Evaluation of $$\int^{\frac{\pi}{2}}_{0}\frac{x\cos x}{1+\sin^2 x}dx$$
Try: Integration By parts
$$I = x\tan^{-1}(\sin x)\bigg|^{\frac{\pi}{2}}_{0}-\int^{\frac{\pi}{2}}_{0}\tan^{-1}(\sin x)dx$$
$$I = \frac{\pi^2}{8}-\int^{\frac{\pi}{2}}_{0}\tan^{-1}(\sin x)dx$$
Let $$I(a)=\int^{\frac{\pi}{2}}_{0}\tan^{-1}(a\sin x)dx$$
Then $$I'(a) = \int^{\frac{\pi}{2}}_{0}\frac{\sin x}{1+a^2\sin^2 x}dx =\int^{\frac{\pi}{2}}_{0}\frac{\sin x}{(1+a^2)-a^2\cos^2 x}dx$$
$$I'(a) = -\frac{1}{2a\sqrt{1+a^2}}\ln\bigg|\frac{\sqrt{1+a^2}-a}{\sqrt{1+a^2}+a}\bigg|$$
How can i solve it after that , Could some help me , Thanks
