Question

Evaluate: ${\int }_0^{\pi }\frac{1}{5+4\cos x}dx$

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

Answer 1

Answer: (A)

We have, $I={\int }_0^{\pi }\frac{1}{5+4\cos x}dx$
$<br/><br/>={\int }_0^{\pi }\frac{1}{5+4\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}dx<br/><br/>$
$<br/><br/>\begin{array}{l}<br/><br/>={\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{5\left(1+{\tan }^2\frac{x}{2}\right)+4\left(1-{\tan }^2\frac{x}{2}\right)}dx\\ <br/><br/>={\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{9+{\tan }^2\frac{x}{2}}dx={\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}}{9+{\tan }^2\frac{x}{2}}dx<br/><br/>\end{array}<br/><br/>$
Let $\tan \frac{x}{2}=t\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$
Also, $x=0\Rightarrow t=0$ and $x=\pi \Rightarrow t=\infty$
$\therefore I={\int }_0^{\infty }\frac{dt}{9+t^2}$
$\therefore I=2{\int }_0^{\infty }\frac{dt}{3^2+t^2}$
$\therefore I=\frac{2}{3}{\left[{\tan }^{-1}\frac{t}{3}\right]}_0^{\infty }=\frac{2}{3}\left[{\tan }^{-1}\infty -{\tan }^{-1}0\right]$
$=\frac{2}{3}\left(\frac{\pi }{2}-0\right)=\frac{\pi }{3}$