Evaluate $\int\limits_0^\pi \frac{x}{1+\sin^2x} \ dx$ How can one evaluate 
$$\int_0^\pi \frac{x}{1+\sin^2x} \ dx\ ?$$

let $x = -u +\dfrac{\pi}{2}$, then $\sin^2x = \sin^2(-u +\frac{\pi}{2}) = \cos^2u$, and also $dx = -du$. So:
$I = \displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{\frac{\pi}{2} - u}{1 + \cos^2u}du = \pi\cdot \displaystyle \int_{0}^\frac{\pi}{2} \dfrac{1}{1 + \cos^2u}du$.
At this point you can introduce $t = \tan\left(\dfrac{u}{2}\right)$, and continue the fraction decomposition to finish it.
