Question

Solve:-
$\int {\cos }^{4}2x$

Answer 1

We have

${\cos }^{4}2x=(co{s}^{2}2x{)}^2=(\frac{1+cos4x}{2}{)}^2=\frac{1}{4}(1+2\cos 4x+co{s}^{2}4x)$

$=\frac{1}{4}(1+2\cos 4x+\frac{1+\cos 8x}{2})=\frac{1}{4}(\frac{3}{2}+2\cos 4x+\frac{1}{2}\cos 8x)$

$Then\int \frac{1}{4}(\frac{3}{2}+2\cos 4x+\frac{1}{2}\cos 8x)dx=\frac{1}{4}(\frac{3}{2}x+\frac{1}{2}sin4x+\frac{1}{16}cos8x)+C$