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Prove That: ∫ π 0 X F ( Sin X ) D X = π 2 ∫ π 0 F ( Sin X ) D X - Mathematics

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Question

Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]
Sum
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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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Definite Integrals
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Chapter 20: Definite Integrals - Exercise 20.5 [Page 96]

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RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.5 | Q 49 | Page 96

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