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$\int_0^\pi x f(\sin x) d x$ is equal to
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$\pi \int_0^{\frac{\pi}{2}} f(\sin x) d x$
$\begin{aligned} & \text { Let } I=\int_0^\pi x f(\sin x) d x \\ & \Rightarrow \quad I=\int_0^\pi(\pi-x) f[\sin (\pi-x)] d x\end{aligned}$
$\begin{aligned} & \Rightarrow \quad I=\int_0^\pi \pi f(\sin x) d x-\int_0^\pi x f(\sin x) d x \\ & \Rightarrow \quad 2 I=\pi \int_0^\pi f(\sin x) d x \Rightarrow 2 I=\pi \int_0^{\left(\frac{\pi}{2}\right) \times 2} f(\sin x) d x \\ & \Rightarrow \quad 2 I=2 \pi \int_0^{\pi / 2} f(\sin x) d x \\ & \quad \quad\left[\because \text { if } f(2 a-x)=f(x), \operatorname{then} \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right] \\ & \Rightarrow \quad I=\pi \int_0^{\pi / 2} f(\sin x) d x\end{aligned}$
$\begin{aligned} & \Rightarrow \quad I=\int_0^\pi \pi f(\sin x) d x-\int_0^\pi x f(\sin x) d x \\ & \Rightarrow \quad 2 I=\pi \int_0^\pi f(\sin x) d x \Rightarrow 2 I=\pi \int_0^{\left(\frac{\pi}{2}\right) \times 2} f(\sin x) d x \\ & \Rightarrow \quad 2 I=2 \pi \int_0^{\pi / 2} f(\sin x) d x \\ & \quad \quad\left[\because \text { if } f(2 a-x)=f(x), \operatorname{then} \int_0^{2 a} f(x) d x=2 \int_0^a f(x) d x\right] \\ & \Rightarrow \quad I=\pi \int_0^{\pi / 2} f(\sin x) d x\end{aligned}$
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