Exercise 7.10 (Revised) - Chapter 7 - Integrals - Ncert Solutions class 12 - Maths
Updated On 26-08-2025 By Lithanya
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NCERT Class 12 Maths Solutions: Chapter 7 Integrals | Free PDF Download
Ex 7.10 Question 1.
$\int_0^{\frac{\pi}{2}} \cos ^2 x d x$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \cos ^2 x d x$
$
\begin{aligned}
& =\int_0^{\frac{\pi}{2}} \cos ^2\left(\frac{\pi}{2}-x\right) d x \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]}
\end{aligned}
$
$
\Rightarrow \int_0^{\frac{\pi}{2}} \sin ^2 x d x
$
Adding eq. (i) and (ii),
$
\begin{aligned}
& 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(\cos ^2 x+\sin ^2 x\right) d x \\
& =\int_0^{\frac{\pi}{2}} 1 d x
\end{aligned}
$
$\begin{aligned}
& =>2 I=(x)_0^{\frac{\pi}{2}} \\
& \Rightarrow 2 \mathrm{I}=\frac{\pi}{2} \\
& \Rightarrow \mathrm{I}=\frac{\pi}{4} \text { Ans. }
\end{aligned}$
Ex 7.10 Question 2.
$\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]} \\
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x+\sqrt{\sin x}}} d x \ldots \ldots . . \text { (ii) }
\end{aligned}
$
Adding eq. (i) and (ii),
$
2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\right) d x
$
$\begin{aligned}
& =\int_0^{\frac{\pi}{2}}\left(\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\right) d x \\
& \Rightarrow 2 \mathrm{I}=\int_0^{\frac{\pi}{2}} 1 d x \\
& \Rightarrow 2 I=(x)_0^{\frac{\pi}{2}} \\
& \Rightarrow 2 \mathrm{I}=\frac{\pi}{2} \\
& \Rightarrow \mathrm{I}=\frac{\pi}{4} \text { Ans. }
\end{aligned}$
Ex 7.10 Question 3.
$\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$ $\qquad$
$\Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)}{\sin ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)+\cos ^{\frac{3}{2}}\left(\frac{\pi}{2}-x\right)} d x$
$\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]$
$\Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\cos ^{\frac{3}{2}} x}{\cos ^{\frac{3}{2}} x+\sin ^{\frac{3}{2}} x} d x$
Adding eq. (i) and (ii),
$
\begin{aligned}
& 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(\frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}+\frac{\cos ^{\frac{3}{2}} x}{\cos ^{\frac{3}{2}} x+\sin ^{\frac{3}{2}} x}\right) d x \\
& =\int_0^{\frac{\pi}{2}}\left(\frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}\right) d x \\
& \Rightarrow 2 \mathrm{I}=\int_0^{\frac{\pi}{2}} 1 d x \\
& \Rightarrow 2 I=(x)_0^{\frac{\pi}{2}} \\
& \Rightarrow 2 \mathrm{I}=\frac{\pi}{2} \\
& \Rightarrow \mathrm{I}=\frac{\pi}{4} \text { Ans. }
\end{aligned}
$
Ex 7.10 Question 4.
$\int_0^{\frac{\pi}{2}} \frac{\cos ^5 x d x}{\sin ^5 x+\cos ^5 x}$
Answer.
Let I $=\int_0^{\frac{\pi}{2}} \frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x} d x$
$
\Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\cos ^5\left(\frac{\pi}{2}-x\right)}{\sin ^5\left(\frac{\pi}{2}-x\right)+\cos ^5\left(\frac{\pi}{2}-x\right)} d x
$
$
\begin{aligned}
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]} \\
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin ^5 x}{\cos ^5 x+\sin ^5 x} d x \cdots \ldots . . \text { (ii) }
\end{aligned}
$
Adding eq. (i) and (ii),
$
\begin{aligned}
& 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(\frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x}+\frac{\sin ^5 x}{\cos ^5 x+\sin ^5 x}\right) d x \\
& =\int_0^{\frac{\pi}{2}}\left(\frac{\cos ^5 x+\sin ^5 x}{\sin ^5 x+\cos ^5 x}\right) d x \\
& \Rightarrow 2 \mathrm{I}=\int_0^{\frac{\pi}{2}} 1 d x \\
& \Rightarrow>2 I=(x)_0^{\frac{\pi}{2}} \\
& \Rightarrow 2 \mathrm{I}=\frac{\pi}{2} \\
& \Rightarrow \mathrm{I}=\frac{\pi}{4} \text { Ans. }
\end{aligned}
$
Ex 7.10 Question 5.
$\int_{-5}^5|x+2| d x$
Answer.
Let $\mathrm{I}=\int_{-5}^5|x+2| d x$
Putting $x+2=0$
$
\Rightarrow x=-2 \in(-5,5)
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int_{-5}^{-2}|x+2| d x+\int_{-2}^5|x+2| d x \\
& =\int_{-5}^{-2}-(x+2) d x+\int_{-2}^5(x+2) d x \\
& =-\left(\frac{x^2}{2}+2 x\right)_{-5}^{-2}+\left(\frac{x^2}{2}+2 x\right)_{-2}^5 \\
& =-\left[\left(\frac{4}{2}-4\right)-\left(\frac{25}{2}-10\right)\right]+\left[\left(\frac{25}{2}+10\right)-\left(\frac{4}{2}-4\right)\right] \\
& =-\left(-2-\frac{5}{2}\right)+\left(\frac{45}{2}+2\right) \\
& =2+\frac{5}{2}+\frac{45}{2}+2 \\
& =4+25=29 \text { Ans. }
\end{aligned}
$
Ex 7.10 Question 6.
$\int_2^8|x-5| d x$
Answer.
Let $\mathrm{I}=\int_2^8|x-5| d x$
Putting $x-5=0$
$\Rightarrow x=5 \in(2,8)$
$\therefore$ From eq. (i),
$\begin{aligned}
\mathrm{I} & =\int_2^5|x-5| d x+\int_5^8|x-5| d x \\
& =\int_2^5-(x-5) d x+\int_5^8(x-5) d x \\
& =-\left(\frac{x^2}{2}-5 x\right)_2^5+\left(\frac{x^2}{2}-5 x\right)_5^8 \\
& =-\left[\left(\frac{25}{2}-25\right)-(2-10)\right]+\left[(32-40)-\left(\frac{25}{2}-25\right)\right] \\
& =-\left(-\frac{25}{2}+8\right)+\left(-8+\frac{25}{2}\right) \\
& =\frac{25}{2}-8-8+\frac{25}{2} \\
& =25-16 \\
& =9 \text { Ans. }
\end{aligned}$
$\begin{aligned}
\mathrm{I} & =\int_2^5|x-5| d x+\int_5^8|x-5| d x \\
& =\int_2^5-(x-5) d x+\int_5^8(x-5) d x \\
& =-\left(\frac{x^2}{2}-5 x\right)_2^5+\left(\frac{x^2}{2}-5 x\right)_5^8 \\
& =-\left[\left(\frac{25}{2}-25\right)-(2-10)\right]+\left[(32-40)-\left(\frac{25}{2}-25\right)\right] \\
& =-\left(-\frac{25}{2}+8\right)+\left(-8+\frac{25}{2}\right) \\
& =\frac{25}{2}-8-8+\frac{25}{2} \\
& =25-16 \\
& =9 \text { Ans. }
\end{aligned}$
Ex 7.10 Question 7.
$\int_0^1 x(1-x)^n d x$
Answer.
Let $\mathrm{I}=\int_0^1 x(1-x)^n d x$
$
\begin{aligned}
& =\int_0^1(1-x)\left\{1-(1-x)^n\right\} d x \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]} \\
& \Rightarrow \mathrm{I}=\int_0^1(1-x)(1-1+x)^n d x
\end{aligned}
$
$\begin{aligned}
& =\int_0^1(1-x) x^n d x \\
& =\int_0^1\left(x^n-x^{n+1}\right) d x \\
& \Rightarrow \mathrm{I}=\left(\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right)_0^1 \\
& =\frac{1}{n+1}-\frac{1}{n+2}-(0-0) \\
& =\frac{n+2-n-1}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)} \text { Ans. }
\end{aligned}$
Ex 7.10 Question 8.
$\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{4}} \log (1+\tan x) d x$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-x\right)\right] d x \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]} \\
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{4}} \log \left[1+\frac{1-\tan x}{1+\tan x}\right] d x
\end{aligned}
$
$
\begin{aligned}
& =\int_0^{\frac{\pi}{4}} \log \left[\frac{1+\tan x+1-\tan x}{1+\tan x}\right] d x \\
& =\int_0^{\frac{\pi}{4}} \log \left[\frac{2}{1+\tan x}\right] d x \ldots . \text { (ii) }
\end{aligned}
$
Adding eq. (i) and (ii),
$
\begin{aligned}
& 2 \mathrm{I}=\int_0^{\frac{\pi}{4}}\left[\log (1+\tan x)+\log \left(\frac{2}{1+\tan x}\right)\right] d x \\
& =\int_0^{\frac{\pi}{4}}\left[\log (1+\tan x)\left(\frac{2}{1+\tan x}\right)\right] d x \\
& \Rightarrow 2 \mathrm{I}=\int_0^{\frac{\pi}{4}}[\log 2] d x \\
& =(\log 2)(x)_0^{\frac{\pi}{4}} \\
& =\frac{\pi}{4} \log 2 \\
& \Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2 \text { Ans }
\end{aligned}
$
Ex 7.10 Question 9.
$\int_0^2 x \sqrt{2-x} d x$
Answer.
Let $\mathrm{I}=\int_0^2 x \sqrt{2-x} d x$
$\begin{aligned}
& =\int_0^2(2-x) \sqrt{2-(2-x)} d x \\
& \left.\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right] \\
& \Rightarrow \mathrm{I}=\int_0^2(2-x) \sqrt{x} d x \\
& =\int_0^2\left(2 x^{\frac{1}{2}}-x^{\frac{3}{2}}\right) d x \\
& =\left[2 \cdot \frac{x^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_0^2 \\
& =\left(\frac{4}{3} \cdot 2^{\frac{3}{2}}-\frac{2}{5} \cdot 2^{\frac{5}{2}}\right)-(0-0) \\
& \Rightarrow \mathrm{I}=\frac{4}{3} \times 2 \sqrt{2}-\frac{2}{5} \times 4 \sqrt{2} \\
& =\left(\frac{8}{3}-\frac{8}{5}\right) \sqrt{2} \\
& =\frac{16 \sqrt{2}}{15} \text { Ans. }
\end{aligned}$
Ex 7.10 Question 10.
$\text {} \int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$
$
\begin{aligned}
& =\int_0^{\frac{\pi}{2}}\left(\log \sin ^2 x-\log \sin 2 x\right) d x \\
& =\int_0^{\frac{\pi}{2}} \log \left(\frac{\sin ^2 x}{\sin 2 x}\right) d x \\
& =\int_0^{\frac{\pi}{2}} \log \left(\frac{\sin ^2 x}{2 \sin x \cos x}\right) d x \\
& =\int_0^{\frac{\pi}{2}} \log \left(\frac{1}{2} \tan x\right) d x \ldots \ldots \ldots . \text { (i) } \\
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \log \left(\frac{1}{2} \tan \left(\frac{\pi}{2}-x\right)\right) d x \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]} \\
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \log \left(\frac{1}{2} \cot x\right) d x
\end{aligned}
$
Adding eq. (i) and (ii),
$
2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left[\log \left(\frac{1}{2} \tan x\right)+\log \left(\frac{1}{2} \cot x\right)\right] d x
$
$\begin{aligned}
& =\int_0^{\frac{\pi}{2}}\left[\log \left(\frac{1}{2} \tan x\right)\left(\frac{1}{2} \cot x\right)\right] d x \\
& \Rightarrow 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left[\log \frac{1}{4}\right] d x \\
& =\log \frac{1}{4}(x)_0^{\frac{\pi}{2}} \\
& =(\log 1-\log 4) \frac{\pi}{2} \\
& =-\frac{\pi}{2} \log 4 \\
& \Rightarrow \mathrm{I}=-\frac{\pi}{4} \log 2^2 \\
& =-\frac{2 \pi}{4} \log 2 \\
& =-\frac{\pi}{2} \log 2 \\
& =\frac{\pi}{2} \log 2^{-1} \\
& =\frac{\pi}{2} \log \frac{1}{2} \text { Ans. }
\end{aligned}$
Ex 7.10 Question 11.
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x d x$
Answer.
Let $\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x d x$
$
\begin{aligned}
& =2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x \ldots \text { (i) } \\
& {\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x, \text { when } f(x) \text { is even function }\right]} \\
& \Rightarrow \mathrm{I}=2 \int_0^{\frac{\pi}{2}} \sin ^2\left(\frac{\pi}{2}-x\right) d x \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]} \\
& \Rightarrow \mathrm{I}=2 \int_0^{\frac{\pi}{2}} \cos ^2 x d x \quad \text {.........(ii) }
\end{aligned}
$
Adding eq. (i) and (ii),
$\begin{aligned}
& 2 \mathrm{I}=2 \int_0^{\frac{\pi}{2}}\left(\sin ^2 x+\cos ^2 x\right) d x \\
&=2 \int_0^{\frac{\pi}{2}} 1 d x \\
&=2(x)_0^{\frac{\pi}{2}} \\
&=2 \cdot \frac{\pi}{2}=\pi \\
& \Rightarrow \mathrm{I}=\frac{\pi}{2} \text { Ans. }
\end{aligned}$
Ex 7.10 Question 12.
$\int_0^\pi \frac{x d x}{1+\sin x}$
Answer.
Let $\mathrm{I}=\int_0^\pi \frac{x}{1+\sin x} d x$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_0^\pi \frac{\pi-x}{1+\sin (\pi-x)} d x \\
& =\int_0^\pi \frac{\pi-x}{1+\sin x} d x \quad \ldots . . . . . . \text { (ii) }
\end{aligned}
$
Adding eq. (i) and (ii),
$
\begin{aligned}
& 2 \mathrm{I}=\int_0^\pi\left(\frac{x}{1+\sin x}+\frac{\pi-x}{1+\sin x}\right) d x \\
& =\int_0^\pi\left(\frac{x+\pi-x}{1+\sin x}\right) d x \\
& =\int_0^\pi\left(\frac{\pi}{1+\sin x}\right) d x \\
& =\pi \int_0^\pi\left(\frac{1}{1+\sin x}\right) d x \\
& \Rightarrow 2 \mathrm{I}=2 \pi \int_0^{\frac{\pi}{2}} \frac{d x}{1+\sin x}
\end{aligned}
$
$\left[\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x, \text { if } f(2 a-x)=f(x)\right]$
$\begin{aligned}
& \Rightarrow 2 \mathrm{I}=2 \pi \int_0^{\frac{\pi}{2}} \frac{d x}{1+\sin \left(\frac{\pi}{2}-x\right)} \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]} \\
& \Rightarrow 2 \mathrm{I}=2 \pi \int_0^{\frac{\pi}{2}} \frac{d x}{1+\cos x} \\
& \Rightarrow \mathrm{I}=\pi \int_0^{\frac{\pi}{2}} \frac{d x}{2 \cos ^2 \frac{x}{2}} \\
& =\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x
\end{aligned}$
$\begin{aligned}
& =\frac{\pi}{2}\left[\frac{\tan \frac{x}{2}}{\frac{1}{2}}\right]_0^{\frac{\pi}{2}} \\
& =\pi\left(\tan \frac{\pi}{4}-\tan 0^{\circ}\right) \\
& =\pi(1-0)=\pi \text { Ans. }
\end{aligned}$
Ex 7.10 Question 13.
$\text {} \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^7 x d x$
Answer.
Let $\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^7 x d x$
Here $f(x)=\sin ^7 x$
$
\begin{aligned}
& \therefore f(-x)=\sin ^7(-x) \\
& =(-\sin x)^7 \\
& =-\sin ^7 x=-f(x)
\end{aligned}
$
$\therefore f(x)$ is an odd function of $x$.
$
\begin{aligned}
& \therefore \mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^7 x d x=0 \\
& {\left[\because \int_{-a}^a f(x) d x=0, \text { when } f(x) \text { is an odd function }\right]}
\end{aligned}
$
Ex 7.10 Question 14.
$\int_0^{2 \pi} \cos ^5 x d x$
Answer.
$\int_0^{2 \pi} \cos ^5 x d x$
$
\begin{aligned}
& =2 \int_0^\pi \cos ^5 x d x \\
& {\left[\because \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x, \text { if } f(2 a-x)=f(x)\right]}
\end{aligned}
$
$\begin{aligned}
& \text { Here } f(x)=\cos ^5 x \\
& \therefore f(2 \pi-x)=\cos ^5(2 \pi-x)=\cos ^5 x \\
& \Rightarrow f(x)=2(0)=0
\end{aligned}$
Ex 7.10 Question 15.
$\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
Answer.
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
$\Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)-\cos \left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)} d x$
$=\int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x \sin x} d x$
$=-\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
Adding eq. (i) and (ii), we have $2 \mathrm{I}=0 \Rightarrow \mathrm{I}=0$ Ans.
Ex 7.10 Question 16.
$\int_0^\pi \log (1+\cos x) d x R$
Answer.
Let $\mathrm{I}=\int_0^\pi \log (1+\cos x) d x$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_0^\pi \log (1+\cos (\pi-x)) d x \\
& =\int_0^\pi \log (1-\cos x) d x \quad \ldots \ldots \ldots . . \text { (ii) }
\end{aligned}
$
Adding eq. (i) and (ii),
$
\begin{aligned}
& 2 \mathrm{I}=\int_0^\pi[\log (1+\cos x)+\log (1-\cos x)] d x \\
= & \int_0^\pi[\log (1+\cos x)(1-\cos x)] d x \\
\Rightarrow & 2 \mathrm{I}=\int_0^\pi\left[\log \left(1-\cos ^2 x\right)\right] d x \\
= & \int_0^\pi\left[\log \sin ^2 x\right] d x
\end{aligned}
$
$
\begin{aligned}
& =2 \int_0^\pi[\log \sin x] d x \\
& \Rightarrow \mathrm{I}=2 \int_0^{\frac{\pi}{2}} \log \sin x d x \\
& \Rightarrow \mathrm{I}=2 \int_0^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) d x \\
& =2 \int_0^{\frac{\pi}{2}} \log \cos x d x \\
&
\end{aligned}
$
Adding eq. (i) and (ii),
$\begin{aligned}
2 \mathrm{I} & =2 \int_0^{\frac{\pi}{2}}(\log \sin x+\log \cos x) d x \\
& =2 \int_0^{\frac{\pi}{2}}(\log \sin x \cos x) d x \\
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \log \left(\frac{2 \sin x \cos x}{2}\right) d x \\
& =\int_0^{\frac{\pi}{2}} \log \left(\frac{\sin 2 x}{2}\right) d x \\
& =\int_0^{\frac{\pi}{2}}(\log \sin 2 x-\log 2) d x
\end{aligned}$
$
\begin{aligned}
& =\int_0^{\frac{\pi}{2}}(\log \sin 2 x-\log 2) d x \\
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \log \sin 2 x d x-\int_0^{\frac{\pi}{2}} \log 2 d x \\
& =\int_0^{\frac{\pi}{2}} \log \sin 2 x d x-\log 2(x)_0^{\frac{\pi}{2}} \\
& =\int_0^{\frac{\pi}{2}} \log \sin 2 x d x-\frac{\pi}{2} \log 2 \\
& \Rightarrow \mathrm{I}=\mathrm{I}_1-\frac{\pi}{2} \log 2 \ldots \ldots \ldots \ldots .(\mathrm{v}) \\
& \text { Where } \mathrm{I}_1=\int_0^{\frac{\pi}{2}} \log \sin 2 x d x \cdots \ldots . \text { (vi) }
\end{aligned}
$
Where $\mathrm{I}_1=\int_0^{\frac{\pi}{2}} \log \sin 2 x d x$ .(vi)
Putting $2 x=t$ in eq. (vi),
$
\begin{aligned}
& \Rightarrow 2=\frac{d t}{d x} \\
& \Rightarrow d x=\frac{d t}{2}
\end{aligned}
$
Limits of integration when $x=0, t=0$ and $x=\frac{\pi}{2}, t=\pi$
$\therefore$ From eq. (vi),
$
\begin{aligned}
& \mathrm{I}_1=\int_0^\pi \log \sin t \frac{d t}{2} \\
& =\frac{1}{2} \int_0^\pi \log \sin t d t \\
& =\frac{1}{2} \times 2 \int_0^{\frac{\pi}{2}} \log \sin t d t \\
& \Rightarrow \mathrm{I}_1=\int_0^{\frac{\pi}{2}} \log \sin t d t \\
& =\int_0^{\frac{\pi}{2}} \log \sin x d x \\
& {\left[\because \int_a^b f(t) d t=\int_a^b f(x) d x\right]}
\end{aligned}
$
$\Rightarrow I_1=\frac{1}{2} \text { [From eq. (iii)] }$
Putting this value in eq. (v), $\mathrm{I}=\frac{1}{2}-\frac{\pi}{2} \log 2$
$
\begin{aligned}
& \Rightarrow 2 \mathrm{I}=\mathrm{I}-\pi \log 2 \\
& \Rightarrow 2 \mathrm{I}-\mathrm{I}=-\pi \log 2 \\
& \Rightarrow \mathrm{I}=-\pi \log 2
\end{aligned}
$
Ex 7.10 Question 17.
$\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$
Answer.
Let $\mathrm{I}=\int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}} d x \\
& \left.\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right] \\
& =\int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x
\end{aligned}
$
Adding eq. (i) and (ii),
$
\begin{aligned}
2 \mathrm{I} & =\int_0^a\left(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}+\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\right) d x \\
& =\int_0^a\left(\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\right) d x \\
& =\int_0^a 1 d x=(x)_0^a=a
\end{aligned}
$
$\Rightarrow \mathrm{I}=\frac{a}{2} \text { Ans. }$
Ex 7.10 Question 18.
$\int_0^4|x-1| d x$
Answer.
Let $\mathrm{I}=\int_0^4|x-1| d x$
Here $x-1=0$
$
\Rightarrow x=1 \in(0,4)
$
$\therefore$ From eq. (i),
$
\begin{aligned}
\mathrm{I} & =\int_0^1|x-1| d x+\int_1^4|x-1| d x \\
& =-\int_0^1(x-1) d x+\int_1^4(x-1) d x \\
& \Rightarrow \mathrm{I}=-\left(\frac{x^2}{2}-x\right)_0^1+\left(\frac{x^2}{2}-x\right)_1^4 \\
& =-\left\{\left(\frac{1}{2}-1\right)-0\right\}+\left\{\frac{16}{2}-4-\left(\frac{1}{2}-1\right)\right\} \\
& =\frac{1}{2}+8-4+\frac{1}{2} \\
& =9-4
\end{aligned}
$
$
=5 \text { Ans. }
$
Ex 7.10 Question 19.
$\text {Show that } \int_0^a f(x) g(x) d x=2 \int_0^a f(x) d x \text { if } \mathbf{f} \text { and } g \text { are defined as } f(x)=f(a-x)$
$
\text { and } g(x)+g(a-x)=4
$
Answer.
Here $f(x)=f(a-x)$ $\qquad$ (i) and $g(x)+g(a-x)=4$ $\qquad$
Let $\mathrm{I}=\int_0^a f(x) g(x) d x$ $\qquad$ (iii)
$
\begin{aligned}
& \therefore \mathrm{I}=\int_0^a f(a-x) g(a-x) d x \\
& {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x=\right]} \\
& =\int_0^a f(x) g(a-x) d x[\text { From eq. (i)]. }
\end{aligned}
$
Adding eq. (iii) and (iv)
$
\begin{aligned}
& 2 \mathrm{I}=\int_0^a(f(x) g(x)+f(x) g(a-x)) d x \\
& =\int_0^a f(x)(g(x)+g(a-x)) d x \\
& \Rightarrow 2 \mathrm{I}=\int_0^a f(x)(4) d x \text { [From eq. (ii)] } \\
& \Rightarrow 2 \mathrm{I}=4 \int_0^a f(x) d x
\end{aligned}
$
$
\Rightarrow \mathrm{I}=2 \int_0^a f(x) d x
$
Hence proved.
Ex 7.10 Question 20.
The value of $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^3+x \cos x+\tan ^5 x+1\right) d x$ is:
(A) 0
(B) 2
(C) $\pi$
(D) 1
Answer.
Let $\mathrm{I}=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^3+x \cos x+\tan ^5 x+1\right) d x$
$
\begin{aligned}
& =\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x^3 d x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} x \cos x d x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \tan ^5 x d x+\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 1 d x \\
& \Rightarrow \mathrm{I}=0+0+0+(x)_{\frac{-\pi}{2}}^{\frac{\pi}{2}}
\end{aligned}
$
Because $x^3, x \cos x$ and $\tan ^5 x$ all are odd functions.
$
=\frac{\pi}{2}-\left(\frac{-\pi}{2}\right)=\pi
$
Therefore, option (C) is correct.
Ex 7.10 Question 21.
The value of $\int_0^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is
(A) 2
(B) $\frac{3}{4}$
(C) 0
(D) -2
Answer.
Let $I=\int_0^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$
$
\begin{aligned}
& \Rightarrow \mathrm{I}=\int_0^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)} d x\right. \\
& =\int_0^{\frac{\pi}{2}} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x \cdots \ldots \ldots . \text { (ii) }
\end{aligned}
$
Adding eq. (i) and (ii),
$
\begin{aligned}
& 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)+\log \left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right) d x \\
& =\int_0^{\frac{\pi}{2}}\left(\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) \cdot\left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right) d x \\
& \Rightarrow 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}(\log 1) d x \\
& =\int_0^{\frac{\pi}{2}} 0 d x=0 \\
& \Rightarrow \mathrm{I}=0
\end{aligned}
$
Option (C) is correct.
