Question
Find the integral of the function $\sin^4 x$

Answer

We can write, $\sin^4x = \sin^2x \sin^2x$
$= \left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1-\cos 2 x}{2}\right)$
$=\frac{1}{4}(1-\cos 2 x)^{2}$
$= \frac{1}{4}\left[1+\cos ^{2} 2 x-2 \cos 2 x\right]$
$= \frac{1}{4}\left[1+\left(\frac{1+\cos 4 x}{2}\right)-2 \cos 2 x\right]$
$= \frac{1}{4}\left[1+\frac{1}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$
$= \frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right]$
Now, $\int \sin ^{4} x d x=\frac{1}{4} \int\left[\frac{3}{2}+\frac{1}{2} \cos 4 x-2 \cos 2 x\right] d x$
$= \frac{1}{4}\left[\frac{3}{2} x+\frac{1}{2}\left(\frac{\sin 4 x}{4}\right)-\frac{2 \sin 2 x}{2}\right]+C$
$= \frac{1}{8}\left[3 \mathrm{x}+\left(\frac{\sin 4 \mathrm{x}}{4}\right)-2 \sin 2 \mathrm{x}\right]+\mathrm{C}$
$= \frac{3 \mathrm{x}}{8}-\frac{1}{4} \sin 2 \mathrm{x}+\frac{1}{32} \sin 4 \mathrm{x}+\mathrm{C}$

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