Question

$I={\int }_0^{\pi /2}\frac{1+2cosx}{(2+cosx{)}^2}$dx

Answer 1

Now,

$\begin{array}{c}I={\int }_0^{\frac{\pi }{2}}\frac{1+2\cos x}{{\left(2+\cos x\right)}^2}dx\\ I={\int }_0^{\frac{\pi }{2}}\frac{\cos e{c}^{2}x+2\cot s\cos ecx}{{\left(2\cos ecx+\cot x\right)}^2}dx\\ t=2\cos ecx+cotx\\ \frac{dt}{dx}=-2\cos ecx\cot x-\cos e{c}^{2}x\\ \int \frac{-dt}{t^2}=\frac{1}{t}=\frac{1}{2\cos ecx+\cot x}\\ ={\left[\frac{\sin x}{2+\cos x}\right]}_0^{\frac{\pi }{2}}\\ =\frac{1}{2}\end{array}$