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Let $f(\sin x)$ be a given function of $\sin x$.

How would I show that $\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx$?

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If you make the substitution $w = \pi-x$, so that $dw = -dx$, you get $$\int_0^\pi xf(\sin x)dx=-\int_\pi^0 (\pi-w)f(\sin(\pi-w))dw$$ $$ = \int_0^\pi (\pi-x)f(\sin(x))dx$$ $$ = \pi\int_0^\pi f(\sin(x))dx - \int_0^\pi xf(\sin(x))dx$$ which gives the result you want.

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  • $\begingroup$ Thanks, most appreciated. :) $\endgroup$ – Steven Jun 17 '12 at 16:01
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$\int_0^\pi xf(\sin x)d$

=$\int_0^\pi (\pi-x)f(\sin (\pi - x))dx$

= $\int_0^\pi (\pi-x)f(\sin x)dx$

= $\pi\int_0^\pi f(\sin x)dx$ - $\int_0^\pi xf(\sin x)dx$

$2\int_0^\pi xf(\sin x)d$ = $\pi\int_0^\pi f(\sin x)dx$

$\int_0^\pi xf(\sin x)d$ = $\frac{\pi}{2}\int_0^\pi f(\sin x)dx$

I am using this formula, $\int_a^b f(x)dx$ =$\int_a^b f(a+b-x)dx$

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  • $\begingroup$ Its ok, thanks for your help! $\endgroup$ – Steven Jun 17 '12 at 16:01
  • $\begingroup$ there should be $\frac{\pi}{2}\int_0^\pi f(\sin x)dx$ instead of $\frac{\pi}{2}\int_0^\pi \sin xdx$. $\endgroup$ – Aang Jul 12 '12 at 8:55
  • $\begingroup$ @avatar: Thanks $\endgroup$ – Prasad G Jul 12 '12 at 8:59

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