NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8

Exercise 7.8

1.Integrate the given integral $\underset{\mathbf{-}\mathbf{1}}{\overset{\mathbf{1}}{\int }}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{dx}$.

Ans: To simplify the question let us suppose, $I=\underset{-1}{\overset{1}{\int }}(x+1)dx$

$\int (x+1)dx={\displaystyle \frac{x^2}{2}}+x$

Again, let us suppose the function is $F(x)={\displaystyle \frac{x^2}{2}}+x$

By second fundamental theorem of calculus, we obtain

$I=F(1)-F(-1)$

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

$={\displaystyle \frac{1}{2}}+1-{\displaystyle \frac{1}{2}}+1$

$=2$ 

2.Integrate the given integral $\underset{\mathbf{2}}{\overset{\mathbf{3}}{\int }}{\displaystyle \frac{\mathbf{1}}{\mathbf{x}}}\mathbf{dx}$. 

Ans: To simplify the question let us suppose, $I=\underset{2}{\overset{3}{\int }}{\displaystyle \frac{1}{x}}dx$

Thus, $\int {\displaystyle \frac{1}{x}}dx=\log \left|x\right|$

Again, let us suppose the function is$F(x)=\log \left|x\right|$

By second fundamental theorem of calculus, we obtain

$I=F(3)-F(2)$

$=\log \left|3\right|-\log \left|2\right|$

$=\log {\displaystyle \frac{3}{2}}$ 

3.Integrate the given integral $\underset{\mathbf{1}}{\overset{\mathbf{2}}{\int }}\left(\mathbf{4}{\mathbf{x}}^{\mathbf{3}}\mathbf{-}\mathbf{5}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{9}\right)\mathbf{dx}$

Ans: To simplify the question let us suppose,

$I=\underset{1}{\overset{2}{\int }}\left(4x^3-5x^2+6x+9\right)dx$

$\underset{1}{\overset{2}{\int }}\left(4x^3-5x^2+6x+9\right)dx=4\left({\displaystyle \frac{x^4}{4}}\right)-5\left({\displaystyle \frac{x^3}{3}}\right)+6\left({\displaystyle \frac{x^2}{2}}\right)+9(x)$ 

Again, let us suppose the function is 

$\Rightarrow x^4-{\displaystyle \frac{5x^3}{3}}+3x^2+9x=F(x)$

By second fundamental theorem of calculus, we obtain

$I=F(2)-F(1)$

$I=\left\{2^4-{\displaystyle \frac{5{(2)}^3}{3}}+3{(2)}^2+9(2)\right\}-\left\{{(1)}^4-{\displaystyle \frac{5{(1)}^3}{3}}+3{(1)}^2+9(1)\right\}$

$=\left(16-{\displaystyle \frac{40}{3}}+12+18\right)-\left(1-{\displaystyle \frac{5}{3}}+3+9\right)$

$=16-{\displaystyle \frac{40}{3}}+12+18-1+{\displaystyle \frac{5}{3}}-3-9$

$=33-{\displaystyle \frac{35}{3}}$

$={\displaystyle \frac{99-35}{3}}$

$={\displaystyle \frac{64}{3}}$

4.Integrate the given integral $\underset{\mathbf{0}}{\overset{{\displaystyle \frac{\pi }{\mathbf{4}}}}{\int }}\left(\sin \mathbf{2}\mathbf{xdx}\right)$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\left(\sin 2xdx\right)$

  $\int \sin 2xdx=\left({\displaystyle \frac{-\cos 2x}{2}}\right)$ 

Again, let us suppose the function is$\left({\displaystyle \frac{-\cos 2x}{2}}\right)=F(x)$

By second fundamental theorem of calculus, we obtain

$I=F\left({\displaystyle \frac{\pi }{4}}\right)-F(0)$

$=-{\displaystyle \frac{1}{2}}\left[\cos 2\left({\displaystyle \frac{\pi }{4}}\right)-\cos 0\right]$

$=-{\displaystyle \frac{1}{2}}\left[\cos 2\left({\displaystyle \frac{\pi }{4}}\right)-\cos 0\right]$

$=-{\displaystyle \frac{1}{2}}\left[\cos 2\left({\displaystyle \frac{\pi }{4}}\right)-\cos 0\right]$

$=-{\displaystyle \frac{1}{2}}\left[0-1\right]$

$={\displaystyle \frac{1}{2}}$ 

5.Integrate the given integral $\underset{\mathbf{0}}{\overset{{\displaystyle \frac{\pi }{\mathbf{2}}}}{\int }}\left(\cos \mathbf{2}\mathbf{xdx}\right)$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{{\displaystyle \frac{\pi }{2}}}{\int }}\left(\cos 2xdx\right)$

$\int \cos 2xdx=\left({\displaystyle \frac{\sin 2x}{2}}\right)$ 

Again, let us suppose the function is$\left({\displaystyle \frac{\sin 2x}{2}}\right)=F(x)$

By second fundamental theorem of calculus, we obtain

$I=F\left({\displaystyle \frac{\pi }{2}}\right)-F(0)$

$={\displaystyle \frac{1}{2}}\left[\sin 2\left({\displaystyle \frac{\pi }{2}}\right)-\sin 0\right]$

$={\displaystyle \frac{1}{2}}\left[\sin \pi -\sin 0\right]$

$={\displaystyle \frac{1}{2}}\left[0-0\right]$

$=0$

6.Integrate the given integral $\underset{\mathbf{4}}{\overset{\mathbf{5}}{\int }}{\mathbf{e}}^{\mathbf{x}}\mathbf{dx}$

Ans: To simplify the question let us suppose,

$I=\underset{4}{\overset{5}{\int }}{e}^{x}dx$

$\underset{4}{\overset{5}{\int }}{e}^{x}dx=e^x=F(x)$ 

By second fundamental theorem of calculus, we obtain

$I=F(5)-F(4)$

$=e^5-e^4$

 $=e^4\left(e-1\right)$ 

7.Integrate the given integral $\underset{\mathbf{0}}{\overset{{\displaystyle \frac{\pi }{\mathbf{4}}}}{\int }}\tan \mathbf{xdx}$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\tan xdx$

 $\int \tan xdx=-\log \left|\cos x\right|$ 

Again, let us suppose the function is$-\log \left|\cos x\right|=F(x)$

By second fundamental theorem of calculus, we obtain

$I=F\left({\displaystyle \frac{\pi }{4}}\right)-F(0)$

$=-\log \left|\cos {\displaystyle \frac{\pi }{4}}\right|+\log \left|\cos 0\right|$

$=-\log \left|{\displaystyle \frac{1}{\sqrt{2}}}\right|+\log \left|1\right|$

$=-\log {\left(2\right)}^{-{\displaystyle \frac{1}{2}}}$

$={\displaystyle \frac{1}{2}}\log 2$ 

8.Integrate the given integral $\underset{{\displaystyle \frac{\pi }{\mathbf{6}}}}{\overset{{\displaystyle \frac{\pi }{\mathbf{4}}}}{\int }}\cos \mathbf{ecxdx}$

Ans: To simplify the question let us suppose,

$I=\underset{{\displaystyle \frac{\pi }{6}}}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\cos ecxdx$

$\int \cos ecxdx=\log \left|\cos ecx{\displaystyle \frac{\pi }{6}}-\cot {\displaystyle \frac{\pi }{6}}\right|$ 

Again, let us suppose the function is$F(x)=\log \left|\cos x\right|$

By second fundamental theorem of calculus, we obtain

$I=F\left({\displaystyle \frac{\pi }{4}}\right)-F\left({\displaystyle \frac{\pi }{6}}\right)$

$=\log \left|\cos ec{\displaystyle \frac{\pi }{4}}-\cot {\displaystyle \frac{\pi }{4}}\right|-\log \left|\cos ec{\displaystyle \frac{\pi }{6}}-\cot {\displaystyle \frac{\pi }{6}}\right|$

$=\log \left|\sqrt{2}-1\right|-\log \left|2-\sqrt{3}\right|$

$=\log \left({\displaystyle \frac{\sqrt{2}-1}{2-\sqrt{3}}}\right)$ 

9.Integrate the given integral $\underset{\mathbf{0}}{\overset{\mathbf{1}}{\int }}{\displaystyle \frac{\mathbf{dx}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}}}$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{1}{\int }}{\displaystyle \frac{dx}{\sqrt{1-x^2}}}$

Again, let us suppose the function is

$I=\int {\displaystyle \frac{dx}{\sqrt{1-x^2}}}=F(x)$

${\sin }^{-1}x=F(x)$ 

By second fundamental theorem of calculus, we obtain

$I=F(1)-F(0)$

$={\sin }^{-1}(1)-{\sin }^{-1}(0)$

$={\displaystyle \frac{\pi }{2}}-0$

$={\displaystyle \frac{\pi }{2}}$ 

10.Integrate the given integral $\underset{\mathbf{0}}{\overset{\mathbf{1}}{\int }}{\displaystyle \frac{\mathbf{dx}}{\sqrt{\mathbf{1}\mathbf{+}{\mathbf{x}}^{\mathbf{2}}}}}$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{1}{\int }}{\displaystyle \frac{dx}{1+x^2}}$

$\int {\displaystyle \frac{dx}{1+x^2}}={\tan }^{-1}$ 

Again, let us suppose the function is

${\tan }^{-1}x=F(x)$

By second fundamental theorem of calculus, we obtain

$I=F(1)-F(0)$

$={\tan }^{-1}(1)-{\tan }^{-1}(0)$

$={\displaystyle \frac{\pi }{4}}$ 

11.Integrate the given integral $\underset{\mathbf{2}}{\overset{\mathbf{3}}{\int }}{\displaystyle \frac{\mathbf{dx}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}}$

Ans: To simplify the question let us suppose,

$I=\underset{2}{\overset{3}{\int }}{\displaystyle \frac{dx}{x^2-1}}$

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

Again, let us suppose the function is

${\displaystyle \frac{1}{2}}\log \left|{\displaystyle \frac{x-1}{x+1}}\right|=F(x)$

By second fundamental theorem of calculus, we obtain

$I=F(3)-F(2)$

$={\displaystyle \frac{1}{2}}\left[\log \left|{\displaystyle \frac{3-1}{3+1}}\right|-\log \left|{\displaystyle \frac{2-1}{2+1}}\right|\right]$

$={\displaystyle \frac{1}{2}}\left[\log \left|{\displaystyle \frac{2}{4}}\right|-\log \left|{\displaystyle \frac{1}{3}}\right|\right]$

$={\displaystyle \frac{1}{2}}\left[\log {\displaystyle \frac{1}{2}}-\log {\displaystyle \frac{1}{3}}\right]$

$={\displaystyle \frac{1}{2}}\left[\log {\displaystyle \frac{3}{2}}\right]$ 

12.Integrate the given integral $\underset{\mathbf{0}}{\overset{{\displaystyle \frac{\pi }{\mathbf{2}}}}{\int }}{\cos }^{\mathbf{2}}\mathbf{xdx}$

Ans: To simplify the question let us suppose,

$\int {\cos }^{2}xdx=\int \left({\displaystyle \frac{1+\cos 2x}{2}}\right)dx$

${\displaystyle \frac{x}{2}}+{\displaystyle \frac{\sin 2x}{4}}={\displaystyle \frac{1}{2}}\left(x+{\displaystyle \frac{\sin 2x}{2}}\right)$ 

Again, let us suppose the function is

${\displaystyle \frac{1}{2}}\left(x+{\displaystyle \frac{\sin 2x}{2}}\right)=F(x)$

By second fundamental theorem of calculus, we obtain

$I=F\left({\displaystyle \frac{\pi }{2}}\right)-F(0)$

$={\displaystyle \frac{1}{2}}\left[\left({\displaystyle \frac{\pi }{2}}+{\displaystyle \frac{\sin \pi }{2}}\right)-\left(0+{\displaystyle \frac{\sin 0}{2}}\right)\right]$

$={\displaystyle \frac{1}{2}}\left[{\displaystyle \frac{\pi }{2}}+0-0+0\right]$

$={\displaystyle \frac{\pi }{4}}$ 

13.Integrate the given integral $\underset{\mathbf{2}}{\overset{\mathbf{3}}{\int }}{\displaystyle \frac{\mathbf{xdx}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}}$

Ans: To simplify the question let us suppose,

$I=\underset{2}{\overset{3}{\int }}{\displaystyle \frac{xdx}{x^2+1}}$

$\int {\displaystyle \frac{xdx}{x^2+1}}$

$={\displaystyle \frac{1}{2}}\int {\displaystyle \frac{2x}{x^2+1}}dx$

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

Again, let us suppose the function is

${\displaystyle \frac{1}{2}}\log \left(1+x^2\right)=F(x)$

By second fundamental theorem of calculus, we obtain

$I=F(3)-F(2)$

$={\displaystyle \frac{1}{2}}\left[\log \left(1+{\left(3\right)}^2\right)-\log \left(1+{(2)}^2\right)\right]$

$={\displaystyle \frac{1}{2}}\left[\log 10-\log 5\right]$

$={\displaystyle \frac{1}{2}}\left[\log {\displaystyle \frac{10}{5}}\right]$

$={\displaystyle \frac{1}{2}}\left[\log 2\right]$ 

14.Integrate the given integral $\underset{\mathbf{0}}{\overset{\mathbf{1}}{\int }}{\displaystyle \frac{\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}}{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}}\mathbf{dx}$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{1}{\int }}{\displaystyle \frac{2x+3}{5x^2+1}}dx$

$\int {\displaystyle \frac{2x+3}{5x^2+1}}dx$

$={\displaystyle \frac{1}{5}}\int {\displaystyle \frac{5\left(2x+3\right)}{5x^2+1}}dx$

$={\displaystyle \frac{1}{5}}\int {\displaystyle \frac{\left(10x+15\right)}{5x^2+1}}dx$

$={\displaystyle \frac{1}{5}}\int {\displaystyle \frac{10x}{5x^2+1}}dx+3\int {\displaystyle \frac{1}{5x^2+1}}dx$

$={\displaystyle \frac{1}{5}}\int {\displaystyle \frac{10x}{5x^2+1}}dx+3\int {\displaystyle \frac{1}{5\left(x^2+{\displaystyle \frac{1}{5}}\right)}}dx$

$={\displaystyle \frac{1}{5}}\log \left(5x^2+1\right)+{\displaystyle \frac{3}{5}}{\displaystyle \frac{1}{{\displaystyle \frac{1}{\sqrt{5}}}}}{\tan }^{-1}{\displaystyle \frac{x}{{\displaystyle \frac{1}{\sqrt{5}}}}}$

$={\displaystyle \frac{1}{5}}\log \left(5x^2+1\right)+{\displaystyle \frac{3}{\sqrt{5}}}{\tan }^{-1}\left(\sqrt{5}\right)x$

$=F(x)$ 

By second fundamental theorem of calculus, we obtain

$I=F(3)-F(2)$

$={\displaystyle \frac{1}{5}}\left[\log \left(1+5\right)-{\displaystyle \frac{3}{\sqrt{5}}}\log \left(\sqrt{5}\right)\right]-\left[{\displaystyle \frac{1}{5}}\log (1)+{\displaystyle \frac{3}{\sqrt{5}}}{\tan }^{-1}(0)\right]$

$={\displaystyle \frac{1}{5}}\left[\log 6\right]+{\displaystyle \frac{3}{\sqrt{5}}}{\tan }^{-1}\sqrt{5}$ 

15.Integrate the given integral $\underset{\mathbf{0}}{\overset{\mathbf{1}}{\int }}\mathbf{x}{\mathbf{e}}^{{\mathbf{x}}^{\mathbf{2}}}\mathbf{dx}$

Ans: To simplify the question let us suppose,

$=\underset{0}{\overset{1}{\int }}x{e}^{x^2}dx$

$x^2=t$

$\Rightarrow 2xdx=dt$ 

As $x\to 0,t\to 0$ ,

Again $x\to 1,t\to 1,$, 

$\therefore I={\displaystyle \frac{1}{2}}\underset{0}{\overset{1}{\int }}{e}^{t}dt$

${\displaystyle \frac{1}{2}}\underset{0}{\overset{1}{\int }}{e}^{t}dt={\displaystyle \frac{1}{2}}{e}^{t}dt$

${\displaystyle \frac{1}{2}}{e}^{t}dt=F(t)$ 

By second fundamental theorem of calculus, we obtain

$I=F(1)-F(0)$

$={\displaystyle \frac{1}{2}}e-{\displaystyle \frac{1}{2}}{e}^0$

$={\displaystyle \frac{1}{2}}\left(e-1\right)$ 

16.Integrate the given integral $\underset{\mathbf{0}}{\overset{\mathbf{1}}{\int }}{\displaystyle \frac{\mathbf{5}{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}}}\mathbf{dx}$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{1}{\int }}{\displaystyle \frac{5x^2}{x^2+4x+3}}dx$

Dividing $5x^2$by $x^2+4x+3$, we obtain

$I=\underset{1}{\overset{2}{\int }}\left\{5-{\displaystyle \frac{20x+15}{x^2+4x+3}}\right\}dx$

$=\underset{1}{\overset{2}{\int }}5dx-\underset{1}{\overset{2}{\int }}{\displaystyle \frac{20x+15}{x^2+4x+3}}dx$

$={\left[5x\right]}_1^2-\underset{1}{\overset{2}{\int }}{\displaystyle \frac{20x+15}{x^2+4x+3}}dx$

$I=5-I_1$ 

 Consider,

Let,

$20x+15=A{\displaystyle \frac{d}{dx}}\left(x^2+4x+3\right)+B$

$=2Ax+(4A+B)$ 

Equating the coefficients of $x$ and constant term, we obtain

$A=10$ and $B=-25$

Let,$x^2+4x+3=t$

$\Rightarrow \left(2x+4\right)dx=dt$

$\Rightarrow I_1=10\int {\displaystyle \frac{dt}{t}}-25\int {\displaystyle \frac{dx}{{\left(x+2\right)}^2-1^2}}$

$=10\log t-25{\left[{\displaystyle \frac{1}{2}}\log \left({\displaystyle \frac{x+1}{x+3}}\right)\right]}_1^2$

$=\left[10\log 15-10\log 18\right]-25\left[{\displaystyle \frac{1}{2}}\log {\displaystyle \frac{3}{5}}-{\displaystyle \frac{1}{2}}\log {\displaystyle \frac{2}{4}}\right]$

$=\left[10\log 5+10\log 3-10\log 4-10\log 2\right]-{\displaystyle \frac{25}{2}}\left[\log 3-\log 5-\log 2+\log 4\right]$

$=\left[10+{\displaystyle \frac{25}{2}}\right]\log 5+\left[-10-{\displaystyle \frac{25}{2}}\right]\log 4+\left[10+{\displaystyle \frac{25}{2}}\right]\log 3+\left[-10+{\displaystyle \frac{25}{2}}\right]\log 2$

$={\displaystyle \frac{45}{2}}\log 5-{\displaystyle \frac{45}{2}}\log 4-{\displaystyle \frac{5}{2}}\log 3+{\displaystyle \frac{5}{2}}\log 2$

$={\displaystyle \frac{45}{2}}\log {\displaystyle \frac{5}{4}}-{\displaystyle \frac{5}{2}}\log {\displaystyle \frac{3}{2}}$ 

Substituting the value of $I_1$in (1), we obtain

$I=5-\left[{\displaystyle \frac{45}{2}}\log {\displaystyle \frac{5}{2}}-{\displaystyle \frac{5}{2}}\log {\displaystyle \frac{3}{2}}\right]$

$=5-{\displaystyle \frac{5}{2}}\left[9\log {\displaystyle \frac{5}{2}}-\log {\displaystyle \frac{3}{2}}\right]$ 

17.Integrate the given integral $\underset{\mathbf{0}}{\overset{{\displaystyle \frac{\pi }{\mathbf{4}}}}{\int }}\left(\mathbf{2}{\sec }^{\mathbf{2}}\mathbf{x}\mathbf{+}{\mathbf{x}}^{\mathbf{3}}\mathbf{+}\mathbf{2}\right)\mathbf{dx}$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{{\displaystyle \frac{\pi }{4}}}{\int }}\left(2{\sec }^{2}x+x^3+2\right)dx$

 $\int \left(2{\sec }^{2}x+x^3+2\right)dx=2\tan x+{\displaystyle \frac{x^4}{4}}+2x$

$2\tan x+{\displaystyle \frac{x^4}{4}}+2x=F(x)$ 

By second fundamental theorem of calculus, we obtain

$I=F\left({\displaystyle \frac{\pi }{4}}\right)+F\left(0\right)$

$=\left\{\left(2\tan {\displaystyle \frac{\pi }{4}}+{\displaystyle \frac{1}{4}}{\left({\displaystyle \frac{\pi }{4}}\right)}^2+2\left({\displaystyle \frac{\pi }{4}}\right)\right)-\left(2\tan 0+0+0\right)\right\}$

  $=2\tan {\displaystyle \frac{\pi }{4}}+{{\displaystyle \frac{\pi }{4^5}}}^4+{\displaystyle \frac{\pi }{2}}$

   $=2+{\displaystyle \frac{\pi }{2}}+{\displaystyle \frac{{\pi }^4}{1024}}$ 

18.Integrate the given integral $\underset{\mathbf{0}}{\overset{\pi }{\int }}\left({\sin }^{\mathbf{2}}{\displaystyle \frac{\mathbf{x}}{\mathbf{2}}}\mathbf{-}{\cos }^{\mathbf{2}}{\displaystyle \frac{\mathbf{x}}{\mathbf{2}}}\right)\mathbf{dx}$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{\pi }{\int }}\left({\sin }^2{\displaystyle \frac{x}{2}}-{\cos }^2{\displaystyle \frac{x}{2}}\right)dx$

 $=-\underset{0}{\overset{\pi }{\int }}\left({\cos }^2{\displaystyle \frac{x}{2}}-{\sin }^2{\displaystyle \frac{x}{2}}\right)dx$

   $=-\underset{0}{\overset{\pi }{\int }}\cos xdx$

   $=-\underset{0}{\overset{\pi }{\int }}\cos xdx=-\sin x$ 

By second fundamental theorem of calculus, we obtain

$I=F\left({\displaystyle \frac{\pi }{4}}\right)+F\left(0\right)$

  $=-\sin \pi +\sin 0$

$=0$ 

19.Integrate the given integral $\underset{\mathbf{0}}{\overset{\mathbf{1}}{\int }}{\displaystyle \frac{\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{3}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}}}\mathbf{dx}$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{2}{\int }}{\displaystyle \frac{6x+3}{x^2+4}}dx$

$\int {\displaystyle \frac{6x+3}{x^2+4}}dx$

$=3\int {\displaystyle \frac{2x+1}{x^2+4}}dx$

$=3\int {\displaystyle \frac{2x+1}{x^2+4}}dx$

$=3\int {\displaystyle \frac{2x}{x^2+4}}dx+3\int {\displaystyle \frac{1}{x^2+4}}dx$

$=3\log (x^2+4)+{\displaystyle \frac{3}{2}}{\tan }^{-1}{\displaystyle \frac{x}{2}}$

$=F(x)$ 

By second fundamental theorem of calculus, we obtain

$I=F(3)-F(2)$

$=3\left[\log \left(2^2+4\right)-{\displaystyle \frac{3}{2}}{\tan }^{-1}\left({\displaystyle \frac{2}{2}}\right)\right]-\left[3\log (0+4)+{\displaystyle \frac{3}{2}}{\tan }^{-1}({\displaystyle \frac{0}{2}})\right]$

$=3\left[\log 8\right]+{\displaystyle \frac{3}{2}}{\tan }^{-1}1-3\log 4-{\displaystyle \frac{3}{2}}{\tan }^{-1}0$

$=3\log 8+{\displaystyle \frac{3}{2}}{\tan }^{-1}1-3\log 4-{\displaystyle \frac{3}{2}}{\tan }^{-1}0$

$=3\log 8+{\displaystyle \frac{3}{2}}\left({\displaystyle \frac{\pi }{4}}\right)-3\log 4-0$

$=3\log \left({\displaystyle \frac{8}{4}}\right)+{\displaystyle \frac{3\pi }{8}}$

$=3\log 2+{\displaystyle \frac{3\pi }{8}}$ 

20.Integrate the given integral $\underset{0}{\overset{1}{\int }}\left(x{e}^x-\sin {\displaystyle \frac{\pi x}{4}}\right)dx$

Ans: To simplify the question let us suppose,

$I=\underset{0}{\overset{1}{\int }}\left(x{e}^x-\sin {\displaystyle \frac{\pi x}{4}}\right)dx$

$\underset{0}{\overset{1}{\int }}\left(x{e}^x-\sin {\displaystyle \frac{\pi x}{4}}\right)dx=x\int e^{x}dx-\int \left\{\left({\displaystyle \frac{d}{dx}}x\right)\int e^{x}dx\right\}dx+\left\{{\displaystyle \frac{-\cos {\displaystyle \frac{\pi x}{4}}}{{\displaystyle \frac{\pi }{4}}}}\right\}$

$=x{e}^x-x\int e^{x}dx-{\displaystyle \frac{4}{\pi }}\cos \pi {\displaystyle \frac{x}{4}}$

$=F(x)$ 

By second fundamental theorem of calculus, we obtain

 $I=F(1)-F(0)$

 $=\left(e^1-e^1-{\displaystyle \frac{4}{\pi }}\cos \pi {\displaystyle \frac{1}{4}}\right)-\left(0.e^0-e^0-{\displaystyle \frac{4}{\pi }}\cos 0\right)$

 $=e-e-{\displaystyle \frac{4}{\pi }}\left({\displaystyle \frac{1}{\sqrt{2}}}\right)+1+{\displaystyle \frac{4}{\pi }}$

$=1+{\displaystyle \frac{4}{\pi }}-{\displaystyle \frac{2\sqrt{2}}{\pi }}$ 

21.Integrate the given integral $\underset{1}{\overset{\sqrt{3}}{\int }}{\displaystyle \frac{dx}{1+x^2}}$

1. $${\displaystyle \frac{\pi }{3}}$$

2. ${\displaystyle \frac{2\pi }{3}}$

3. ${\displaystyle \frac{\pi }{6}}$

4. ${\displaystyle \frac{\pi }{12}}$

Ans: To simplify the question let us suppose,

$\int {\displaystyle \frac{dx}{1+x^2}}={\tan }^{-1}x$

By second fundamental theorem of calculus, we obtain

$\underset{1}{\overset{\sqrt{3}}{\int }}{\displaystyle \frac{dx}{1+x^2}}=F(\sqrt{3})-F(1)$

$={\tan }^{-1}\sqrt{3}-{\tan }^{{}^{-1}}1$

$={\displaystyle \frac{\pi }{3}}-{\displaystyle \frac{\pi }{4}}$

$={\displaystyle \frac{\pi }{12}}$ 

Hence, the correct Answer is ${\displaystyle \frac{\pi }{12}}$

22.Integrate the given integral$\int {\displaystyle \frac{dx}{4+9x^2}}=\int {\displaystyle \frac{dx}{{(2)}^2+{(3x)}^2}}$ 

1. ${\displaystyle \frac{\pi }{6}}$

2. ${\displaystyle \frac{\pi }{12}}$

3. ${\displaystyle \frac{\pi }{24}}$

4. ${\displaystyle \frac{\pi }{4}}$ 

equals

Ans: To simplify the question let us suppose,

 3x = t

$\Rightarrow 3dx=dt$ 

$\int {\displaystyle \frac{dx}{{(2)}^2+{(3x)}^2}}={\displaystyle \frac{1}{3}}\int {\displaystyle \frac{dt}{{(2)}^2+{(t)}^2}}$

$={\displaystyle \frac{1}{3}}\left[{\displaystyle \frac{1}{2}}{\tan }^{-1}{\displaystyle \frac{t}{2}}\right]$

$={\displaystyle \frac{1}{6}}{\tan }^{-1}\left({\displaystyle \frac{3x}{2}}\right)$

$=F(x)$ 

By second fundamental theorem of calculus,

$\underset{0}{\overset{{\displaystyle \frac{2}{3}}}{\int }}{\displaystyle \frac{dx}{4+9x^2}}=F\left({\displaystyle \frac{2}{3}}\right)-F(0)$

$={\displaystyle \frac{1}{6}}{\tan }^{-1}\left({\displaystyle \frac{3}{2}}\times {\displaystyle \frac{2}{3}}\right)-{\displaystyle \frac{1}{6}}{\tan }^{{}^{-1}}0$

 $={\displaystyle \frac{1}{6}}{\tan }^{-1}1-0$

 $={\displaystyle \frac{1}{6}}\times {\displaystyle \frac{\pi }{4}}$

 $={\displaystyle \frac{\pi }{24}}$ 

Hence, the correct answer is C.${\displaystyle \frac{\pi }{24}}$

Conclusion

NCERT Solutions for Exercise 7.8 Maths Class 12 Chapter 7 - Integrals provide detailed explanations to all of the questions in the NCERT textbook. This exercise contains 22 questions with fully solved solutions. It includes a range of exercises that help students understand how integrals can be applied in real-world circumstances. With a focus on practical integration methods and their applications, Exercise 7.8 provides students with the knowledge they require to succeed in mathematics exams.

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