Integral finding - $u$-substitution

Question

I have a continuous function $f:[-1,1]\rightarrow\mathbb{R}$ and I want to prove that: $$\int_0^\pi xf(\sin x) \, \text{d}x = \frac \pi 2 \int_0^\pi{f(\sin x) \, \text{d}x} = \pi\int_0^{\frac{\pi}{2}} f(\sin x) \, \text{d}x$$ Using the substitution $\pi-x=u$ I proved easily that: $$\int_0^\pi xf(\sin x ) \, \text{d}x = \frac \pi 2 \int_0^\pi f(\sin x) \, \text{d}x$$ But I don't know what to do in order to prove the other part. Any ideas?

Answer 1

use the property that if $f(x)=f(a+b-x)$ where a and b are the limits then $\int _a^b f(x) \, dx = 2\int_a^{\frac{b}{2}} f(x) \, dx$. Forgot to add here $a=0$ is generally the case. The proof can now be as follows $\int _0^b f (x)dx=\int _0^{\frac {b}{2}}f (x)dx+\int _{\frac {b}{2}}^b f (x)dx$. Now for second integral put $x=b-u $ and from that conclude the proof.