Em resposta ao Gabriel Sousa
Se $u=\pi -x$ então $x=\pi -u$ e ${x}'=-u$
Por outro lado $x=0\,\,\,\,\Rightarrow \,\,\,u=\pi $ e $x=\pi
\,\,\,\Rightarrow \,\,\,u=0$
O integral fica
$\int\limits_{0}^{\pi }{x\,f(\operatorname{sen}x)dx}=-\int\limits_{\pi
}^{0}{(\pi -u)\,f(\operatorname{sen}(\pi -u))\,du}$
mas $\operatorname{sen}(\pi
-u)=\operatorname{sen}u$ logo
$\int\limits_{0}^{\pi
}{x\,f(\operatorname{sen}x)dx}=\int\limits_{0}^{\pi }{(\pi
-u)\,f(\operatorname{sen}u)\,du}=\pi \int\limits_{0}^{\pi }{f(\operatorname{sen}u)\,du}-\int\limits_{0}^{\pi
}{u\,f(\operatorname{sen}u)\,du}$
$\int\limits_{0}^{\pi }{x\,f(\operatorname{sen}x)dx}=\pi
\int\limits_{0}^{\pi }{f(\operatorname{sen}u)\,du}-\int\limits_{0}^{\pi
}{u\,f(\operatorname{sen}u)\,du}$
Como a variável de qualquer integral
é muda podemos escrever
$\int\limits_{0}^{\pi }{x\,f(\operatorname{sen}x)dx}=\pi
\int\limits_{0}^{\pi }{f(\operatorname{sen}x)\,dx}-\int\limits_{0}^{\pi
}{x\,f(\operatorname{sen}x)\,dx}$
$2\,\int\limits_{0}^{\pi }{x\,f(\operatorname{sen}x)dx}=\pi
\int\limits_{0}^{\pi }{f(\operatorname{sen}x)\,dx}$
$\int\limits_{0}^{\pi }{x\,f(\operatorname{sen}x)dx}=\frac{\pi
}{2}\int\limits_{0}^{\pi }{f(\operatorname{sen}x)\,dx}$ q. e. d.
