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

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प्रश्न

Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]
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उत्तर

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

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पाठ 19: Definite Integrals - Exercise 20.5 [पृष्ठ ९६]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.5 | Q 49 | पृष्ठ ९६

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