co{{s}^{2}}x+~{{b}^{2}}si{{n}^{2}}x \right)}^{2}}}{co{{s}^{4}}x}}$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{se{{c}^{4}}x~dx}{{{\left( {{a}^{2}}+~{{b}^{2}}ta{{n}^{2}}x \right)}^{2}}}$

⇒ $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\left( ~1+~ta{{n}^{2}}x \right)~se{{c}^{2}}x~dx}{{{\left( {{a}^{2}}+~{{b}^{2}}ta{{n}^{2}}x \right)}^{2}}}$

Put $b\tan x=a\tan t$

⇒ $~b~se{{c}^{2}}x~dx=a~se{{c}^{2}}t~dt$

⇒ $~~se{{c}^{2}}x~dx=\frac{a}{b}~se{{c}^{2}}t~dt$

When $x\to 0,~t\to 0$ and When $x\to \frac{\pi }{2},\tan t\to 0~,~t\to \frac{\pi }{2}$

Therefore, 

⇒ $I=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{\left( 1+~{{\left( \frac{a}{b}\tan t \right)}^{2}} \right)}{{{a}^{4}}se{{c}^{4}}t}\frac{a}{b}~se{{c}^{2}}t~dt$

⇒ $I=\frac{1}{{{a}^{3}}{{b}^{3}}}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( {{b}^{2}}+{{a}^{2}}ta{{n}^{2}}t \right)co{{s}^{2}}t~dt$

⇒ $I=\frac{1}{{{a}^{3}}{{b}^{3}}}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( {{b}^{2}}co{{s}^{2}}t+{{a}^{2}}si{{n}^{2}}t \right)~dt$

⇒ $I=\frac{1}{{{a}^{3}}{{b}^{3}}}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( {{b}^{2}}~\frac{1+\cos 2t}{2}+{{a}^{2}}~\frac{1-\cos 2t}{2} \right)~dt$

⇒ $I=\frac{1}{2{{a}^{3}}{{b}^{3}}}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left[ ({{a}^{2}}+{{b}^{2}})+\left( ~{{b}^{2}}-{{a}^{2}} \right)\cos 2t \right]dt$

⇒ $I=\frac{1}{2{{a}^{3}}{{b}^{3}}}\left[ ({{a}^{2}}+{{b}^{2}})x-\frac{1}{2}\left( ~{{b}^{2}}-{{a}^{2}} \right)\sin 2t~ \right]_{0}^{\frac{\pi }{2}}$

⇒ $I=\frac{1}{2{{a}^{3}}{{b}^{3}}}\left[ ({{a}^{2}}+{{b}^{2}})\frac{\pi }{2}-\frac{1}{2}\left( ~{{b}^{2}}-{{a}^{2}} \right)\sin \pi  \right]$

⇒ $I=\frac{1}{2{{a}^{3}}{{b}^{3}}}\left[ ({{a}^{2}}+{{b}^{2}})\frac{\pi }{2} \right]$

⇒ $I=\frac{\pi ({{a}^{2}}+{{b}^{2}})}{4{{a}^{3}}{{b}^{3}}~}$

Hence, $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\frac{dx}{{{\left( {{a}^{2}}co{{s}^{2}}x+~{{b}^{2}}si{{n}^{2}}x \right)}^{2}}}=\frac{\pi ({{a}^{2}}+{{b}^{2}})}{4{{a}^{3}}{{b}^{3}}~}.$

45. $\underset{0}{\overset{1}{\mathop \int }}\,x\log \left( 1+2x \right)~dx$

Ans: Let $I=\underset{0}{\overset{1}{\mathop \int }}\,x\log \left( 1+2x \right)~dx$

⇒ $I=\left[ \log \left( 1+2x \right)\underset{0}{\overset{1}{\mathop \int }}\,x~dx \right]_{0}^{1}-\underset{0}{\overset{1}{\mathop \int }}\,\left( \frac{d}{dx}\log \left( 1+2x \right).\underset{0}{\overset{1}{\mathop \int }}\,x~dx \right)dx$

⇒ $I=\left[ \frac{{{x}^{2}}}{2}\log \left( 1+2x \right) \right]_{0}^{1}-\underset{0}{\overset{1}{\mathop \int }}\,\left( \frac{2}{1~+~2x}.\frac{{{x}^{2}}}{2} \right)dx$

⇒ $I=\left( \frac{1}{2}\log 3-0 \right)-\underset{0}{\overset{1}{\mathop \int }}\,\left( \frac{{{x}^{2}}}{1~+~2x} \right)dx$

⇒ $I=\left( \frac{1}{2}\log 3-0 \right)-\underset{0}{\overset{1}{\mathop \int }}\,\left( \frac{x}{2}-\frac{\frac{x}{2}}{1~+~2x} \right)dx$

⇒ $I=\left( \frac{1}{2}\log 3 \right)-\frac{1}{2}\underset{0}{\overset{1}{\mathop \int }}\,xdx+\frac{1}{2}\underset{0}{\overset{1}{\mathop \int }}\,\frac{x}{1+2x}dx$

⇒ $I=\frac{1}{2}\log 3-\frac{1}{2}\left( \frac{1}{2}-0 \right)+\frac{1}{4}\underset{0}{\overset{1}{\mathop \int }}\,\frac{1~+~2x-1}{1+2x}dx$

⇒ $I=\frac{1}{2}\log 3-\frac{1}{4}+\frac{1}{4}\underset{0}{\overset{1}{\mathop \int }}\,\frac{1+~2x}{1+2x}dx-\frac{1}{4}\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{1+2x}$

⇒ $I=\frac{1}{2}\log 3-\frac{1}{4}+\frac{1}{4}\underset{0}{\overset{1}{\mathop \int }}\,dx-\frac{1}{4}\left[ \frac{1}{2}\log \left( 1+2x \right) \right]_{0}^{1}$

⇒ $I=\frac{1}{2}\log 3-\frac{1}{4}+\frac{1}{4}\left( x \right)_{0}^{1}-\frac{1}{8}\left[ \log \left( 3 \right)-0 \right]$

⇒ $I=\frac{1}{2}\log 3-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\left[ \log \left( 3 \right) \right]$

⇒ $I=\frac{1}{2}\log 3-\frac{1}{8}\log 3$

⇒ $I=\frac{3}{8}\log 3$

Hence, $\underset{0}{\overset{1}{\mathop \int }}\,x\log \left( 1+2x \right)~dx=\frac{3}{8}\log 3.$

46. $\underset{0}{\overset{\pi }{\mathop \int }}\,x\log \sin x~dx$

Ans: Let $I=\underset{0}{\overset{\pi }{\mathop \int }}\,x\log \sin x~dx$ ……………….. (i)

⇒ $I=\underset{0}{\overset{\pi }{\mathop \int }}\,\left( \pi -x \right)\log \sin \left( \pi -x \right)~dx$

⇒ $I=\underset{0}{\overset{\pi }{\mathop \int }}\,\left( \pi -x \right)\log \sin x~dx$ …………..(ii)

Adding eq (i) and (ii), we get

⇒ $2I=\pi \underset{0}{\overset{\pi }{\mathop \int }}\,\log \sin x~dx$

⇒ $2I=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin x~dx$

⇒ $I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin x~dx$ ………………..(iii)

⇒ $I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin \left( \frac{\pi }{2}-x \right)~dx$

⇒ $I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \cos x~dx$ ……………(iv)

Adding eq (iii) and (iv), we get

⇒ $2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,(\log \sin x+\log \cos x)~dx$

⇒ $2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( \log \sin x.\cos x \right)dx$

⇒ $2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( \log \frac{2}{2}\sin x.\cos x \right)dx$

⇒ $2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( \log \frac{\sin 2x}{2} \right)dx$

⇒ $2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( \log \sin 2x-\log 2 \right)dx$

⇒ $2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin 2x~dx-\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log 2dx$

⇒ $2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin 2x~dx-\pi \log 2~\times \frac{\pi }{2}$

Put $2x=t~\Rightarrow dx=\frac{dt}{2}$

Here, at $x=0\to t=0$ and at $x=\frac{\pi }{2}\to t=\pi $

⇒ $2I=\frac{\pi }{2}\underset{0}{\overset{\pi }{\mathop \int }}\,\log \sin t~dt-\frac{{{\pi }^{2}}}{2}~\log 2~$

⇒ $2I=\frac{\pi }{2}\times 2\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin t~dt-\frac{{{\pi }^{2}}}{2}~\log 2$

⇒ $2I=\pi \underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin t~dt-\frac{{{\pi }^{2}}}{2}~\log 2$

⇒ $2I=I-\frac{{{\pi }^{2}}}{2}~\log 2$                           [from eq (iii)]

⇒ $I=-\frac{{{\pi }^{2}}}{2}~\log 2$

Hence, $\underset{0}{\overset{\pi }{\mathop \int }}\,x\log \sin x~dx=-\frac{{{\pi }^{2}}}{2}~\log 2.$

47. $\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left( \sin x+\cos x \right)dx$ 

Ans: Let $I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left( \sin x+\cos x \right)dx~$……….. (i)

⇒ $I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left[ \sin \left( \frac{\pi }{4}-\frac{\pi }{4}-x \right)+\cos \left( \frac{\pi }{4}-\frac{\pi }{4}-x \right) \right]dx$

⇒ $I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left[ \sin \left( -x \right)+\cos \left( -x \right) \right]dx$

⇒ $I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left( -\sin x+\cos x \right)dx$ 

⇒ $I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left( \cos x-\sin x \right)dx$ …………..(ii)

Adding eq (i) and (ii),we get

⇒ 2$I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left( \sin x+\cos x \right)dx+\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left( \cos x-\sin x \right)dx$ 

⇒ 2$I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left[ \left( \sin x+\cos x \right).\left( \cos x-\sin x \right) \right]dx$ 

⇒ 2$I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left[ co{{s}^{2}}x-si{{n}^{2}}x \right]dx$

⇒ 2$I=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \cos 2xdx$

⇒ 2$I=2\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \cos 2xdx$

⇒ $I=\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \cos 2xdx$  ……………(iii)

Now, put $2x=t\Rightarrow dx=\frac{dt}{2}$

Here, at $x=0\to t=0$ and at $x=\frac{\pi }{4}\to t=\frac{\pi }{2}$

Therefore, 

⇒ $I=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \cos t~dt$ ………… (iv)

⇒ $I=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \cos \left( \frac{\pi }{2}-t \right)~dt$

⇒ $I=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin t~dt$ …………..(v)

Adding eq (iv) and (v), we get

⇒ $2I=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \cos t~dt+\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \sin t~dt$

⇒ $2I=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\text{log }\!\!~\!\!\text{ sin}t\cos t~dt$

⇒ $4I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \frac{2}{2}\sin t\cos t~dt$

⇒ $4I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log \frac{\sin 2t}{2}~dt$

⇒ $4I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( \log \sin 2t-\log 2 \right)~dt$

⇒ $4I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( \log \sin \left( \frac{\pi }{2}-2t \right) \right)~dt-\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\log 2~dt$

⇒ $4I=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\left( \log \cos 2t \right)~dt-\frac{\pi }{2}\log 2$

⇒ $4I=2\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\left( \log \cos 2t \right)~dt-\frac{\pi }{2}\log 2$

⇒ $4I=2I-\frac{\pi }{2}\log 2$  

⇒ $2I=-\frac{\pi }{2}\log 2$

⇒ $I=-\frac{\pi }{4}\log 2$

Hence, $\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}}{\mathop \int }}\,\log \left( \sin x+\cos x \right)dx=-\frac{\pi }{4}\log 2.$

Objective Type Questions 

Choose the correct option from given four options in each of the Exercises from 48 to 63.

48. $\int \frac{\cos 2x-\cos 2\theta ~}{\cos x-\cos \theta }dx$ is equal to

(A) $2(\sin x+x\cos \theta )+C$                    

(B) $2(\sin x-x\cos \theta )+C$       

(C)  $2(\sin x+2x\cos \theta )+C$                   

(D) $2(\sin x-2x\cos \theta )+C$

Ans: Let $\text{I}=~\int \frac{\cos 2x-\cos 2\theta ~}{\cos x-\cos \theta }dx~~$

⇒ $\text{I}=~\int \frac{2~co{{s}^{2}}x-1-\left( 2~co{{s}^{2}}\theta -1 \right)}{\cos x-\cos \theta }dx$

⇒ $\text{I}=~\int \frac{2~co{{s}^{2}}x-1-2~co{{s}^{2}}\theta +1}{\cos x-\cos \theta }dx$

⇒ $\text{I}=~\int \frac{2~co{{s}^{2}}x-2~co{{s}^{2}}\theta }{\cos x-\cos \theta }dx$

⇒ $\text{I}=~\int \frac{2\left( ~co{{s}^{2}}x-~co{{s}^{2}}\theta  \right)}{(\cos x-\cos \theta )}dx$

⇒ $\text{I}=~\int \frac{2\left( ~co{{s}^{2}}x-~co{{s}^{2}}\theta  \right)}{(\cos x-\cos \theta )}dx$

⇒ $\text{I}=~2\int (\cos x+~\cos \theta )dx$

⇒ $\text{I}=~2\left( \sin x+x\cos \theta  \right)+C$

Hence, option (A) is correct answer.

49. $\int \frac{dx}{\sin \left( x-a \right)\sin \left( x-b \right)}~$is equal to 

(A) $\sin \left( b-a \right)\log \left| \frac{\sin \left( x-b \right)}{\sin \left( x-a \right)} \right|+C$    

(B) $\text{cosec}\left( b-a \right)\log \left| \frac{\sin \left( x-a \right)}{\sin \left( x-b \right)} \right|+C$

(C) $\text{cosec}\left( b-a \right)\log \left| \frac{\sin \left( x-b \right)}{\sin \left( x-a \right)} \right|+C$            

(D) $\sin \left( b-a \right)\log \left| \frac{\sin \left( x-b \right)}{\sin \left( x-a \right)} \right|+C$

Ans: Let $\text{I}=~\int \frac{dx}{\sin \left( x-a \right)\sin \left( x-b \right)}~$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\int \frac{\sin \left( b-a \right)~dx}{\sin \left( x-a \right)\sin \left( x-b \right)}~$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\int \frac{\sin \left( x-a-~x~+~b \right)~}{\sin \left( x-a \right)\sin \left( x-b \right)}dx~$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\int \frac{\sin \left\{ \left( x-a \right)-\left( ~x-~b \right) \right\}~}{\sin \left( x-a \right)\sin \left( x-b \right)}~dx$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\int \frac{\sin \left( x-a \right).\cos \left( x-b \right)-~\sin \left( x-b \right).\cos \left( x-a \right)~~}{\sin \left( x-a \right)\sin \left( x-b \right)}~dx$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\int \frac{\cos \left( x-b \right)}{\sin \left( x-b \right)}dx-\frac{1}{\sin \left( b-a \right)}\int \frac{\cos \left( x-a \right)}{\sin \left( x-a \right)}dx$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\int \cot \left( x-b \right)dx-\frac{1}{\sin \left( b-a \right)}\int \cot \left( x-a \right)dx$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\log \left| \sin \left( x-b \right) \right|-\frac{1}{\sin \left( b-a \right)}\log \left| \sin \left( x-a \right) \right|+C$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\left[ \log \left| \sin \left( x-b \right) \right|-\log \left| \sin \left( x-a \right) \right| \right]+C$

⇒ $\text{I}=\frac{1}{\sin \left( b-a \right)}\log \left| \frac{\left( x-b \right)}{\left( x-a \right)} \right|+C$

⇒ $\text{I}=\text{cosec}\left( b-a \right)\log \left| \frac{\left( x-b \right)}{\left( x-a \right)} \right|+C$

Hence, option (C) is correct answer.

50. $\int ta{{n}^{-1}}\sqrt{x}~dx$ is equal to 

(A) $\left( x+1 \right)ta{{n}^{-1}}\sqrt{x}-\sqrt{x}+C$       

(B) $~xta{{n}^{-1}}\sqrt{x}-\sqrt{x}+C$      

(C) $\sqrt{x}-xta{{n}^{-1}}\sqrt{x}+C$                 

(D) $\sqrt{x}-\left( x+1 \right)ta{{n}^{-1}}\sqrt{x}+C$

Ans: Let $\text{I}=\int ta{{n}^{-1}}\sqrt{x}~dx$

⇒ $\text{I}=\int 1.~ta{{n}^{-1}}\sqrt{x}~dx$

⇒ $\text{I}=~ta{{n}^{-1}}\sqrt{x}~\int 1~dx-\int \left( \frac{d}{dx}ta{{n}^{-1}}\sqrt{x}~.\int 1~dx \right)dx$

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-\int \left( \frac{1}{1+x}.\frac{1}{2\sqrt{x}}~.x \right)dx$

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-\frac{1}{2}\int \left( \frac{\sqrt{x}}{1+x} \right)dx$

Now, put  $x={{t}^{2}}$

⇒ $dx=2t~dt$

Therefore, 

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-\frac{1}{2}\int \frac{t}{1+{{t}^{2}}}.~2t~dt$

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-\int \frac{{{t}^{2}}}{1+{{t}^{2}}}~dt$

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-\int \frac{1+{{t}^{2}}-1}{1+{{t}^{2}}}~dt$

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-\int \left( 1-\frac{1}{1+{{t}^{2}}} \right)~dt$

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-\int 1~dt+\int \left( \frac{1}{1+{{t}^{2}}} \right)~dt$

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-t+ta{{n}^{-1}}t+C$

⇒ $\text{I}=~xta{{n}^{-1}}\sqrt{x}~-\sqrt{x}+ta{{n}^{-1}}\sqrt{x}+C$

⇒ $\text{I}=\left( x+1 \right)ta{{n}^{-1}}\sqrt{x}~-\sqrt{x}+C$

Hence, option (A) is correct answer.

51. $\int {{e}^{x}}{{\left( \frac{1-x~}{1+{{x}^{2}}} \right)}^{2}}dx$ is equal to

(A)  $\frac{{{e}^{x}}}{1+{{x}^{2}}}+C~~$                                         

(B) $\frac{-{{e}^{x}}}{1+{{x}^{2}}}+C~~$          

(C) $\frac{{{e}^{x}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}+C~~$                                    

(D) $\frac{-{{e}^{x}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}+C~~$                                                                                            

Ans: Let $\text{I}=\int {{e}^{x}}{{\left( \frac{1-x~}{1+{{x}^{2}}} \right)}^{2}}dx$

⇒ $\text{I}=\int {{e}^{x}}\left[ \frac{1+{{x}^{2}}-2x~}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right]dx$

⇒ $\text{I}=\int {{e}^{x}}\left[ \frac{1+{{x}^{2}}~}{{{\left( 1+{{x}^{2}} \right)}^{2}}}-\frac{2x~}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right]dx$

⇒ $\text{I}=\int {{e}^{x}}\left[ \frac{1~}{\left( 1+{{x}^{2}} \right)}-\frac{2x~}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \right]dx$

⇒ $\text{I}=\frac{{{e}^{x}}~}{\left( 1+{{x}^{2}} \right)}+C$

Hence, option (A) is correct answer.

52. $\int \frac{{{x}^{9}}}{{{\left( 4{{x}^{2}}+1 \right)}^{6}}}dx$ is equal to

(A) $\frac{1}{5x}{{\left( 4+\frac{1}{{{x}^{2}}} \right)}^{-5}}+C$                   

(B) $\frac{1}{5}{{\left( 4+\frac{1}{{{x}^{2}}} \right)}^{-5}}+C$        

(C) $\frac{1}{10x}{{\left( 1+4 \right)}^{-5}}+C$                    

(D) $\frac{1}{10}{{\left( 4+\frac{1}{{{x}^{2}}} \right)}^{-5}}+C$                                  

Ans: Let $\text{I}=\int \frac{{{x}^{9}}}{{{\left( 4{{x}^{2}}+1 \right)}^{6}}}dx$

⇒ $\text{I}=\int \frac{{{x}^{9}}}{{{\left[ {{x}^{2}}\left( 4+\frac{1}{{{x}^{2}}} \right) \right]}^{6}}}dx$

⇒ $\text{I}=\int \frac{{{x}^{9}}}{{{x}^{12}}{{\left( 4+\frac{1}{{{x}^{2}}} \right)}^{6}}}dx$

⇒ $\text{I}=\int \frac{dx}{{{x}^{3}}{{\left( 4+\frac{1}{{{x}^{2}}} \right)}^{6}}}$

Now, put $4+\frac{1}{{{x}^{2}}}=t$

⇒ $-\frac{2}{{{x}^{3}}}dx=dt$

⇒ $\frac{dx}{{{x}^{3}}}=-\frac{dt}{2}$

Therefore, 

⇒ $\text{I}=-\frac{1}{2}\int \frac{dt}{{{\left( t \right)}^{6}}}$

⇒ $\text{I}=-\frac{1}{2}\int {{t}^{-6}}~dt$

⇒ $\text{I}=-\frac{1}{2}~\times ~\frac{{{t}^{-5}}}{-5}+C$

⇒ $\text{I}=\frac{1}{10}{{\left( 4+\frac{1}{{{x}^{2}}} \right)}^{-5}}+C$

Hence, option (D) is correct answer.

53. $\int \frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}=a\log \left| 1+{{x}^{2}} \right|+b~ta{{n}^{-1}}x+\frac{1}{5}\log \left| x+2 \right|+C,~~$ then

(A). $a=-\frac{1}{10}~,~b=-\frac{2}{5}$                    

(B) $a=\frac{1}{10}~,~b=-\frac{2}{5}$

(C) $a=-\frac{1}{10}~,~b=\frac{2}{5}$                          

(D) $a=\frac{1}{10}~,~b=\frac{2}{5}$

Ans: Let $I=\int \frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}~$

Now, $\frac{1}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{\left( x+2 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+1 \right)}$

⇒ $\frac{1}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}=\frac{A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)~\left( x+2 \right)}{\left( {{x}^{2}}+1 \right)}$

⇒ $1=A\left( {{x}^{2}}+1 \right)+\left( Bx+C \right)~\left( x+2 \right)$

⇒ $1=A{{x}^{2}}+A+B{{x}^{2}}+2Bx+Cx+2C$

⇒ $1={{x}^{2}}\left( A+B \right)+x\left( 2B+C \right)+\left( A+2C \right)$

On comparing both sides, we get

⇒ $A+B=0,~~2B+C=0$ and $A+2C=1$

From above equations, we get

⇒ $A=\frac{1}{5}~,~B=-\frac{1}{5}$  and $C=\frac{2}{5}$

Therefore, $\frac{1}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}=\frac{1}{5\left( x+2 \right)}+\frac{-\frac{x}{5}+\frac{2}{5}}{\left( {{x}^{2}}+1 \right)}$

⇒ $\int \frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}=\int \frac{dx}{5\left( x+2 \right)}+\int \frac{-\frac{x}{5}+\frac{2}{5}}{\left( {{x}^{2}}+1 \right)}dx$

⇒ $\int \frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}=\int \frac{dx}{5\left( x+2 \right)}-\frac{1}{5}\int \frac{x}{\left( {{x}^{2}}+1 \right)}dx+\frac{2}{5}\int \frac{1}{\left( {{x}^{2}}+1 \right)}dx$

$=\frac{1}{5}\log \left| x+5 \right|-\frac{1}{10}\log \left| 1+{{x}^{2}} \right|+\frac{2}{5}ta{{n}^{-1}}x+C$ 

Here, $a=-\frac{1}{10}$ and $b=\frac{2}{5}$

Hence, option (C) is correct answer.

54. $\int \frac{{{x}^{3}}}{x+1}dx$ is equal to 

(A) $x+\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\log \left| 1-x \right|+C$    

(B) $x+\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\log \left| 1-x \right|+C$

(C) $x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\log \left| 1+x \right|+C$

(D) $x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\log \left| 1+x \right|+C$

Ans: Let $\text{I}=\int \frac{{{x}^{3}}}{x+1}dx$

⇒ $\text{I}=\int \frac{{{x}^{3}}+1-1}{x+1}dx$

⇒ $\text{I}=\int \frac{{{x}^{3}}+1}{x+1}dx-\int \frac{1}{x+1}dx$

⇒ $\text{I}=\int \left( {{x}^{2}}-x+1 \right)dx-\int \frac{1}{x+1}dx$

⇒ $\text{I}=\frac{{{x}^{3}}}{3}-\frac{{{x}^{2}}}{2}+x-\log \left| x+1 \right|+C$

Hence, option (D) is correct answer.

55. $\int \frac{x+\sin x}{1+\cos x}~dx$ is equal to 

(A) $\log \left| 1+\cos x \right|+C$                

(B) $\log \left| x+\sin x \right|+C$

(C) $x-\tan \frac{x}{2}+C$                           

(D) $x\tan \frac{x}{2}+C$                           

Ans: Let $\text{I}=\int \frac{x+\sin x}{1+\cos x}~dx$

⇒ $\text{I}=\int \frac{x}{1+\cos x}~dx+\int \frac{\sin x}{1+\cos x}~dx$

⇒ $\text{I}={{\text{I}}_{1}}+{{\text{I}}_{2}}$   ………(i), where, ${{\text{I}}_{1}}=\int \frac{x}{1+\cos x}~dx$ & ${{\text{I}}_{2}}=\int \frac{\sin x}{1+\cos x}~dx$

Now, ${{\text{I}}_{1}}=\int \frac{x}{1+\cos x}~dx$

⇒ ${{\text{I}}_{1}}=\int \frac{x}{2~co{{s}^{2}}\frac{x}{2}}~dx$

⇒ ${{\text{I}}_{1}}=\frac{1}{2}\int x~se{{c}^{2}}\frac{x}{2}~dx$

⇒ ${{\text{I}}_{1}}=\frac{1}{2}\left[ x\int ~se{{c}^{2}}\frac{x}{2}~dx-~\int \left( \frac{d}{dx}x.\int ~se{{c}^{2}}\frac{x}{2}~dx~~ \right)dx \right]$

⇒ ${{\text{I}}_{1}}=\frac{1}{2}\left[ 2x\tan \frac{x}{2}-~2\int \tan \frac{x}{2}~dx \right]+C$

⇒ ${{\text{I}}_{1}}=x\tan \frac{x}{2}-~\int \tan \frac{x}{2}~dx+C$ …………. (ii)

And also we have,

⇒ ${{\text{I}}_{2}}=\int \frac{\sin x}{1+\cos x}~dx$

⇒ ${{\text{I}}_{2}}=\int \frac{2\sin \frac{x}{2}~.\cos \frac{x}{2}}{2~co{{s}^{2}}\frac{x}{2}}~dx$

⇒ ${{\text{I}}_{2}}=\int \frac{\sin \frac{x}{2}~}{\cos \frac{x}{2}}~dx$

⇒ ${{\text{I}}_{2}}=\int \tan \frac{x}{2}~dx$ …………..(iii)

Now, put the value of ${{\text{I}}_{1}}$ and ${{\text{I}}_{2}}$ in eq (i), 

⇒ $\text{I}=x\tan \frac{x}{2}-~\int \tan \frac{x}{2}~dx+C+\int \tan \frac{x}{2}~dx$ 

⇒ $\text{I}=x\tan \frac{x}{2}+C$

Hence, option (D) is correct answer.

56. $\int \frac{{{x}^{3}}~dx}{\sqrt{1+~{{x}^{2}}}}=a{{\left( 1+{{x}^{2}} \right)}^{\frac{3}{2}}}+b\sqrt{1+{{x}^{2}}}+C,$ then 

(A) $a=\frac{1}{3}~,~b=1$                          

(B) $a=\frac{-1}{3}~,~b=1$               

(C) $a=\frac{-1}{3}~,~b=-1$                    

(D) $a=\frac{1}{3}~,~b=-1$                                                         

Ans: Let $\text{I}=\int \frac{{{x}^{3}}~dx}{\sqrt{1+~{{x}^{2}}}}$

⇒ $\text{I}=\int \frac{{{x}^{2}}.x~dx}{\sqrt{1+~{{x}^{2}}}}$

Put $1+~{{x}^{2}}={{t}^{2}}$

⇒ $2x~dx=2t~dt$

⇒ $x~dx=t~dt$

Therefore, 

⇒ $\text{I}=\int \frac{\left( {{t}^{2}}-1 \right)t~dt}{t}$

⇒ $\text{I}=\int \left( {{t}^{2}}-1 \right)dt$

⇒ $\text{I}=\frac{{{t}^{3}}}{3}-t+C$

⇒ $\text{I}=\frac{{{\left( \sqrt{1+{{x}^{2}}} \right)}^{3}}}{3}-\sqrt{1+{{x}^{2}}}+C$

⇒ $\text{I}=\frac{{{\left( 1+{{x}^{2}} \right)}^{\frac{3}{2}}}}{3}-\sqrt{1+{{x}^{2}}}+C$

Here, $a=\frac{1}{3}~,~b=-1$

Hence, option (D) is correct answer.

57 $\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}~}{\mathop \int }}\,~\frac{dx}{1+\cos 2x}$ is equal to

(A) 1

(B) 2

(C) 3

(D)  4

Ans: Let $\text{I}=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}~}{\mathop \int }}\,~\frac{dx}{1+\cos 2x}$

⇒ $\text{I}=\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}~}{\mathop \int }}\,~\frac{dx}{2~co{{s}^{2}}x}$

⇒ $\text{I}=\frac{1}{2}\underset{-\frac{\pi }{4}}{\overset{\frac{\pi }{4}~}{\mathop \int }}\,~se{{c}^{2}}xdx$

⇒ $\text{I}=\frac{1}{2}~\left[ \tan x \right]_{-\frac{\pi }{4}~}^{\frac{\pi }{4}~}$

⇒ $\text{I}=\frac{1}{2}~\left[ \tan \frac{\pi }{4}~-\tan \left( -\frac{\pi }{4} \right)~ \right]$

⇒ $\text{I}=\frac{1}{2}~\left[ 1+1 \right]$

⇒ $\text{I}=1$

Hence, option (A) is correct answer.

58. $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\sqrt{1-\sin 2x}~dx$ is equal to

(A) $2\sqrt{2}$            

(B) $2\left( \sqrt{2}+1 \right)$        

(C) 2            

(D) $2\left( \sqrt{2}-1 \right)$        

Ans: Let $\text{I}=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\sqrt{1-\sin 2x}~dx$

⇒ $\text{I}=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\sqrt{si{{n}^{2}}x+co{{s}^{2}}x-2\sin x.\cos x~}~dx$

⇒ $\text{I}=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\sqrt{{{\left( \sin x-\cos x \right)}^{2}}~}~dx$

⇒ $\text{I}=~\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,~\left| \sin x-\cos x \right|~dx$

Since, $(\sin x-\cos x)\le 0$, $x\in \left[ 0,~\frac{\pi }{4} \right]$ and $(\sin x-\cos x)\ge 0$, $x\in ~\left[ \frac{\pi }{4},\frac{\pi }{2} \right]$

Therefore, 

⇒ $\text{I}=~-\underset{0}{\overset{\frac{\pi }{4}}{\mathop \int }}\,~\left( \sin x-\cos x \right)dx+\underset{\frac{\pi }{4}}{\overset{\frac{\pi }{2}}{\mathop \int }}\,~\left( \sin x-\cos x \right)dx$

⇒ $\text{I}=~-\left[ -\cos x-\sin x \right]_{0}^{\frac{\pi }{4}}+\left[ -\cos x-\sin x \right]_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

⇒ $\text{I}=~\left[ \cos x+\sin x \right]_{0}^{\frac{\pi }{4}}-\left[ \cos x+\sin x \right]_{\frac{\pi }{4}}^{\frac{\pi }{2}}$

⇒ $\text{I}=~\left[ \cos \frac{\pi }{4}+\sin \frac{\pi }{4}-1-0 \right]-\left[ \cos \frac{\pi }{2}+\sin \frac{\pi }{2}-\cos \frac{\pi }{4}-\sin \frac{\pi }{4} \right]$

⇒ $\text{I}=~\left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 \right]-\left[ 1-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \right]$

⇒ $\text{I}=~\left[ \sqrt{2}-1 \right]-\left[ 1-\sqrt{2} \right]$

⇒ $\text{I}=~\sqrt{2}-1-1+\sqrt{2}$

⇒ $\text{I}=~2\sqrt{2}-2$

⇒ $\text{I}=~2(\sqrt{2}-1)$

Hence, option (D) is correct answer.

59. $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\cos x~{{e}^{\sin x}}dx$ is equal to …………..

Ans: Let $\text{I}=\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\cos x~{{e}^{\sin x}}~dx$

Put $\sin x=t$

⇒ $\cos xdx=dt$

Now, at $x=0\Rightarrow t=0$ and at $x=\frac{\pi }{2}\Rightarrow t=1$

Therefore, 

⇒ $\text{I}=\underset{0}{\overset{1}{\mathop \int }}\,{{e}^{t}}~dx~$

⇒ $\text{I}=\left[ {{e}^{t}} \right]_{0}^{1}$

⇒ $\text{I}={{e}^{1}}-{{e}^{0}}$

⇒ $\text{I}=e-1$

Hence, $\underset{0}{\overset{\frac{\pi }{2}}{\mathop \int }}\,\cos x~{{e}^{\sin x}}~dx=e-1.$

60. $\int \frac{x+3~}{{{\left( x+4 \right)}^{2}}}~{{e}^{x}}dx$ = ………

Ans: Let $\text{I}=\int \frac{x+3~}{{{\left( x+4 \right)}^{2}}}~{{e}^{x}}dx$

⇒ $\text{I}=\int {{e}^{x}}\left[ \frac{x+4-1~}{{{\left( x+4 \right)}^{2}}} \right]~dx$

⇒ $\text{I}=\int {{e}^{x}}\left[ \frac{x+4~}{{{\left( x+4 \right)}^{2}}}-\frac{1~}{{{\left( x+4 \right)}^{2}}} \right]~dx$

⇒ $\text{I}=\int {{e}^{x}}\left[ \frac{1~}{x+4}-\frac{1~}{{{\left( x+4 \right)}^{2}}} \right]~dx$

⇒ $\text{I}=\frac{{{e}^{x}}~}{x+4}+C$

Hence, $\int \frac{x+3~}{{{\left( x+4 \right)}^{2}}}~{{e}^{x}}dx=\frac{{{e}^{x}}~}{x+4}+C.$

Fill in the blanks in each of the following Exercise 60 to 63. 

61. $\underset{0}{\overset{a}{\mathop \int }}\,\frac{1}{1+4{{x}^{2}}}dx=\frac{\pi }{8}$, then $a=~$………..

Ans: here, we have $\underset{0}{\overset{a}{\mathop \int }}\,\frac{1}{1+4{{x}^{2}}}dx=\frac{\pi }{8}$

⇒ $\frac{1}{4}\underset{0}{\overset{a}{\mathop \int }}\,\frac{1}{\frac{1}{4}+{{x}^{2}}}dx=\frac{\pi }{8}$

⇒ $\underset{0}{\overset{a}{\mathop \int }}\,\frac{1}{{{\left( \frac{1}{2} \right)}^{2}}+{{x}^{2}}}dx=\frac{\pi }{2}$

⇒ $\left[ \frac{1}{\frac{1}{2}}~ta{{n}^{-1}}\left( \frac{x}{\frac{1}{2}} \right) \right]_{0}^{a}=\frac{\pi }{2}$

⇒ $\left[ 2~ta{{n}^{-1}}\left( 2x \right) \right]_{0}^{a}=\frac{\pi }{2}$

⇒ $ta{{n}^{-1}}\left( 2a \right)-ta{{n}^{-1}}\left( 0 \right)=\frac{\pi }{4}$

⇒ $ta{{n}^{-1}}\left( 2a \right)=\frac{\pi }{4}$

⇒ $2a=\tan \frac{\pi }{4}$

⇒ $2a=1$

⇒ $a=\frac{1}{2}$

Hence, value of a is $\frac{1}{2}$ .

Therefore, $\underset{0}{\overset{a}{\mathop \int }}\,\frac{1}{1+4{{x}^{2}}}dx=\frac{\pi }{8}$, then $a=\frac{1}{2}~.$

62. $\int \frac{\sin x}{3+4~co{{s}^{2}}x}dx=$

Ans: Let $\text{I}=\int \frac{\sin x}{3+4~co{{s}^{2}}x}dx$

Put $\cos x=t$ 

⇒ $-\sin x~dx=dt$

⇒ $\sin x~dx=-dt$

Therefore, 

⇒ $\text{I}=-\int \frac{dt}{3+4~{{t}^{2}}}$

⇒ $\text{I}=-\frac{1}{4}\int \frac{dt}{\frac{3}{4}+~{{t}^{2}}}$

⇒ $\text{I}=-\frac{1}{4}\int \frac{dt}{{{\left( \frac{\sqrt{3}}{2} \right)}^{2}}+~{{t}^{2}}}$

⇒ $\text{I}=-\frac{1}{4}~.~\frac{1}{\frac{\sqrt{3}}{2}}ta{{n}^{-1}}\left( \frac{t}{\frac{\sqrt{3}}{2}} \right)+C$

⇒ $\text{I}=-~\frac{1}{2\sqrt{3}}ta{{n}^{-1}}\left( \frac{2t}{\sqrt{3}} \right)+C$

⇒ $\text{I}=-~\frac{1}{2\sqrt{3}}ta{{n}^{-1}}\left( \frac{2\cos x}{\sqrt{3}} \right)+C$

Hence, $\int \frac{\sin x}{3+4~co{{s}^{2}}x}dx=-~\frac{1}{2\sqrt{3}}ta{{n}^{-1}}\left( \frac{2\cos x}{\sqrt{3}} \right)+C.$

63. The value of $\mathop{\int }_{-\pi }^{\pi }si{{n}^{3}}x.co{{s}^{2}}x~dx$ is …………..

Ans: Let $\text{I}=\mathop{\int }_{-\pi }^{\pi }si{{n}^{3}}x.co{{s}^{2}}x~dx~$…………. (i)

⇒ $\text{I}=\mathop{\int }_{-\pi }^{\pi }si{{n}^{3}}\left( \pi -\pi -x \right).co{{s}^{2}}\left( \pi -\pi -x \right)~dx$

⇒ $\text{I}=\mathop{\int }_{-\pi }^{\pi }si{{n}^{3}}\left( -x \right).co{{s}^{2}}\left( -x \right)~dx$

⇒ $\text{I}=-\mathop{\int }_{-\pi }^{\pi }si{{n}^{3}}x.co{{s}^{2}}x~dx$ …………….(ii)

Now, adding eq (i) and (ii), we get

⇒ $2\text{I}=0$

⇒ $\text{I}=0$

Hence, the value of $\mathop{\int }_{-\pi }^{\pi }si{{n}^{3}}x.co{{s}^{2}}x~dx$ is 0.

Importance of Integrals

Students will get a brief knowledge about the concepts that are mentioned in the PDF. The Exemplar PDF of Chapter 7 - Integrals is designed as per the latest syllabus pattern advised by the Central Board of Secondary Education. The PDF is structured by the expert faculty of Vedantu, specifically structured to enhance the learning and understanding capacity of the students. The students following the Exemplar with full dedication will surely achieve good results. 

Class 12 is the most important academic year for students. All the Chapters in all the subjects are very important, students do have pressure to perform well in the Exams. Chapter 7 Integrals is such an important Chapter that focuses on the concepts used in other Chapters as well. This Chapter increases the thinking process of the students, questions asked from this Chapter are not difficult so this Chapter can be quite scoring for all students. 

Students neglecting the concepts of this Chapter face trouble as the concepts present in this Chapter also help to solve other Chapter questions as well. NCERT Exemplar of Chapter 7 consists of a textbook solution for Class 12. There are many important concepts but the most important concepts are mentioned below. This Chapter basically focuses on the questions asked on topics like -

1. Integrations

2. Properties of Integrals

3. Geometrical representation

4. Comparison between integration and differentiation 

5. Definite Integral - limit of the sum

6. Different

7.  methods of integration

8. Properties of definite Integral and some other important question

Those students who want to have a good grip on the concepts must focus on the structured solutions mentioned in the Exemplar. Students can register themselves on Vedantu’s website for other important topics. Vedantu also offers online study support to the students, the online present material consists of additional questions, notes, sample papers, Exemplars. All the study material is designed and approved by the community of faculty that consists of a group of teachers that focuses on the quality of the content present from the Vedantu’s end.