calculate $\int_o^\pi \frac{x\sin x}{1+\cos^2x} \,dx$ Given $$I = \int\limits_0^\pi \frac{x\sin x}{1+\cos^2\!x}dx$$


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*prove without calculating $ \frac{\pi}{2} \le I \le \pi $

*calulate $I$
so far I know $I = \int\limits_0^\pi \frac{x\sin x}{1+\cos^2\!x}dx \le \int\limits_0^\pi x\sin x =\pi$
 A: We have $$I = \int\limits_0^\pi \frac{x\sin x}{1+\cos^2\!x}dx\tag 1$$
Using property of definite integral $\int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$, we get 
$$I = \int\limits_0^\pi \frac{(\pi-x)\sin (\pi-x)}{1+\cos^2(\pi-x)}dx$$
$$I = \int\limits_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2 x}dx\tag 2$$
Now, adding (1) & (2), we get $$I+I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^2 x}dx$$
$$I=\frac{\pi}{2}\int_{0}^{\pi}\frac{\sin x}{1+\cos^2 x}dx$$
$$I=-\frac{\pi}{2}\left[\tan^{-1}(\cos x)\right]_{0}^{\pi}$$
$$I=-\frac{\pi}{2}\left[\tan^{-1}(\cos \pi)-\tan^{-1}(\cos 0)\right]$$
$$I=-\frac{\pi}{2}\left[\tan^{-1}(-1)-\tan^{-1}(1)\right]$$
$$I=-\frac{\pi}{2}\left[\frac{-\pi}{4}-\frac{\pi}{4}\right]=\frac{\pi^2}{4}$$
A: $I=\int_0 ^\pi$$\frac{xsinx}{1+cos^2 x}$$dx$....(i)
$\Rightarrow I= \int_0 ^\pi \frac{(\pi-x)sinx}{1+cos^2x} dx$...(ii)
Summing (i) and (ii) we have,
$2I=\pi\int_0 ^\pi \frac{sinx}{1+cos^2x}dx$
$\Rightarrow I=\frac \pi2 \int_0 ^\pi \frac{sinx}{1+cos^2x}dx$
Now assume that $cosx=z$
$\Rightarrow -sinx=dz$
Now replace the variables and the according limits in your integral 
i.e. $I= \frac \pi2 \int_0 ^\pi \frac{-z}{1+z^2}dz$
Then it is nothing difficult to find the result.
