Question

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

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

Answer 1

The correct option is C $\frac{\pi }{3}$

Putting $\cos x=\frac{1-{\tan }^2\left(x/2\right)}{1+{\tan }^2\left(x/2\right)}$,

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

$={\int }_0^{\pi }\frac{\{1+{\tan }^2\left(x/2\right)\}dx}{9+{\tan }^2\left(x/2\right)}$

$={\int }_0^{\pi }\frac{{\sec }^2\left(x/2\right)dx}{{\tan }^2\left(x/2\right)+3^2}$

Let , $z=\tan \left(x/2\right)⟹dz=\frac{1}{2}{\sec }^2\left(x/2\right)dx⟹{\sec }^2\left(x/2\right)dx=2dz$

And when, $x=0,z=0$ and when $x=\pi ,z=\infty$

Hence, integration becomes:-

${\int }_0^{\infty }\frac{2dz}{z^2+3^2}$

$=\frac{2}{3}{\left[{\tan }^{-1}\frac{z}{3}\right]}_0^{\infty }$

$=\frac{2}{3}[{\tan }^{-1}\infty -{\tan }^{-1}0]$

$=\frac{2}{3}(\frac{\pi }{2}-0)$

$=\frac{\pi }{3}$