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प्रश्न
उत्तर
\[Let\ I = \int_0^\frac{\pi}{4} \sin^3 2t\ \cos 2t\ d\ t . Then, \]
\[Let\ \sin 2t = u . Then, 2 \cos\ 2t\ dt = du\]
\[When\ t = 0, u = 0\ and\ t\ = \frac{\pi}{4}, u = 1\]
\[ \therefore I = \frac{1}{2} \int_0^1 u^3 du\]
\[ \Rightarrow I = \frac{1}{2} \left[ \frac{u^4}{4} \right]_0^1 \]
\[ \Rightarrow I = \frac{1}{2}\left( \frac{1}{4} - 0 \right)\]
\[ \Rightarrow I = \frac{1}{8}\]
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