Question

Evaluate:
${\int }_0^{\pi /2}\frac{dx}{(3+2\cos x)}$

Answer 1

Consider the given integral.

$I={\int }_0^{\pi /2}\frac{1}{3+2\cos x}dx$

$I={\int }_0^{\pi /2}\frac{1}{3+2(1-2{\sin }^2\frac{x}{2})}dx$

$I={\int }_0^{\pi /2}\frac{1}{3+2-4{\sin }^2\frac{x}{2}}dx$

$I={\int }_0^{\pi /2}\frac{{\sec }^2\frac{x}{2}}{5{\sec }^2\frac{x}{2}-4{\tan }^2\frac{x}{2}}dx$

$I={\int }_0^{\pi /2}\frac{{\sec }^2\frac{x}{2}}{5({\tan }^2\frac{x}{2}+1)-4{\tan }^2\frac{x}{2}}dx$

$I={\int }_0^{\pi /2}\frac{{\sec }^2\frac{x}{2}}{5+{\tan }^2\frac{x}{2}}dx$

Let $t=\tan \frac{x}{2}$

$\frac{dt}{dx}={\sec }^2\frac{x}{2}\left(\frac{1}{2}\right)$

$2dt={\sec }^2\frac{x}{2}dx$

Therefore,

$I=2{\int }_0^1\frac{1}{{\left(\sqrt{5}\right)}^2+t^2}dt$

$I=2{\left[\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{t}{\sqrt{5}}\right)\right]}_0^1$

$I=2[\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{1}{\sqrt{5}}\right)-\frac{1}{\sqrt{5}}{\tan }^{-1}\left(0\right)]$

$I=2[\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{1}{\sqrt{5}}\right)-0]$

$I=\frac{2}{\sqrt{5}}{\tan }^{-1}\left(\frac{1}{\sqrt{5}}\right)$

Hence, this is the answer.