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This question already has an answer here:

To prove that $\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx$ is true, first I started calculating the integral of the left indefinitely $$ \int {x\,f(\sin x)\,\,dx} $$ using substitution: $$ \sin x = t,\quad x = \arcsin t, \quad {dx = \frac{{dt}}{{\sqrt {1 - {t^2}} }}}$$ is obtained: $$ \int {x\,f(\sin x)\,\,dx} = \int {\arcsin t \cdot f(t) \cdot \frac{{dt}}{{\sqrt {1 - {t^2}} }}}$$ $$ \qquad\quad = \int {\frac{{\arcsin t\,dt}}{{\sqrt {1 - {t^2}} }} \cdot f(t)} $$ Then using integration by parts: $$ \begin{array}{*{20}{c}} {u = f(t)},&{dv = \frac{{\arcsin t\,dt}}{{\sqrt {1 - {t^2}} }}} \\ {du = f'(t)\,dt},&{v = \frac{{{{(\arcsin t)}^2}}}{2}} \end{array} $$ then: \begin{align*} \int {x\,f(\sin x)\,\,dx} &= f(t) \cdot \frac{{{{(\arcsin t)}^2}}}{2} - \int {\frac{{{{(\arcsin t)}^2}}}{2}} \cdot f'(t)\,dt \\ &= f(\sin x) \cdot \frac{{{x^2}}}{2} - \int {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx} \\ \end{align*} Now, evaluating from 0 to $\pi$ \begin{align} \int_0^\pi {x\,f(\sin x)} \,dx & = \left[ {f(t) \cdot \frac{{{x^2}}}{2}} \right]_0^\pi - \int_0^\pi {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx} \\ \int_0^\pi {x\,f(\sin x)} \,dx & = f(0) \cdot \frac{{{\pi ^2}}}{2} - \int_0^\pi {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx} \qquad ..[1] \\ \end{align} On the other hand, doing the same process with the integral on the right side I get: \begin{equation}\int_0^\pi {f(\sin x)} \,dx = f(0) \cdot \pi - \int_0^\pi {x \cdot f'(\sin x)\,\cos x\,dx} \qquad ..[2] \end{equation} And even here I do not have enough data to say that equality $\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx$ is true.

Can anyone suggest me what to do with the equalities [1] and [2]?

Thanks in advance.

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marked as duplicate by Hans Lundmark, user98602, Davide Giraudo real-analysis Jan 7 '15 at 8:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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You can maybe use the substitution $t=\pi-x$. Then you'll get $$ \int_0^\pi x f(\sin x)\, dx = \int_0^\pi (\pi -t)f(\sin (\pi-t))\, dt. $$ Can you finish the argument?

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  • $\begingroup$ I integrate by parts the integral $\int_0^\pi (\pi -t)f(\sin (\pi-t))\, dt$? $\endgroup$ – mathsalomon Jan 6 '15 at 11:05
  • $\begingroup$ No, it is easier than that. Use that $\sin(\pi-t)=\sin t$ and split the integral into two by multiplying in the parenthesis $(\pi -t)$. Do you recognize the two integrals? $\endgroup$ – mickep Jan 6 '15 at 11:06
  • $\begingroup$ I get it, thanks, your answer I found very useful. $\endgroup$ – mathsalomon Jan 6 '15 at 11:09
  • $\begingroup$ To you who wanted a $-dt$ instead of $dt$: It should either be $\int_0^\pi (\pi-t)f(\sin(\pi-t))\,dt$ or $\int_\pi^0 (\pi-t)f(\sin(\pi-t))\,(-dt)$. I chose the first. $\endgroup$ – mickep Jan 7 '15 at 7:30
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In general $$\int^b_a f(x) dx\,=\int^b_a f(b+a-x)dx\,$$

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