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

integration

Solution 1:

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

$$I=\int_0^\pi xf(\sin x)dx=\int_0^\pi(\pi+0-x)f\{\sin(\pi+0-x)\}dx$$

As $\sin(\pi-x)=\sin x,$

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

Please let me know if this is not enough hint

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