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Question
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Solution
\[\int \sin^4 \text{ 2x dx }\]
\[ \Rightarrow \int \left( \sin^2 2x \right)^2 dx\]
\[ \Rightarrow \int \left[ \frac{1 - \cos 4x}{2} \right]^2 dx\]
\[ \Rightarrow \frac{1}{4}\int \left( 1 - \cos 4x \right)^2 \]
\[ \Rightarrow \frac{1}{4}\int\left( 1 + \cos^2 4x - 2 \cos 4x \right)dx\]
\[ \Rightarrow \frac{1}{4}\int\left[ 1 + \left( \frac{1 + \cos 8x}{2} \right) - 2 \cos 4x \right]dx\]
\[ \Rightarrow \frac{1}{4}\int\left[ \frac{3}{2} + \frac{\cos 8x}{2} - 2 \cos 4x \right]dx\]
\[ \Rightarrow \frac{1}{4}\left[ \frac{3x}{2} + \frac{\sin 8x}{16} - \frac{2 \sin 4x}{4} \right] + C\]
\[ \Rightarrow \frac{3x}{8} + \frac{\sin 8x}{64} - \frac{\sin 4x}{8} + C\]
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