NCERT Solutions For Class 12 Maths Chapter 7 Integrals Exercise 7.10 - 2025-26

Exercise 7.10

Solve the following integrals.

1. $\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}$                     …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}\left( \dfrac{\pi }{2}-x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$                          …(2)

Adding equation (1) and (2),

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

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

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}xdx}=\dfrac{\pi }{4}$

2. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$                     …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}}{\sqrt{\sin \left( \dfrac{\pi }{2}-x \right)}+\sqrt{\cos \left( \dfrac{\pi }{2}-x \right)}}dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

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

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}=\dfrac{\pi }{4}$

3. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}$                        …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)dx}{{{\sin }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)+{{\cos }^{\dfrac{3}{2}}}\left( \dfrac{\pi }{2}-x \right)}}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

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

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{\dfrac{3}{2}}}xdx}{{{\sin }^{\dfrac{3}{2}}}x+{{\cos }^{\dfrac{3}{2}}}x}}=\dfrac{\pi }{4}$

4. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}$

Ans: Given $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}$                        …(1)

We know that,

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, the integral becomes

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}\left( \dfrac{\pi }{2}-x \right)dx}{{{\sin }^{5}}\left( \dfrac{\pi }{2}-x \right)+{{\cos }^{5}}\left( \dfrac{\pi }{2}-x \right)}}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}$                                         …(2)

Adding equation (1) and (2),

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\sin }^{5}}x+{{\cos }^{5}}x}{{{\sin }^{5}}x+{{\cos }^{5}}x}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{1dx}$

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

$\Rightarrow 2I=\dfrac{\pi }{2}$

$\Rightarrow I=\dfrac{\pi }{4}$

$\therefore \int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{{{\cos }^{5}}xdx}{{{\sin }^{5}}x+{{\cos }^{5}}x}}=\dfrac{\pi }{4}$

5. $\int\limits_{-5}^{5}{\left| x+2 \right|}dx$

Ans:  Let $I=\int\limits_{-5}^{5}{\left| x+2 \right|}dx$ 

Since, $\left( x+2 \right)\le 0$ for interval $\left[ -5,-2 \right]$ .

Therefore, $\left( x+2 \right)\ge 0$ for interval $\left[ -2,5 \right]$.

As, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{c}{f\left( x \right)}dx+\int\limits_{c}^{b}{f\left( x \right)dx}$

Hence, $\int\limits_{-5}^{-2}{-\left( x+2 \right)}dx+\int\limits_{-2}^{5}{\left( x+2 \right)dx}$ .

Thus, 

$I=\int\limits_{-5}^{-2}{-\left( x+2 \right)}dx+\int\limits_{-2}^{5}{\left( x+2 \right)dx}$

$=-\left[ \dfrac{{{x}^{2}}}{2}+2x \right]_{-5}^{2}+\left[ \dfrac{{{x}^{2}}}{2}+2x \right]_{-2}^{5}$

$=-\left[ \dfrac{{{\left( 2 \right)}^{2}}}{2}+2\left( 2 \right)-\dfrac{{{\left( -5 \right)}^{2}}}{2}-2\left( -5 \right) \right]+\left[ \dfrac{{{\left( 5 \right)}^{2}}}{2}+2\left( 5 \right)-\dfrac{{{\left( -2 \right)}^{2}}}{2}-2\left( -2 \right) \right]$$=-\left[ 2-4-\dfrac{25}{2}+10 \right]+\left[ \dfrac{25}{2}+10-2+4 \right]$

$=-2+4+\dfrac{25}{2}+10+\dfrac{25}{2}+10-2+4$

$=29$

6. $\int\limits_{2}^{8}{\left| x-5 \right|}dx$

Ans: Let $I=\int\limits_{2}^{8}{\left| x-5 \right|}dx$ 

Since, $\left( x-5 \right)\le 0$ for interval $\left[ 2,5 \right]$ .

Therefore, $\left( x-5 \right)\ge 0$ for interval $\left[ 5,8 \right]$.

As, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{c}{f\left( x \right)}dx+\int\limits_{c}^{b}{f\left( x \right)dx}$

Hence, $\int\limits_{2}^{5}{-\left( x-5 \right)}dx+\int\limits_{5}^{8}{\left( x-5 \right)dx}$ .

Thus, 

$I=\int\limits_{2}^{5}{-\left( x-5 \right)}dx+\int\limits_{5}^{8}{\left( x-5 \right)dx}$

$=-\left[ \dfrac{{{x}^{2}}}{2}-5x \right]_{2}^{5}+\left[ \dfrac{{{x}^{2}}}{2}-5x \right]_{5}^{8}$

$=-\left[ \dfrac{{{\left( 5 \right)}^{2}}}{2}-5\left( 5 \right)-\dfrac{{{\left( 2 \right)}^{2}}}{2}+5\left( 2 \right) \right]+\left[ \dfrac{{{\left( 8 \right)}^{2}}}{2}-5\left( 8 \right)-\dfrac{{{\left( 5 \right)}^{2}}}{2}+5\left( 5 \right) \right]$

$=-\left[ \dfrac{25}{2}-25-2+10 \right]+\left[ 32-40-\dfrac{25}{2}+25 \right]$

$=-\dfrac{25}{2}+25+2-10+32-40-\dfrac{25}{2}+25$

$=9$

7. $\int\limits_{0}^{1}{x{{\left( 1-x \right)}^{n}}dx}$

Ans: Let  $I=\int\limits_{0}^{1}{x{{\left( 1-x \right)}^{n}}dx}$

Thus, $I=\int\limits_{0}^{1}{\left( 1-x \right){{\left( 1-\left( 1-x \right) \right)}^{n}}dx}$

Since, $\int\limits_{1}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$.

Therefore, 

$\int\limits_{0}^{1}{\left( 1-x \right){{\left( x \right)}^{n}}}dx$

$=\int\limits_{0}^{1}{\left( {{x}^{n}}-{{x}^{n+1}} \right)dx}$

$=\left[ \dfrac{{{x}^{n+1}}}{n+1}-\dfrac{{{x}^{n+2}}}{n+2} \right]_{0}^{1}$

$=\left[ \dfrac{1}{n+1}-\dfrac{1}{n+2} \right]$

$=\dfrac{\left( n+2 \right)-\left( n+1 \right)}{\left( n+1 \right)\left( n+2 \right)}$

$=\dfrac{1}{\left( n+1 \right)\left( n+2 \right)}$

8. $\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}$

Ans:  Let  $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)dx}$

Since, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, $I=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ 1+\tan \left( \dfrac{\pi }{4}-x \right) \right]dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left[ 1+\dfrac{\tan \dfrac{\pi }{4}-\tan x}{1+\tan \dfrac{\pi }{4}\tan x} \right]dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left\{ 1+\dfrac{1-\tan x}{1+\tan x} \right\}dx}$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log \dfrac{2}{1+\tan x}}dx$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx}-\int\limits_{0}^{\dfrac{\pi }{4}}{\log \left( 1+\tan x \right)}dx$

$=\int\limits_{0}^{\dfrac{\pi }{4}}{\log 2dx}-I$

$2I=\left[ x\log 2 \right]_{0}^{\dfrac{\pi }{4}}$

$2I=\dfrac{\pi }{4}\log 2$

$I=\dfrac{\pi }{8}\log 2$

9. $\int\limits_{0}^{2}{x\sqrt{2-x}dx}$

Ans: Let  $I=\int\limits_{0}^{2}{x\sqrt{2-x}dx}$

Since, $\int\limits_{0}^{a}{f\left( x \right)}dx=\int\limits_{0}^{a}{f\left( a-x \right)}dx$

Therefore, $I=\int\limits_{0}^{2}{\left( 2-x \right)}\sqrt{x}dx$

$=\int\limits_{0}^{2}{\left\{ 2{{x}^{{1}/{2}\;}}-{{x}^{{3}/{2}\;}} \right\}}dx$

$=\left[ 2\left( \dfrac{{{x}^{{3}/{2}\;}}}{{3}/{2}\;} \right)-\dfrac{{{x}^{{5}/{2}\;}}}{{5}/{2}\;} \right]_{0}^{2}$

$=\left[ \dfrac{4}{3}{{x}^{{3}/{2}\;}}-\dfrac{2}{5}{{x}^{{5}/{2}\;}} \right]_{0}^{2}$

$=\dfrac{4}{3}{{\left( 2 \right)}^{{3}/{2}\;}}-\dfrac{2}{5}{{\left( 2 \right)}^{{5}/{2}\;}}$

$=\dfrac{8\sqrt{2}}{3}-\dfrac{8\sqrt{2}}{5}$

$=\dfrac{40\sqrt{2}-24\sqrt{2}}{15}$

$=\dfrac{16\sqrt{2}}{15}$

10. $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \sin 2x \right)dx}$

Ans: Let  $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \sin 2x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \left( 2\sin x\cos x \right) \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\log \sin x-\log \sin x-\log \cos x-\log 2 \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \sin x-\log \cos x-\log 2 \right)dx}$                      …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \cos x-\log \sin x-\log 2 \right)dx}$ …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( -\log 2-\log 2 \right)dx}$

$\Rightarrow 2I=-2\log 2\int\limits_{0}^{\dfrac{\pi }{2}}{dx}$

$\Rightarrow I=-\log 2\left[ \dfrac{\pi }{2} \right]$

$\Rightarrow I=-\dfrac{\pi }{2}\left[ \log 2 \right]$

$\Rightarrow I=\dfrac{\pi }{2}\left[ -\log 2 \right]$

$\Rightarrow I=\dfrac{\pi }{2}\left[ \log \dfrac{1}{2} \right]$

$\Rightarrow I=\dfrac{\pi }{2}\log \dfrac{1}{2}$

11. $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$

Ans: Let  $I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$

Since, ${{\sin }^{2}}x$ is an even function.

Therefore , $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}=2\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$

As if $f\left( x \right)$ is an even function, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$.

Hence,

$I=2\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}$

$=2\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1-\cos 2x}{2}dx}$

$=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1-\cos 2x \right)dx}$

$=\left[ x-\dfrac{sin2x}{2} \right]_{0}^{\dfrac{\pi }{2}}$

$=\dfrac{\pi }{2}$

12. $\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}$

Ans: Let  $I=\int\limits_{0}^{\pi }{\dfrac{x}{1+\sin x}dx}$               …….(1)

Since, $\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, $I=\int\limits_{0}^{\pi }{\dfrac{\left( \pi -x \right)}{1+\sin x\left( \pi -x \right)}dx}$

$I=\int\limits_{0}^{\pi }{\dfrac{\left( \pi -x \right)}{1+\sin x}dx}$                   …..(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\pi }{\dfrac{\pi }{1+\sin x}dx}$

$\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{\dfrac{\left( 1-\sin x \right)}{\left( 1+\sin x \right)+\left( 1-\sin x \right)}dx}$

$\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{\dfrac{\left( 1-\sin x \right)}{{{\cos }^{2}}x}dx}$

$\Rightarrow 2I=\pi \int\limits_{0}^{\pi }{\left\{ {{\sec }^{2}}x-\tan x\sec x \right\}dx}$

$\Rightarrow 2I=\pi \left[ 2 \right]$

$\Rightarrow I=\pi $

13. $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{7}}xdx}$.

Ans: Let  $I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{7}}xdx}$

Since, ${{\sin }^{7}}x$ is an even function.

Therefore , $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{{{\sin }^{2}}xdx}=0$

As if $f\left( x \right)$ is an odd function, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=0$.

Hence, $I=0$

14. $\int\limits_{0}^{2\pi }{{{\cos }^{5}}xdx}$.

Ans: Let  

$I=\int\limits_{0}^{2\pi }{{{\cos }^{5}}xdx}$      

${{\cos }^{5}}\left( 2\pi -x \right)={{\cos }^{5}}x$                   …..(1)

If $f\left( 2a-x \right)=f\left( x \right)$ then $\int\limits_{0}^{2a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)dx}$.

If $f\left( 2a-x \right)=-f\left( x \right)$ then $\int\limits_{0}^{2a}{f\left( x \right)dx}=0$

Since, ${{\cos }^{5}}\left( \pi -x \right)=-{{\cos }^{5}}x$

Therefore, 

$I=2\int\limits_{0}^{2\pi }{{{\cos }^{5}}xdx}$

$\Rightarrow I=2\left( 0 \right)$

$\Rightarrow I=0$

15. $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{1+\sin x\cos x}dx}$

Ans:   Let $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin x-\cos x}{1+\sin x\cos x}dx}$                     ……(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\sin \left( \dfrac{\pi }{2}-x \right)-\cos \left( \dfrac{\pi }{2}-x \right)}{1+\sin \left( \dfrac{\pi }{2}-x \right)\cos \left( \dfrac{\pi }{2}-x \right)}dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x-\sin x}{1+\cos x\sin x}dx}$                               ……(2)

On adding equation 1 and 2

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{0}{1+\cos x\sin x}dx}$

$\Rightarrow I=0$

16. $\int\limits_{0}^{\pi }{\log \left( 1+\cos x \right)dx}$

Ans: Let  

$I=\int\limits_{0}^{\pi }{\log \left( 1+\cos x \right)dx}$       …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow I=\int\limits_{0}^{\pi }{\log \left( 1+\cos \left( \pi -x \right) \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\pi }{\log \left( 1-\cos x \right)dx}$                 …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\pi }{\left\{ \log \left( 1-\cos x \right)+\log \left( 1-\cos x \right) \right\}dx}$

$\Rightarrow 2I=\int\limits_{0}^{\pi }{\log \left( 1-{{\cos }^{2}}x \right)dx}$

 $\Rightarrow 2I=\int\limits_{0}^{\pi }{\log {{\sin }^{2}}xdx}$

$\Rightarrow 2I=2\int\limits_{0}^{\pi }{\log \sin xdx}$

$\Rightarrow I=\int\limits_{0}^{\pi }{\log \sin xdx}$                                     …..(3)

Since, $\sin \left( \pi -x \right)=\sin x$

Therefore, $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log \sin xdx}$                        ……(4)

$\Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{2}}{\log \sin \left( \dfrac{\pi }{2}-x \right)dx}$

$\Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{2}}{\log \cos xdx}$                                   …….(5)

On adding equation 4 and 5-

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \sin x+\log \cos x \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log \sin x+\log \cos x+\log 2-\log 2 \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log 2\sin x\cos x-\log 2 \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \log 2\sin x\cos x \right)dx}-\int\limits_{0}^{\dfrac{\pi }{2}}{\log 2dx}$

Let $2x=t$

On differentiating-

$2dx=dt$

If $x=0$ then $t=0$.

Thus,

$\Rightarrow I=\dfrac{I}{2}-\dfrac{\pi }{2}\log 2$

$\Rightarrow \dfrac{I}{2}=-\dfrac{\pi }{2}\log 2$

$\Rightarrow I=-\pi \log 2$

17. $\int\limits_{0}^{a}{\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx}$

Ans:  Let $I=\int\limits_{0}^{a}{\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx}$                              …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$I=\int\limits_{0}^{a}{\dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx}$                                     …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{a}{\dfrac{\sqrt{x}+\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx}$

$\Rightarrow 2I=\int\limits_{0}^{a}{dx}$

$\Rightarrow 2I=\left[ x \right]_{0}^{a}$

$\Rightarrow 2I=a$

$\Rightarrow I=\dfrac{a}{2}$

18. $\int\limits_{0}^{4}{\left| x-1 \right|dx}$

Ans: Let $I=\int\limits_{0}^{4}{\left| x-1 \right|dx}$

Thus, $\left( x-1 \right)\le 0$ when $0\le x\le 1$  and $\left( x-1 \right)\ge 0$ when $1\le x\le 4$

Since, $\int\limits_{a}^{b}{f\left( x \right)}dx=\int\limits_{a}^{c}{f\left( x \right)dx}+\int\limits_{c}^{b}{f\left( x \right)}$

Therefore, $I=\int\limits_{0}^{1}{\left| x-1 \right|}dx+\int\limits_{1}^{4}{\left| x-1 \right|dx}$

$\Rightarrow I=\int\limits_{0}^{1}{-\left( x-1 \right)}dx+\int\limits_{1}^{4}{\left( x-1 \right)dx}$

$=\left[ x-\dfrac{{{x}^{2}}}{2} \right]_{0}^{1}+\left[ \dfrac{{{x}^{2}}}{2}-x \right]_{1}^{4}$

$=1-\dfrac{1}{2}+\dfrac{{{\left( 4 \right)}^{2}}}{2}-4-\dfrac{1}{2}+1$

$=1-\dfrac{1}{2}+8-4-\dfrac{1}{2}+1$

$=5$

19. Show that $\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)}dx=2\int\limits_{0}^{a}{f\left( x \right)}dx$ , if $f$and $g$ are defined as $f\left( x \right)=f\left( a-x \right)$ and $g\left( x \right)+g\left( a-x \right)=4$ .

Ans: Let $\int\limits_{0}^{a}{f\left( x \right)}g\left( x \right)dx$              …….(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow \int\limits_{0}^{a}{f\left( a-x \right)}g\left( a-x \right)dx$

$\Rightarrow \int\limits_{0}^{a}{f\left( x \right)}g\left( a-x \right)dx$   ……(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{a}{\left\{ f\left( x \right)g\left( x \right)+f\left( x \right)g\left( a-x \right) \right\}dx}$

$\Rightarrow 2I=\int\limits_{0}^{a}{f\left( x \right)\left\{ g\left( x \right)+g\left( a-x \right) \right\}dx}$

As, $g\left( x \right)+g\left( a-x \right)=4$.

Thus,

 $\Rightarrow 2I=\int\limits_{0}^{a}{4f\left( x \right)dx}$

$\Rightarrow I=2\int\limits_{0}^{a}{f\left( x \right)dx}$

Hence, $\int\limits_{0}^{a}{f\left( x \right)g\left( x \right)}dx=2\int\limits_{0}^{a}{f\left( x \right)}dx$ , if $f$and $g$ are defined as $f\left( x \right)=f\left( a-x \right)$ and $g\left( x \right)+g\left( a-x \right)=4$ .

20. The value of $\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( {{x}^{3}}+x\cos x+{{\tan }^{5}}x+1 \right)dx}$ is 

1. $0$

2. $2$

3. $\pi $

4. $1$

Ans: Let $I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( {{x}^{3}}+x\cos x+{{\tan }^{5}}x+1 \right)dx}$

$\Rightarrow I=\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( {{x}^{3}} \right)dx}+\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( x\cos x \right)dx}+\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( {{\tan }^{5}}x \right)dx}+\int\limits_{\dfrac{-\pi }{2}}^{\dfrac{\pi }{2}}{\left( 1 \right)dx}$

If $f\left( x \right)$ is an even function, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=2\int\limits_{0}^{a}{f\left( x \right)}dx$

And $f\left( x \right)$ is an odd function, then $\int\limits_{-a}^{a}{f\left( x \right)dx}=0$

Thus,

$I=0+0+0+2\int\limits_{0}^{\dfrac{\pi }{2}}{dx}$

$=2\left[ x \right]_{0}^{\dfrac{\pi }{2}}$

$=2\left[ \dfrac{\pi }{2} \right]$

$=\pi $

Hence, the correct option is C.

21. The value of $\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin x}{4+3\cos x} \right)dx}$ is

1. $2$

2. $\dfrac{3}{4}$

3. $0$

4. $-2$

Ans: Let $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin x}{4+3\cos x} \right)dx}$                …..(1)

Since, 

$\int\limits_{0}^{a}{f\left( x \right)dx}=\int\limits_{0}^{a}{f\left( a-x \right)dx}$

Therefore, 

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\sin \left( \dfrac{\pi }{2}-x \right)}{4+3\cos \left( \dfrac{\pi }{2}-x \right)} \right)dx}$

$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{4+3\cos x}{4+3\sin x} \right)dx}$                       …….(2)

On adding equation 1 and 2-

$2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\log 1dx}$

$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{0dx}$

$\Rightarrow I=0$

Hence, the correct option is C.

Formulas Used in Class 12 Chapter 7 Exercise 7.10

$\begin{array}{ll}\mathbf{P}_{0}: & \int_{a}^{b} f(x) d x=\int_{a}^{b} f(t) d t \\ \mathbf{P}_{1}: & \int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x . \text { In particular, } \int_{a}^{a} f(x) d x=0 \\ \mathbf{P}_{2}: & \int_{a}^{b} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{b} f(x) d x\end{array} $

\[ \begin{aligned} \mathbf{P}_3: & \quad \int_a^b f(x) \,dx = \int_a^b f(a + b - x) \,dx \\ \mathbf{P}_4: & \quad \int_0^a f(x) \,dx = \int_0^a f(a - x) \,dx \end{aligned} \]

${{\mathbf{P}}_5}:\quad \int_0^{2a} f (x)dx = \int_0^a f (x)dx + \int_0^a f (2a - x)dx$

${{\mathbf{P}}_6}:\quad \int_0^{2a} f (x)dx = 2\int_0^a f (x)dx$, if $f(2a - x) = f(x)$ and $0$ if  $f(2a - x) =  - f(x)$

${{\mathbf{P}}_7}:\quad $ (i) $\int_{ - a}^a f (x)dx = 2\int_0^a f (x)dx$, if $f$ is an even function, i.e., if $f( - x) = f(x)$.

(ii) $\int_{ - a}^a f (x)dx = 0$, if $f$ is an odd function, i.e., if $f( - x) =  - f(x)$.

Conclusion

NCERT Solutions for Maths Exercise 7.10 in Class 12 Chapter 7 - Integrals offer a comprehensive exploration of the properties of definite integrals. With clear, step-by-step solutions provided by Vedantu, students can deepen their understanding of integral properties and enhance their problem-solving skills. This practice not only builds a solid foundation for tackling complex integrals but also boosts confidence for CBSE exams.

Class 12 Maths Chapter 7: Exercises Breakdown

CBSE Class 12 Maths Chapter 7 Other Study Materials

NCERT Solutions for Class 12 Maths | Chapter-wise List

Given below are the chapter-wise NCERT 12 Maths solutions PDF. Using these chapter-wise class 12th maths ncert solutions, you can get clear understanding of the concepts from all chapters.

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