# Find the value of the integral (1+2cosx)/(2+cosx)^2 dx from 0 to ?/2.

$I = int_{0}^{pi /2}frac{(1+2cosx)}{(2+cosx)^{2}}dx t = tanfrac{x}{2} dt = frac{1}{2}sec^{2}frac{x}{2}dx sinx = frac{2t}{1+t^{2}}, cosx = frac{1-t^{2}}{1+t^{2}} I = int_{0}^{1} frac{-2(t^{2}-3)}{t^{4}+6t^{2}+9}dt I = -2int_{0}^{1} ((frac{1}{t^{2}+3})-(frac{6}{(t^{2}+3)^{2}}))dt I = (frac{2t}{t^{2}+3})_{0}^{1} = frac{1}{2}$