<span class="math-tex">\(\rm\int_0^\pi x\ f(\sin x)\ dx\)</span> is equal to:
Let \(I = \int_0^\pi x f(\sin x)\,dx\). Replace \(x\) by \(\pi - x\), noting \(\sin(\pi - x) = \sin x\):
\(I = \int_0^\pi (\pi - x) f(\sin x)\,dx\). Adding the two forms: \(2I = \pi\int_0^\pi f(\sin x)\,dx\).
\(\sin x\) is symmetric about \(\frac{\pi}{2}\), so \(\int_0^\pi f(\sin x)\,dx = 2\int_0^{\pi/2} f(\sin x)\,dx\).
\(I = \pi\int_0^{\pi/2} f(\sin x)\,dx\)
Answer: (A)
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