Evaluate
$\displaystyle I = \int_{0}^{\pi} \frac{x sin x}{1+cos^2 x} dx$
Hello,
Let $\displaystyle t=\pi-x$ :
$\displaystyle I=-\int_\pi^0 \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt=\int_0^\pi \frac{(\pi-t) \sin(\pi-t)}{1+\cos^2(\pi-t)} ~ dt$
But $\displaystyle \sin(\pi-t)=\sin(t)$ and $\displaystyle \cos(\pi-t)=-\cos(t)$
Hence :
$\displaystyle I=\int_0^\pi \frac{(\pi-t) \sin(t)}{1+\cos^2(t)}~dt$
$\displaystyle I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~dt-\underbrace{\int_0^\pi \frac{t \sin(t)}{1+\cos^2(t)} ~ dt}_{=I}$
$\displaystyle 2I=\pi \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt$
$\displaystyle I=\frac{\pi}{2} \int_0^\pi \frac{\sin(t)}{1+\cos^2(t)} ~ dt$
Now substitute $\displaystyle \varphi=\cos(t)$ and it'll be easy.
The straight dashed line diff's / anti-diff's with respect to the dashed balloon expression.
Don't integrate - balloontegrate! http://www.ballooncalculus.org/examples]