Question

Evaluate: $ \int_{0}^{\pi} \frac{1}{5+4 \cos x} d x $

• A. $\frac{\pi}{3}$
• B. $\frac{\pi}{2}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (A)

We have, $I=\int_{0}^{\pi} \frac{1}{5+4 \cos x} d x$
$
=\int_{0}^{\pi} \frac{1}{5+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x
$
$
\begin{array}{l}
=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} d x \\
=\int_{0}^{\pi} \frac{1+\tan ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} d x=\int_{0}^{\pi} \frac{\sec ^{2} \frac{x}{2}}{9+\tan ^{2} \frac{x}{2}} d x
\end{array}
$
Let $\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^{2} \frac{x}{2} d x=d t$
Also, $x=0 \Rightarrow t=0$ and $x=\pi \Rightarrow t=\infty$
$\therefore I=\int_{0}^{\infty} \frac{d t}{9+t^{2}}$
$\therefore I=2 \int_{0}^{\infty} \frac{d t}{3^{2}+t^{2}}$
$\therefore I=\frac{2}{3}\left[\tan ^{-1} \frac{t}{3}\right]_{0}^{\infty}=\frac{2}{3}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right]$
$=\frac{2}{3}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{3}$