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प्रश्न
उत्तर
Consider
\[f\left( \pi - x \right) = \cos\left( \pi - x \right)\left| \cos\left( \pi - x \right) \right| = - \cos x\left| - \cos x \right| = - \cos x\left| \cos x \right| = - f\left( x \right)\]
\[\therefore \int_0^\pi \cos x\left| \cos x \right|dx = 0 ..................\left[ \int_0^{2a} f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( 2a - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( 2a - x \right) = - f\left( x \right)\end{cases} \right]\]
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