Prove
Question:

$\sin ^{4} x$

Solution:

$\sin ^{4} x=\sin ^{2} x \sin ^{2} x$

$=\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)$

$=\frac{1}{4}(1-\cos 2 x)^{2}$

$=\frac{1}{4}\left[1+\cos ^{2} 2 x-2 \cos 2 x\right]$

$=\frac{1}{4}\left[1+\left(\frac{1+\cos 4 x}{2}\right)-2 \cos 2 x\right]$

$=\frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$

$=\frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$

$\therefore \int \sin ^{4} x d x=\frac{1}{4} \int\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right] d x$

$=\frac{1}{4}\left[\frac{3}{2} x+\frac{1}{2}\left(\frac{\sin 4 x}{4}\right)-\frac{2 \sin 2 x}{2}\right]+\mathrm{C}$

$=\frac{1}{8}\left[3 x+\frac{\sin 4 x}{4}-2 \sin 2 x\right]+\mathrm{C}$

$=\frac{3 x}{8}-\frac{1}{4} \sin 2 x+\frac{1}{32} \sin 4 x+\mathrm{C}$

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