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