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

My progress: nothing cause I don't know where start.

I accept hints, (explicits). And the exercise is from Apostol's book. Ahh... The chapter I'm reading is about "integration by substitution", the book gives this Hint: $u=\pi-x$.

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  • $\begingroup$ Well what do you get if you perform the substitution? $\endgroup$ – TheOscillator Oct 7 '15 at 22:19
  • $\begingroup$ $sen(\pi-x)= sen(x)$ ? $-\int_{-\pi}^{0} (\pi-u)f(\sin(u))\,du$ $\endgroup$ – Elll Oct 7 '15 at 22:24
  • $\begingroup$ I understand ... but why my question was edited, if it looks equals as before? $\endgroup$ – Elll Oct 8 '15 at 1:09
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If you accept the book's hint, you have \begin{align} \int_0^{\pi}xf(\sin(x))\,dx &=\int_{\pi}^0(\pi-u)f(\sin(\pi-u))\cdot(-1)\,du\\ &=\int_{0}^\pi(\pi-u)f(\sin(u))\,du\\ &=\int_{0}^\pi\pi f(\sin(u))\,du-\int_{0}^\pi uf(\sin(u))\,du\\ &=\pi\int_{0}^\pi f(\sin(u))\,du-\int_{0}^\pi uf(\sin(u))\,du\\ \end{align} Now set $$ I=\int_0^{\pi}xf(\sin(x))\,dx $$ and…

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The key to solving this is to notice that since $\sin x$ is symmetric about $x=\pi/2$, then so is $ f(\sin x)$. This means that $$\int_0^\pi xf(\sin x) dx=\int_0^\pi (\pi-x)f(\sin (\pi-x)) dx=\int_0^\pi (\pi-x)f(\sin x) dx,$$ after which you can proceed as in the other answers.

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If you make the substitution that the book suggests, you have $x = \pi-u$ and $dx = -du$, so the integral becomes \begin{align} \int_{0}^{\pi} x\,f(\sin(x))\,dx &= -\int_{\pi}^0 (\pi-u)f(\sin(\pi-u))\,du \\ &= \int_0^{\pi} (\pi-u)f(\sin u)\,du \\ &= \pi\int_0^{\pi} f(\sin u)\,du - \int_0^{\pi} u\,f(\sin u)\,du. \end{align} Collect terms and simplify.

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