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प्रश्न
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \sin x \sin 2x\ dx\ . Then, \]
\[I = \int_0^\frac{\pi}{2} 2 \sin^2 x \cos\ x\ dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} 2\left( 1 - \cos^2 x \right) \cos\ x\ dx\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 2 \cos x - 2 \cos^3 x \right) dx\]
\[ \Rightarrow I = \left[ 2\sin x - 2\left( \sin x - \frac{\sin^3 x}{3} \right) \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \left[ 2 - 2\left( 1 - \frac{1}{3} \right) \right] - 0\]
\[ \Rightarrow I = \frac{2}{3}\]
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