Exercise 9.4-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: $\int_0^{2 \pi} \frac{\cos x}{\sqrt{4+3 \sin x}} d x$
Solution:
$
I=\int_0^{2 \pi} \frac{\cos x}{\sqrt{4+3 \sin x}} d x
$
Put $4+3 \sin x=t$, so that $3 \cos x d x=d t \Rightarrow \cos x d x=\frac{1}{3} d t$ When $x=0, t=4+3(0)=4$ and when $x=2 \pi, t=4+3(0)=4$
$
\begin{aligned}
\therefore \quad \quad \quad & =\frac{1}{3} \int_4^4 \frac{d t}{t^{\frac{1}{2}}}=\frac{1}{3} \int_4^4 t^{-\frac{1}{2}} d t=\frac{1}{3}\left[\frac{t^{\frac{1}{2}}}{\frac{1}{2}}\right]_4^4 \\
& =\frac{1}{3}[2 \sqrt{t}]_4^4=\frac{1}{3}[2 \sqrt{4}-2 \sqrt{4}]=\frac{1}{3}(0)=0
\end{aligned}
$

Question 2.
Evaluate: $\int_0^{\pi / 4} \frac{\sin ^3 x}{\cos ^5 x} d x$
Solution:
$
\begin{aligned}
I=\int_0^{\pi / 4} \frac{\sin ^3 x}{\cos ^5 x} d x & =\int_0^{\pi / 4} \frac{\sin ^3 x}{\cos ^3 x} \cdot \frac{1}{\cos ^2 x} d x \\
& =\int_0^{\pi / 4} \tan ^3 x \sec ^2 x d x
\end{aligned}
$
Put $\tan x=t$, so that $\sec ^2 x d x=d t$
When $x=0, t=0$ and when $x=\frac{\pi}{4}, t=\tan \frac{\pi}{4}=1$
$
\therefore \quad \mathrm{I}=\int_0^1 t^3 d t=\left[\frac{t^4}{4}\right]_0^1=\frac{1}{4}-0=\frac{1}{4} \text {. }
$

Question 3.
Evaluate: $\int_0^{\pi / 2} \sqrt{\sin \theta} \cos ^5 \theta d \theta$
Solution:
$
\begin{aligned}
I & =\int_0^{\pi / 2} \sqrt{\sin \theta} \cos ^5 \theta d \theta=\int_0^{\pi / 2} \sqrt{\sin \theta} \cos ^4 \theta \cos \theta d \theta \\
& =\int_0^{\pi / 2} \sqrt{\sin \theta}\left(1-\sin ^2 \theta\right)^2 \cos \theta d \theta
\end{aligned}
$
Put $\sin \theta=t$, then $\cos \theta d \theta=d t$
When $\theta=0, t=\sin 0=0$ and when $\theta=\frac{\pi}{2}, t=\sin \frac{\pi}{2}=1$
$
\begin{aligned}
& \therefore \quad \mathrm{I}=\int_0^1 \sqrt{t}\left(1-t^2\right)^2 d t=\int_0^1 t^{\frac{1}{2}}\left(1-2 t^2+t^4\right) d t \\
& =\int_0^1\left(t^{\frac{1}{2}}-2 t^{\frac{5}{2}}+t^{\frac{9}{2}}\right) d t=\left[\frac{t^{\frac{3}{2}}}{\frac{3}{2}}-2 \cdot \frac{t^{\frac{7}{2}}}{\frac{7}{2}}+\frac{t^{\frac{11}{2}}}{\frac{11}{2}}\right]_0^1 \\
& =\left(\frac{2}{3}-\frac{4}{7}+\frac{2}{11}\right)-(0)=\frac{154-132+42}{231}=\frac{64}{231} \text {. } \\
&
\end{aligned}
$

Question 4.
Evaluate: $\int_0^{\pi / 3} \frac{\sec x \tan x}{1+\sec ^2 x} d x$
Solution:
$
I=\int_0^{\pi / 3} \frac{\sec x \tan x}{1+\sec ^2 x} d x
$
Put $\sec x=t$, so that $\sec x \tan x d x=d t$
When $x=0, t=1$ and when $x=\frac{\pi}{3}, t=2$
$
\begin{aligned}
\therefore \quad \mathrm{I} & =\int_0^2 \frac{d t}{1+t^2}=\left[\tan ^{-1} t\right]_1^2 \\
& =\tan ^{-1} 2-\tan ^{-1} 1=\tan ^{-1} 2-\frac{\pi}{4} .
\end{aligned}
$

Question 5.
Evaluate: $\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}$
Solution:
$
\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}
$
Put $\tan \frac{x}{2}=t, \therefore \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t \Rightarrow\left(1+\tan ^2 \frac{x}{2}\right) d x=2 d t \Rightarrow d x=\frac{2 d t}{1+t^2}$
When $x=0, t=0$ and when $x=\frac{x}{2}, t=1$
$
\begin{aligned}
\therefore \quad \mathrm{I} & =\int_0^1 \frac{\frac{2 d t}{1+t^2}}{5+4\left(\frac{2 t}{1+t^2}\right)}=2 \int_0^1 \frac{d t}{5\left(1+t^2\right)+8 t} \\
& =\frac{2}{5} \int_0^1 \frac{d t}{t^2+\frac{8}{5} t+1}=\frac{2}{5} \int_0^1 \frac{d t}{\left(t^2+\frac{8}{5} t+\frac{16}{25}\right)+\frac{9}{25}} \\
& =\frac{2}{5} \int_0^1 \frac{d t}{\left(t+\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2}=\left[\frac{2}{5} \times \frac{1}{\frac{3}{5}} \tan ^{-1} \frac{t+\frac{4}{5}}{\frac{3}{5}}\right]_0^1 \\
& =\frac{2}{3}\left[\tan ^{-1} \frac{5 t+4}{3}\right]_0^1=\frac{2}{3}\left[\tan ^{-1} 3-\tan ^{-1} \frac{4}{3}\right] \\
& =\frac{2}{3} \tan ^{-1} \frac{3-\frac{4}{3}}{1+3\left(\frac{4}{3}\right)}=\frac{2}{3} \tan ^{-1} \frac{5}{15}=\frac{2}{3} \tan ^{-1} \frac{1}{3} .
\end{aligned}
$