Exercise 9.3 - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $9.3$
Question 1.
Evaluate the following definite integrals :
$
\int_3^4 \frac{d x}{x^2-4}
$
Solution:
$
\begin{aligned}
\text { Let I } & =\int_3^4 \frac{d x}{x^2-4} \\
& =\int_3^4 \frac{d x}{x^2-2^2}\left[\because \frac{d x}{x^2-2^2}=\frac{1}{2 a} \log \left(\frac{x-a}{x+a}\right)\right] \\
& =\left[\frac{1}{2(2)} \log \left(\frac{x-2}{x+2}\right)\right]_3^4 \\
& =\frac{1}{4}\left[\log \left(\frac{2}{6}\right)-\log \left(\frac{1}{5}\right)\right] \\
& =\frac{1}{4}\left[\log \left(\frac{2}{6} \times \frac{5}{1}\right)\right]=\frac{1}{4} \log \left(\frac{5}{3}\right)
\end{aligned}
$
(ii) $\int_{-1} \frac{d x}{x^2+2 x+5}$
Solution:

Let $\mathrm{I}=\int_{-1}^1 \frac{d x}{x^2+2 x+5}$
$
\begin{aligned}
& =\int_{-1}^1 \frac{d x}{(x+1)^2-1+5} \\
& =\int_{-1}^1 \frac{d x}{(x+1)^2+(2)^2}\left[\int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right] \\
& =\left[\frac{1}{2} \tan ^{-1}\left(\frac{x+1}{2}\right)\right]_{-1}^1 \\
& =\frac{1}{2}\left[\tan ^{-1}\left(\frac{2}{2}\right)-\tan ^{-1}\left(\frac{0}{2}\right)\right]=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \\
& =\frac{1}{2}\left[\frac{\pi}{4}-0\right]=\frac{\pi}{8} .
\end{aligned}
$
(iii) $\int_0^1 \sqrt{\frac{1-x}{1+x}} d x$
Solution:
Let $\mathrm{I}=\int_0^1 \sqrt{\frac{1-x}{1+x}} d x$
Put $x=\cos 2 \theta$

Let $\mathrm{I}=\int_0^1 \sqrt{\frac{1-x}{1+x}} d x$
Put $x=\cos 2 \theta$

Differentiate with respect to $\theta$
$
d x=-2 \sin 2 \theta d \theta
$

Consider:
$
\begin{aligned}
& \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}(\sin 2 \theta) \\
& =\sqrt{\frac{2 \sin ^2 \theta}{2 \cos ^2 \theta}}(\sin 2 \theta) \\
& =\frac{\sin \theta}{\cos \theta}(2 \sin \theta \cos \theta) \\
& =2 \sin ^2 \theta \\
& =2 \frac{(1-\cos 2 \theta)}{2}=1-\cos 2 \theta .
\end{aligned}
$
Substitute (2) and (3) in (1), we get
$
\begin{aligned}
(1) \Rightarrow \mathrm{I} & =\int_{\frac{\pi}{4}}^0 \sqrt{\frac{1-\cos 2 \theta}{1+\cos 2 \theta}}(-2 \sin 2 \theta) d \theta \\
& =2 \int_0^{\frac{\pi}{4}}(1-\cos 2 \theta) d \theta \\
& =2\left[\theta-\frac{\sin 2 \theta}{2}\right]_0^{\frac{\pi}{4}} \\
& =2\left[\frac{\pi}{4}-\frac{1}{2}\right]=\frac{\pi}{2}-1
\end{aligned}
$

(iv) $\int_0^{\frac{\pi}{2}}\left(\frac{1+\sin x}{1+\cos x}\right) d x$
Solution:

Let I
$
\begin{aligned}
I & =\int_0^{\frac{\pi}{2}}\left(\frac{1+\sin x}{1+\cos x}\right) d x \\
& =\int_0^{\frac{\pi}{2}} \frac{1+\frac{2 \tan x / 2}{1+\tan ^2 x / 2}}{1+\frac{1-\tan ^2 x / 2}{1+\tan ^2 x / 2}} \\
& =\int_0^{\frac{\pi}{2}} \frac{1+\tan ^2 x / 2+2 \tan x / 2}{1+\tan ^2 x / 2+1-\tan 2 x / 2} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{1}{2}\left[\sec ^2 x / 2+2 \tan x / 2\right] d x=\frac{1}{2}\left[\frac{\tan x / 2}{(1 / 2)}+2 \frac{\log \sec (x / 2)}{(1 / 2)}\right]_0^{\frac{\pi}{2}} \\
& =\tan \frac{\pi}{4}+2 \log \sec \left(\frac{\pi}{4}\right)-\tan 0-2 \log \sec 0 . \\
& =1+2 \log \sqrt{2}=1+\log 2 .
\end{aligned}
$

(v) $\int_0^{\pi / 2} \sqrt{\cos \theta} \sin ^3 \theta d \theta$
Solution:
$
I=\int_0^{\frac{\pi}{2}} \sqrt{\cos \theta} \sin ^2 \theta \sin \theta d \theta
$

Put $t=\cos \theta$
Differentiate with respect to ' $\theta$ '
$
\begin{gathered}
\frac{d t}{d \theta}=-\sin \theta \\
-d t=\sin \theta d \theta
\end{gathered}
$

Substitute (2) and (3) in (1), we get,
$
\begin{aligned}
(1) \Rightarrow \mathrm{I} & =\int_1^0 \sqrt{t}\left(1-t^2\right)(-d t) \\
& =\int_0^1\left(t^{1 / 2}-t^{5 / 2}\right) d t \\
& =\left(\frac{t^{3 / 2}}{3 / 2}-\frac{t^{7 / 2}}{7 / 2}\right)_0^1=\frac{2}{3}-\frac{2}{7} \\
& =\frac{14-6}{21}=\frac{8}{21} .
\end{aligned}
$

(vi) $
\int_0^1 \frac{1-x^2}{\left(1+x^2\right)^2} d x
$
Solution:

$\begin{aligned}
& \text { Let } I=\int_0^1 \frac{1-x^2}{\left(1+x^2\right)^2} d x \\
& =\int_0^1\left[\frac{2}{\left(1+x^2\right)^2}-\frac{1+x^2}{\left(1+x^2\right)^2}\right] d x \\
& I=\int_0^1\left[\frac{2}{\left(1+x^2\right)^2}-\frac{1}{\left(1+x^2\right)}\right] d x \\
& I_1=\int_0^1 \frac{2}{\left(1+x^2\right)^2} d x \\
& \text { Put } x=\tan \theta \\
& d x=\sec ^2 \theta d \theta \\
& =2 \int_0^{\frac{\pi}{4}} \frac{\sec ^2 \theta d \theta}{\left(1+\tan ^2 \theta\right)^2} \\
& =2 \int_0^{\frac{\pi}{4}} \frac{\sec ^2 \theta}{\left(\sec ^2 \theta\right)^2} d \theta=2 \int_0^{\frac{\pi}{4}} \cos ^2 \theta d \theta \\
& =2 \int_0^{\frac{\pi}{4}}\left(\frac{1+\cos 2 \theta}{2}\right) d \theta=\left(\theta+\frac{\sin 2 \theta}{2}\right)_0^{\frac{\pi}{4}} \\
& =\frac{\pi}{4}+\frac{1}{2} \\
& I_2=\int_0^1 \frac{1}{1+x^2} d x=\left[\tan ^{-1} x\right]_0^1=\frac{\pi}{4} \\
& \text { (1) } \Rightarrow I=\frac{\pi}{4}+\frac{1}{2}-\frac{\pi}{4} \\
&
\end{aligned}$

$I=\frac{1}{2}$

Question 2.
Evaluate the following integrals using properties of integration :
$
\int_{-5}^5 x \cos \left(\frac{e^x-1}{e^x+1}\right) d x
$
Solution:
$
\begin{aligned}
\text { Let } \mathrm{I} & =\int_{-5}^5 x \cos \left(\frac{e^x-1}{e^x+1}\right) d x \\
f(x) & =x \cos \left(\frac{e^x-1}{e^x+1}\right) \\
f(-x) & =-x \cos \left(\frac{\frac{1}{e^x}-1}{\frac{1}{e^x}+1}\right)=-x \cos \left(\frac{1-e^x}{1+e^x}\right) \\
& =-x \cos \left(\frac{e^x-1}{1+e^x}\right) \\
f(-x) & =-f(x)
\end{aligned}
$
$\therefore f(x)$ is an odd functions.
$
\therefore \int_{-5}^5 x \cos \left(\frac{e^x-1}{e^x+1}\right) d x=0
$

(ii) $
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^5+x \cos x+\tan ^3 x+1\right) d x
$
Solution:

$\text { Let } \begin{aligned}
\mathrm{I} & =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(x^5+x \cos x+\tan ^3 x+1\right) d x \\
& =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^5 d x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \cos x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan ^3 x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d x \\
& =0+0+0+[x]_{-\pi / 2}^{\pi / 2} \\
& =\frac{\pi}{2}+\frac{\pi}{2}=\pi
\end{aligned}$

(iii)
$
\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x
$
Solution:

$
\begin{aligned}
f(x) & =\sin ^2 x \\
f(-x) & =\sin ^2(-x) \sin ^2 x=f(x)
\end{aligned}
$
$\therefore f(x)$ is an even function.
$
\begin{aligned}
\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x & =2 \int_0^{\frac{\pi}{4}} \sin ^2 x d x \\
& =2 \int_0^{\frac{\pi}{4}}\left(\frac{1-\cos 2 x}{2}\right) d x \\
& =\left(x-\frac{\sin 2 x}{2}\right)_0^{\frac{\pi}{4}}=\frac{\pi}{4}-\frac{\sin 2\left(\frac{\pi}{4}\right)}{2} \\
& =\frac{\pi}{4}-\frac{1}{2}=\frac{\pi-2}{4} .
\end{aligned}
$

(iv)
$
\int_0^{2 \pi} x \log \left(\frac{3+\cos x}{3-\cos x}\right) d x
$
Solution:
Let $\mathrm{I}=\int_0^{2 \pi} x \log \left(\frac{3+\cos x}{3-\cos x}\right) d x$
Consider:
$
\begin{aligned}
f(x) & =x \log \left(\frac{3+\cos x}{3-\cos x}\right) \\
f(2 \pi-x) & =(2 \pi-x) \log \left[\frac{3+\cos (2 \pi-x)}{3-\cos (2 \pi-x)}\right] \\
& =(2 \pi-x) \log \left[\frac{3+\cos x}{3-\cos x}\right] \\
f(2 \pi-x) & =2 \pi \log \left(\frac{3+\cos x}{3-\cos x}\right)-x \log \left(\frac{3+\cos x}{3-\cos x}\right) \\
f(2 \pi-x)+f(x) & =2 \pi \log \left(\frac{3+\cos x}{3-\cos x}\right) \\
\therefore \int_0^{2 \pi} x \log \left(\frac{3+\cos x}{3-\cos x}\right) d x & =0 .
\end{aligned}
$

(v) $\int_0^{2 \pi} \sin ^4 x \cos ^3 d x$
Solution:
$
\begin{aligned}
f(x) & =\sin ^4 x \cos ^3 x \\
f(2 \pi-x) & =\sin ^4(2 \pi-x) \cos ^3(2 \pi-x) \\
& =\sin ^4 x \cos ^3 x=f(x) \\
& =2 \int_0^{2 a} f(x) d x=0 \text { if } f(2 a-x)=-f(x) . \\
& =2 \int_0^a f(x) d x \text { if } f(2 a-x)=f(x) . \\
\text { Again } f(x) & =\sin ^4 x \cos ^3 x d x \\
f(\pi-x) & =\sin ^4(\pi-x) \cos ^3(\pi-x) \\
f(\pi-x) & =-\sin ^4(\pi-x) \cos (\pi-x)=-f(x)=0 .
\end{aligned}
$

(vi) $\int_0^1|5 x-3| d x$
Solution:
$
\begin{aligned}
& \text { Let } I=\int_0^1|5 x-3| d x \\
& |5 x-3|=\left\{\begin{array}{l}
5 x-3 ; x \geq 3 / 5 \\
3-5 x ; x<3 / 5
\end{array}\right. \\
& I=\int_0^{3 / 5}(3-5 x) d x+\int_{3 / 5}^1(5 x-3) d x \\
& =\left(3 x-\frac{5 x^2}{2}\right)_0^{3 / 5}+\left[\frac{5 x^2}{2}-3 x\right]_{3 / 5}^1 \\
& =\frac{9}{5}-\frac{5}{2}\left(\frac{9}{25}\right)+\left(\frac{5}{2}-3\right)-\left[\frac{5}{2}\left(\frac{9}{25}\right)-\frac{9}{5}\right] \\
& =\frac{9}{5}-\frac{9}{10}-\frac{1}{2}-\frac{9}{10}+\frac{9}{5} \\
& =\frac{18-9-5-9+18}{10}=\frac{13}{10} . \\
&
\end{aligned}
$

(vii) $\int_0^{\sin ^2 x} \sin ^{-1} \sqrt{t} d t+\int_0^{\cos ^2 x} \cos ^{-1} \sqrt{t} d t$
Solution:
Let $\mathrm{I}=\int_0^{\sin ^2 x} \sin ^{-1} \sqrt{t} d t+\int_0^{\cos ^2 x} \cos ^{-1} \sqrt{t} d t$
Consider:
$
I_1=\int_0^{\sin ^2 x} \sin ^{-1} \sqrt{t} d t
$
Put $t=\sin ^2 x$
Differentiate with respect to $x$
$
\begin{aligned}
d t & =2 \sin x \cos x d x \\
d t & =\sin 2 x d x
\end{aligned}
$

$\begin{aligned}
& I_1=\int_0^x \sin ^{-1} \sqrt{\sin ^2 x} \sin 2 x d x \\
& =\int_0^x \sin ^{-1}(\sin x) \sin 2 x d x=\int_0^x x \sin 2 x d x \\
& I_2=\int_0^{\cos ^2 x} \cos ^{-1} \sqrt{t} d t \\
& \text { Put } t=\cos ^2 x \\
& \text { Differentiate with respect to } x \\
& d t=-2 \cos x \sin x d x \\
& d t=-\sin 2 x d x \\
& I_2=\int_{\frac{\pi}{2}}^x \cos ^{-1} \sqrt{\cos ^2 x}(-\sin 2 x) d x \\
& =-\int_{\frac{\pi}{2}}^x \cos ^{-1}(\cos x) \sin 2 x d x \\
& I_2=\int_x^{\frac{\pi}{2}} x \sin 2 x d x \\
& (1) \Rightarrow \quad I=\int_0^x x \sin 2 x d x+\int_x^{\frac{\pi}{2}} x \sin 2 x d x \\
& I=\int_x^{\frac{\pi}{2}} x \sin 2 x d x \\
&
\end{aligned}$

Apply Bernoulli's formula
$
\begin{aligned}
\int u d v & =\left[u v-u^1 v_1\right] \\
& =\left(x\left[\frac{-\cos 2 x}{2}\right]-\left[\frac{-\sin 2 x}{4}\right]\right)_0^{\frac{\pi}{2}} \\
& =\left(-\frac{x \cos 2 x}{2}+\frac{\sin 2 x}{4}\right)_0^{\frac{\pi}{2}} \\
& =-\frac{\pi}{2} \frac{\cos \pi}{2}+\frac{\sin \pi}{4} \\
& =-\frac{\pi}{2}\left(\frac{-1}{2}\right)+0=\frac{\pi}{4}
\end{aligned}
$

(viii) $\int_0^1 \frac{\log (1+x)}{1+x^2} d x$
Solution:
Let $\mathrm{I}=\int_0^1 \frac{\log (1+x)}{1+x^2} d x$ Put $x=\tan \theta$
Differentiate with respect to $\theta$

$
d x=\sec ^2 \theta d \theta
$
Substitute (2) and (3) in (1), we get
$
\begin{aligned}
(1) \Rightarrow \quad I & =\int_0^{\frac{\pi}{4}} \frac{\log (1+\tan \theta)}{1+\tan ^2 \theta} \sec ^2 \theta d \theta \\
\mathrm{I} & =\int_0^{\frac{\pi}{4}} \frac{\log (1+\tan \theta)}{\sec ^2 \theta} \sec ^2 \theta d \theta \\
\mathrm{I} & =\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta
\end{aligned}
$

$\begin{aligned}
\mathrm{I} & =\int_0^{\frac{\pi}{4}} \log \left[1+\tan \left(\frac{\pi}{4}-\theta\right)\right] d \theta=\int_0^{\frac{\pi}{4}} \log \left[1+\frac{1-\tan \theta}{1+\tan \theta}\right] d \theta \\
& =\int_0^{\frac{\pi}{4}} \log \left[\frac{1+\tan \theta+1-\tan \theta}{1+\tan \theta}\right] d \theta=\int_0^{\frac{\pi}{4}} \log \left[\frac{1}{1+\tan \theta}\right] d \theta \\
\mathrm{I} & =\int_0^{\frac{\pi}{4}} \log 2 d \theta-\underbrace{\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta}_{\mathrm{I}} \\
2 \mathrm{I} & =\int_0^{\frac{\pi}{4}} \log 2 d \theta=\log 2[\theta]_0^{\frac{\pi}{4}} \\
2 \mathrm{I} & =\frac{\pi}{4} \log 2 \Rightarrow \mathrm{I}=\frac{\pi}{8} \log 2 .
\end{aligned}$

(ix)
$
\int_0^\pi \frac{x \sin x}{1+\sin x} d x
$
Solution:
Let $I=\int_0^\pi \frac{x \sin x}{1+\sin x} d x$
$
\begin{aligned}
\int^a f(x) d x & =\int^a f(a-x) d x \\
\mathrm{I} & =\int_0^\pi \frac{(\pi-x) \sin x}{1+\sin x} d x
\end{aligned}
$
$
[\because \sin (\pi-x)=\sin x]
$

$\begin{aligned}
& \text { (1)+(2) } 2 \mathrm{I}=\int_0^\pi \frac{x \sin x}{1+\sin x} d x+\int_0^\pi \frac{(\pi-x) \sin x}{1+\sin x} d x \\
& \text { 2I }=\int_0^\pi \frac{(x+\pi-x) \sin x}{1+\sin x} d x=\pi \int_0^\pi \frac{\sin x}{1+\sin x} d x \\
& =\pi \int_0^\pi \frac{\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} d x=\pi \int_0^\pi \frac{\sin x-\sin ^2 x}{1-\sin ^2 x} d x \\
& =\pi \int_0^\pi \frac{\sin x-\sin ^2 x}{\cos ^2 x} d x=\pi \int_0^\pi\left(\frac{\sin x}{\cos ^2 x}-\frac{\sin ^2 x}{\cos ^2 x}\right) d x \\
& =\pi \int_0^\pi\left(\sec x \tan x-\tan ^2 x\right) d x \quad\left(\because \tan ^2 x=\sec ^2 x-1\right) \\
& =\pi \int_0^\pi\left(\sec x \tan x-\sec ^2 x+1\right) d x=\pi[\sec x-\tan x+x]_0^\pi \\
& =\pi[(\sec \pi-\tan \pi+\pi)-(\sec 0-\tan 0+0)] \\
& =\pi[-1-0+\pi-1]=\pi[\pi-2] \\
& 2 \mathrm{I}=\pi^2-2 \pi \\
& I=\frac{\pi^2}{2}-\pi \\
&
\end{aligned}$

$\begin{aligned}
& \int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{1}{1+\sqrt{\tan x}} d x \\
& \text { Solution: } \\
& \text { Let } I=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{1}{1+\sqrt{\tan x}} d x \\
& \int_a^b f(x) d x=\int_a^b f(a+b-x) d x \\
& I=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{1}{1+\sqrt{\tan \left(\frac{\pi}{8}+\frac{3 \pi}{8}-x\right)}} d x=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{1}{1+\sqrt{\tan \left(\frac{\pi}{2}-x\right)}} d x \\
& I=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{1}{1+\sqrt{\cot x}} d x=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{1}{1+\frac{1}{\sqrt{\tan x}}} d x \\
& I=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+1} d x \\
& \text { (1) }+(2) \Rightarrow \\
& 2 I=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{1}{1+\sqrt{\tan x}} d x+\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+1} d x \\
& =\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}} d x=\int_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} d x=[x]_{\frac{\pi}{8}}^{\frac{3 \pi}{8}} \\
&
\end{aligned}$

$\begin{aligned}
& =\frac{3 \pi}{8}-\frac{\pi}{8}=\frac{2 \pi}{8} \\
2 I & =\frac{\pi}{4} \\
I & =\frac{\pi}{8} .
\end{aligned}$

(xi) $\int_0^\pi x\left[\sin ^2(\sin x)+\cos ^2(\cos x)\right] d x$
Solution:
$
\begin{aligned}
\text { Let } \mathrm{I} & =\int_0^\pi\left[\sin ^2(\sin x)+\cos ^2(\cos x)\right] d x \\
\int_0^a f(x) d x & =\int_0^a f(a-x) d x \\
\mathrm{I} & =\int_0^\pi(\pi-x)\left[\sin ^2(\sin (\pi-x))+\cos ^2(\cos (\pi-x))\right] d x \\
\mathrm{I} & =\int_0^\pi(\pi-x)\left[\sin ^2(\sin x)+\cos ^2(\cos x)\right] d x
\end{aligned}
$

$
\begin{aligned}
&(1)+(2) \Rightarrow \\
& 2 I=\pi \int_0^\pi\left[\sin ^2(\sin x)+\cos ^2(\cos x)\right] d x \\
& \therefore I=\frac{\pi}{2} \int_0^\pi\left[\sin ^2(\sin x)+\cos ^2(\cos x)\right] d x \\
& \text { But } \int_0^{2 a} f(2 a-x) d x=2 \int_0^a f(x) d x
\end{aligned}
$
So, $\mathrm{I}=\frac{\pi}{2} \times 2 \int_0^{\pi / 2}\left[\sin ^2(\sin x)+\cos ^2(\cos x)\right] d x$
But, I $=\pi \int_0^{\pi / 2}\left[\sin ^2(\cos x)+\cos ^2(\sin x)\right] d x$
$
\because \int_0^a f(x) d x=\int_0^a f(a-x) d x
$
$
\begin{aligned}
& \text { (3) }+(4) \Rightarrow 2 \mathrm{I}=\pi(2) \int_0^{\pi / 2} d x=2 \pi(x)_0^{\pi / 2}=2 \pi\left(\frac{\pi}{2}\right)=\pi^2 \\
& \Rightarrow \quad \mathrm{I}=\frac{\pi^2}{2} \text {. } \\
&
\end{aligned}
$