Exercise 9.5-Additional Problems - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Evaluate: $\int_0^{\pi / 4} \frac{d x}{4+5 \cos ^2 x}$
Solution:
$
\begin{aligned}
I & =\int_0^{\pi / 4} \frac{d x}{4+5 \cos ^2 x}=\int_0^{\pi / 4} \frac{\sec ^2 x}{4 \sec ^2 x+5} d x \\
& =\int_0^{\pi / 4} \frac{\sec ^2 x}{4\left(1+\tan ^2 x\right)+5} d x=\int_0^{\pi / 4} \frac{\sec ^2 x}{9+4 \tan ^2 x} d x
\end{aligned}
$
Put $2 \tan \mathrm{x}, \therefore 2 \mathrm{sc}^2 \mathrm{x} \mathrm{dx}=\mathrm{dt}$
When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=2$
$
\begin{aligned}
\therefore \quad \mathrm{I} & =\int_0^2 \frac{\frac{1}{2} d t}{9+t^2}=\frac{1}{2} \int_0^2 \frac{d t}{3^2+t^2}=\frac{1}{2}\left[\frac{1}{3} \tan ^{-1} \frac{t}{3}\right]_0^2 \\
& =\frac{1}{6}\left(\tan ^{-1} \frac{2}{3}-\tan ^{-1} 0\right)=\frac{1}{6} \tan ^{-1} \frac{2}{3} .
\end{aligned}
$

Question 2.
Evaluate: $\int_0^{\pi / 2} \frac{d x}{4+9 \cos ^2 x}$

Solution:
$
\begin{aligned}
I & =\int_0^{\pi / 2} \frac{d x}{4+9 \cos ^2 x}=\int_0^{\pi / 2} \frac{d x}{4\left(\cos ^2 x+\sin ^2 x\right)+9 \cos ^2 x} \\
& =\int_0^{\pi / 2} \frac{d x}{13 \cos ^2 x+4 \sin ^2 x}
\end{aligned}
$
Dividing Numerator and Denominator by $\cos ^2 x$, we get
$
I=\int_0^{\pi / 2} \frac{\sec ^2 x}{13+4 \tan ^2 x} d x=\int_0^{\pi / 2} \frac{\sec ^2 x}{13+(2 \tan x)^2} d x
$
Put $2 \tan x=t$, then $2 \cdot \sec ^2 x d x=d t$
When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\infty$
$
\begin{aligned}
\therefore \quad \mathrm{I} & =\frac{1}{2} \int_0^{\infty} \frac{d t}{13+t^2}=\frac{1}{2} \int_0^{\infty} \frac{d t}{t^2+(\sqrt{13})^2}=\frac{1}{(2 \sqrt{13})}\left[\tan ^{-1} \frac{t}{\sqrt{13}}\right]_0^{\infty} \\
& =\frac{1}{2 \sqrt{13}}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)=\frac{1}{2 \sqrt{13}}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4 \sqrt{13}} .
\end{aligned}
$