Question
Find the integral of the function $\cos^4 2x$

Answer

$\cos^42x = (\cos^22x)^2$
$= \left(\frac{1+\cos 4 x}{2}\right)^{2}$
$= \frac{1}{4}\left[1+\cos ^{2} 4 x-2 \cos 4 x\right]$
$= \frac{1}{4}\left[1+\left(\frac{1+\cos 8 x}{2}\right)+2 \cos 4 x\right]$
$= \frac{1}{4}\left[\frac{3}{2}+\frac{1}{2} \cos 8 x+2 \cos 4 x\right]$
Now, $\int \cos ^{4} 2 x d x=\int\left[\frac{3}{8}+\frac{1}{8} \cos 8 x+\frac{1}{2} \cos 4 x\right] d x$
$= \frac{3 x}{8}+\frac{1}{64} \sin 8 x+\frac{1}{8} \sin 4 x+C$
 

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