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$\int_0^\pi x f(\sin x) d x$ is equal to
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The correct answer is:
$\pi \int_0^{\pi / 2} f(\cos x) d x$
$\pi \int_0^{\pi / 2} f(\cos x) d x$
$I=\int_0^\pi x f(\sin x) d x=\int_0^\pi(\pi-x) f(\sin x) d x$
$=\pi \int_0^\pi f(\sin x) d x-1$
$2 I=\pi \int_0^\pi f(\sin x) d x$
$I=\frac{\pi}{2} \int_0^\pi f(\sin x) d x=\pi \int_0^{\pi / 2} f(\sin x) d x$
$=\pi \int_0^{\pi / 2} f(\cos x) d x$
$=\pi \int_0^\pi f(\sin x) d x-1$
$2 I=\pi \int_0^\pi f(\sin x) d x$
$I=\frac{\pi}{2} \int_0^\pi f(\sin x) d x=\pi \int_0^{\pi / 2} f(\sin x) d x$
$=\pi \int_0^{\pi / 2} f(\cos x) d x$
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