$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
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.
 A: You may observe that: $$\frac{2 \cos (x)+1}{(\cos (x)+2)^2}=\frac{\cos (x)}{\cos (x)+2}+\frac{\sin ^2(x)}{(\cos (x)+2)^2}=\frac{\frac{d}{dx}\sin(x)}{\cos (x)+2}-\frac{\sin(x)\frac{d}{dx}(\cos x +2)}{(\cos (x)+2)^2}$$
A: 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 $\displaystyle I = -\int\frac{1}{t^2}dt = \frac{1}{t} = \left[\frac{\sin x}{2+\cos x }\right]_{0}^{\frac{\pi}{2}} = \frac{1}{2}$
A: Considering $$I=\int\frac{1+2\cos (x)}{(2+\cos (x))^2}\,dx$$ you can do several things.
First, considering the squared denominator, you could suppose that the result of integration would be something like $$\frac {a+b\sin(x)+c\cos(x)}{2+\cos(x)}$$ which, one differentiated, would give $$\frac{(a-2 c) \sin (x)+2 b \cos (x)+b}{(2+\cos (x))^2}$$ So, $$\frac{(a-2 c) \sin (x)+2 b \cos (x)+b}{(2+\cos (x))^2}=\frac{1+2\cos (x)}{(2+\cos (x))^2}$$ Removing the denominator let us with  $$(a-2 c) \sin (x)+2 (b-1) \cos (x)+b-1=0$$ So, $b=1$, $a-2c=0$ makes $$I=\frac{\sin(x)}{2+\cos (x)}$$
The second way would use the tangent half-angle substitution $t=\tan(\frac x2)$ which would lead to $$I=\int\frac{1+2\cos (x)}{(2+\cos (x))^2}\,dx=-\int\frac{2 \left(t^2-3\right)}{\left(t^2+3\right)^2}\,dt$$ which is not very difficult to integrate.
I am sure that you can takle from here.
