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Question
Prove that:
Solution
\[\int_0^\pi xf\left( \sin x \right)dx = \int_0^\pi \left( \pi - x \right)f\left[ \sin\left( \pi - x \right) \right]dx .................\left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \int_0^\pi \left( \pi - x \right)f\left( \sin x \right)dx\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \pi \int_0^\pi f\left( \sin x \right)dx - \int_0^\pi xf\left( \sin x \right)dx\]
\[ \Rightarrow 2 \int_0^\pi xf\left( \sin x \right)dx = \pi \int_0^\pi f\left( \sin x \right)dx\]
\[ \Rightarrow \int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]
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