# Integration of sin^4 2x

$\\ \text{Use the trigonometric identities:}\\ \sin ^{2} \alpha=\frac{1-\cos 2 \alpha}{2} \\ \cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2} \\ Now: \\ \sin ^{4}(2 x)=\left(\sin ^{2}(2 x)\right)^{2}=\frac{(1-\cos (4 x))^{2}}{4} \\ Or, \sin ^{4}(2 x)=\frac{1-2 \cos (4 x)+\cos ^{2}(4 x)}{4} \\ Or, \sin ^{4}(2 x)=\frac{1}{4}-\frac{\cos (4 x)}{2}+\frac{1}{8}+\frac{\cos (8 x)}{8} \\ Or, \sin ^{4}(2 x)=\frac{3}{8}-\frac{\cos (4 x)}{2}+\frac{\cos (8 x)}{8}$

\\ Now,\\ \begin{aligned} \int \sin ^{4}(2 x) d x &=\frac{3}{8} \int d x-\frac{1}{2} \int \cos (4 x) d x+\frac{1}{8} \int \cos (8 x) d x \\ \\ Or, \int \sin ^{4}(2 x) d x &=\frac{3 x}{8}-\frac{1}{8} \int \cos (4 x) d(4 x)+\frac{1}{64} \int \cos (8 x) d(8 x) \\ \\ Or, \int \sin ^{4}(2 x) d x &=\frac{3 x}{8}-\frac{\sin (4 x)}{8}+\frac{\sin (8 x)}{64}+C \end{aligned}

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