Question

${\int }_0^{\frac{\pi }{2}}\frac{1+2\cos x}{{(2+\cos x)}^2}dx=k$ then, the value of $6k$ is

Answer 1

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

$\Rightarrow {\int }_0^{\frac{\pi }{2}}\frac{(3-{\tan }^2\frac{x}{2})(1+{\tan }^2\frac{x}{2})}{{(3+{\tan }^2\frac{x}{2})}^2}dx$

Put $\tan \frac{x}{2}=t$
$\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$

$\Rightarrow {\int }_0^1\frac{2(3-t^2)}{{(3+t^2)}^2}dt=2{\int }_0^1\frac{\frac{3}{t^2}-1}{{(\frac{3}{t}+t)}^2}dt$

Put $t+\frac{3}{t}=z$
$\Rightarrow (1-\frac{3}{t^2})dt=dz$

$\Rightarrow -2{\int }_{\infty }^4\frac{dz}{z^2}={\left[\frac{2}{z}\right]}_{\infty }^4=\frac{1}{2}=k$

$\Rightarrow 6k=3$