Exercise 9.5 - 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 $9.5$
Question 1.
Evaluate the following:
(i) $\int_0^{\frac{\pi}{2}} \frac{d x}{1+5 \cos ^2 x}$
Solution:
Let $I=\int_0^{\frac{\pi}{2}} \frac{d x}{1+5 \cos ^2 x}$
Dividing both Numerator and Denominator by $\cos ^2 x$.
$
\begin{aligned}
& =\int_0^{\frac{\pi}{2}} \frac{\frac{1}{\cos ^2 x}}{\frac{1}{\cos ^2 x}+5} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x+5} d x=\int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{1+\tan ^2 x+5} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\tan ^2 x+6} d x \\
& =\tan ^x
\end{aligned}
$
Consider, $t=\tan x$

Differentiate with respect to ' $x$ '
$
d t=\cdot \sec ^2 x d x
$

Substitute (2) and (3) in (1), we get
$
\begin{aligned}
(1) \Rightarrow \quad & =\int_0^{\infty} \frac{d t}{t^2+6} \quad\left\{\begin{array}{l}
\int \frac{d x}{a^2+x^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right) \\
\tan ^{-1}(\infty)=\frac{\pi}{2}
\end{array}\right\} \int_0^{\infty} \frac{d t}{t^2+(\sqrt{6})^2} \\
& =\left[\frac{1}{\sqrt{6}} \tan ^{-1} \frac{t}{\sqrt{6}}\right]_0^{\infty}=\frac{1}{\sqrt{6}} \tan ^{-1}(\infty)-0 \\
& =\frac{1}{\sqrt{6}}\left(\frac{\pi}{2}\right)=\frac{\pi}{2 \sqrt{6}} .
\end{aligned}
$

(ii) $
\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \sin ^2 x}
$
Solution:
Let $I=\int_0^{\frac{\pi}{2}} \frac{d x}{5+4 \sin ^2 x}$
Dividing both Numerator and Denominator by $\cos ^2 \mathrm{x}$

$
\begin{aligned}
& =\int_0^{\frac{\pi}{2}} \frac{\frac{1}{\cos ^2 x}}{\frac{5}{\cos ^2 x}+\frac{4 \sin ^2 x}{\cos ^2 x}} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{5 \sec ^2 x+4 \tan ^2 x} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{5\left(1+\tan ^2 x\right)+4 \tan ^2 x} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{5+9 \tan ^2 x} d x
\end{aligned}
$
Consider, $t=\tan x$
Differentiate with respect to ' $x$ ' $d t=\sec ^2 x d x$

Substitute (2) and (3) in (1), we get

$\begin{aligned}
(1) & \Rightarrow \quad=\int_0^{\infty} \frac{d t}{5+9 t^2}=\frac{1}{9} \int_0^{\infty} \frac{d t}{\frac{5}{9}+t^2} \\
& =\frac{1}{9} \int_0^{\infty} \frac{d t}{\left(\frac{\sqrt{5}}{3}\right)^2+t^2}=\frac{1}{9}\left[\frac{1}{\frac{\sqrt{5}}{3}} \tan ^{-1} \frac{t}{\frac{\sqrt{5}}{3}}\right]_0^{\infty} \\
& =\frac{1}{9}\left[\frac{3}{\sqrt{5}}\left(\frac{\pi}{2}\right)-0\right] \\
& =\frac{1}{9}\left[\frac{3 \pi}{2 \sqrt{5}}\right]=\frac{\pi}{6 \sqrt{5}} .
\end{aligned}$