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Good day. Suppose we're given that real-valued function $f$ is continuous on segment $[0;1]$. I've come across the following formula which i can't prove: $\int_{0}^{\pi}xf(\sin x)dx = \frac{\pi}{2}\int_{0}^{\pi}f(\sin x)dx$

I tried to use integration by parts but failed since we know nothing about the antiderrivative of $f(\sin(x))$. Now there are no ideas so any tips will be highly appreciated. Thank you.

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Write $$\int_0^\pi x f(\sin x) dx = \int_0^{\pi/2} x f(\sin x) dx + \int_{\pi/2}^\pi x f(\sin x) dx.$$

Now use a substitution $u=\pi-x$ in the second integral to see

$$\int_{\pi/2}^\pi x f(\sin x) dx = \int_0^{\pi/2} (\pi-x)f(\sin (\pi-x)) dx.$$

So we have

$$\int_0^\pi x f(\sin x) dx=\pi\int_0^{\pi/2} f(\sin (\pi-x)) dx=\frac\pi2 \int_0^\pi f(\sin x) dx.$$

In the last equality we used $\sin(\pi-x)=\sin(x)$ and $\int_0^{\pi/2} f(\sin(x)) dx=\int_{\pi/2}^\pi f(\sin(x)) dx$.

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$$\int_0^\pi(x-\frac\pi2)f(\sin x)\,dx=\int_{-\pi/2}^{\pi/2}xf(\cos x)\,dx=0$$ as the integrand is an odd function.

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