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I need to prove this identity:

$$ \int _0^{\pi }xf\left(\sin\left(x\right)\right)dx=\:\frac{\pi }{2}\int _0^{\pi }\:f\left(\sin\left(x\right)\right)dx. $$

This is what I tried:

I called I:$ \int _0^{\pi }xf\left(\sin\left(x\right)\right)dx$ and so with this change of variable $$ x=\pi-t \rightarrow t=\pi-x$$ $$ I=\int _0^{\pi }\left(\pi -t\right)f\left(\sin\left(\pi -t\right)\right)d(\pi-t)=-\int _0^{\pi }\left(\pi -t\right)f\left(\sin\left(\pi -t\right)\right)d(t)$$

I know that $\sin(\pi-t)=\sin(t)$ and so $$I=-\int _0^{\pi }\left(\pi -t\right)f\left(\sin\left(t\right)\right)d(t)=-\int _0^{\pi }\pi f\left(\sin\left(t\right)\right)d(t)-t f\left(\sin\left(t\right)\right)d(t)$$ $$I=\int _0^{\pi} t f\left(\sin\left(t\right)\right)d(t)-\int _0^{\pi} \pi f\left(\sin\left(t\right)\right)d(t)$$

At this point I don't know what to do. Can you explain how to solve this and/or where are my mistakes? Thanks in advance.

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    @JeanMarie I don't think so because the OP is wondering where they went wrong not what the answer is – 2017-01-18
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    You are right. I delete the verb "duplicate" ! – 2017-01-18
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    A certain connection with (math.stackexchange.com/q/1856872) – 2017-01-18

2 Answers 2

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Your mistake is immediately after the substitution. With the substitution you get $-\int_{\pi}^{0}(\pi-t)f(\sin t)\text{d}t$ from there you just add $I$ to itself and you get some things that cancel nicely

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    So can I write $$- \int_{\pi}^{0} (\pi-t)f(sin(t))dt = \int_0^{\pi} (\pi-t)f(sin(t))dt$$ right? – 2017-01-18
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    Exactly, and then you can add them and prove your statement – 2017-01-18
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    Oh, yeah now I see it. Many thanks! – 2017-01-18
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Hint:

Use $$I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$$

$$I+I=?$$

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    So $$2I= \int_0^{\pi} \pi f(sin(x))dx$$ $$ 2I= \pi \int_0^{\pi} f(sin(x)dx$$ $$ I= \frac{\pi}{2} \int_0^{\pi} f(sin(x)dx$$ – 2017-01-18