Question

The value of ${\int }_0^{\pi }\frac{dx}{5+3\cos x}$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{8}$
• C. $\frac{\pi }{2}$
• D. Zero

Answer 1

The correct option is A $\frac{\pi }{4}$

Consider, $\frac{1}{5+3\cos x}$

$=\frac{1}{5+\frac{3-3{\tan }^2\left(\frac{x}{2}\right)}{1+{\tan }^2\left(\frac{x}{2}\right)}}$ ...... $[\because \cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}]$

$=\frac{1+{\tan }^2\frac{x}{2}}{5+5{\tan }^2\frac{x}{2}+3-3{\tan }^2\frac{x}{2}}$

$=\frac{{\sec }^2\frac{x}{2}}{8+2{\tan }^2\frac{x}{2}}$

Now, ${\int }_0^{\pi }\frac{1}{5+3\cos x}$

$={\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}}{8+2{\tan }^2\frac{x}{2}}dx$

Take $\tan \frac{x}{2}=t$$⟹\frac{1}{2}.{\sec }^2\frac{x}{2}dx=dt$

Then, we get

$\int \frac{dt}{4+t^2}dx$

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

$=\frac{1}{2}{\left[{\tan }^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)}{2}\right)\right]}_0^{\pi }$

$=\frac{1}{2}\left(\frac{\pi }{2}\right)$

$=\frac{\pi }{4}$