0

How can I prove the following claim for any given continues function:

$$ \int_{0}^{\pi} xf(\sin(x))dx = \pi \int_{0}^{\frac{\pi}{2}}f(\sin(x))dx $$

Thanks!

3SAT
  • 7,627
superuser123
  • 929
  • See also : http://math.stackexchange.com/questions/159381/evaluate-int-0-pi-xf-sin-xdx – lab bhattacharjee Jul 12 '16 at 09:39
  • @OlivierOloa, I didn't change the Question. What title u r talking about? – lab bhattacharjee Jul 12 '16 at 09:42
  • The duplicate only gives the first part of the answer, right? – Olivier Oloa Jul 12 '16 at 09:43
  • 1
    @lab bhattacharjee: IMO this is no duplicate because the RHS reads $\pi \int_{0}^{\frac{\pi}{2}}$ and not $\frac{\pi}{2} \int_{0}^{\pi}$ – gammatester Jul 12 '16 at 09:49
  • 1
    Please add context to the post: what is the background of the question? Where did you encounter it? Is it useful for solving other mathematical problems? Posts that contain nothing but a problem statement are discouraged, and are often put on hold for improvement. – Carl Mummert Jul 12 '16 at 11:19

2 Answers2

1

$$2\int_0 ^ {\frac{\pi}{2}} f(\sin x)dx = \int_0 ^{\pi} f(\sin x)dx $$ Let $$I = \int_0^ {\pi} \left(xf(\sin x) - \frac{\pi}{2} f(\sin x)\right)dx $$ $$I = \int_0 ^{\frac{\pi}{2}} (x-\frac{\pi}{2}) f(\sin x)dx$$ Using the substitution $u = \pi - x $ and the fact that $\sin(\pi-x) =\sin x$ gives $I = -I$ Thus $I =0$ and the result follows.

Behrouz Maleki
  • 11,342
Burrrrb
  • 1,566
0

$$ \int_{0}^{\pi} xf(\sin(x))dx = \int_{0}^{\pi/2} xf(\sin(x))dx +\int_{\pi/2}^{\pi} xf(\sin(x))dx \\ = \int_{0}^{\pi/2} xf(\sin(x))dx +\int_{0}^{\pi/2} (\pi-x)f(\sin(x))dx . $$