Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
add comment

2 Answers

up vote 16 down vote accepted

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.

share|improve this answer
    
Thanks, most appreciated. :) –  Steven Jun 17 '12 at 16:01
add comment

$\int_0^\pi xf(\sin x)d$

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

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

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

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

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

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

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.