$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
I tried to solve it.
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac{3}{(2+\cos x)^2}dx$
But i could not solve further.Please help me in completing.

Given $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1+2\cos x}{(2+\cos x)^2}dx.\;,$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\sin^2 x\;,$ we get
$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{\csc^2 x+2\cot x \cdot \csc x}{\left(2\csc x+\cot x\right)^2}dx\;,$ Now Put $\displaystyle \left(2\csc x+\cot x \right) = t\;,$
Then $\left(\csc^2 x+2\cot x \cdot \csc x \right)dx = -dt.$
So Integral Convert into $\displaystyle I = -\int\frac{1}{t^2}dt = \frac{1}{t} = \left[\frac{\sin x}{2+\cos x }\right]_{0}^{\frac{\pi}{2}} = \frac{1}{2}$
