Question

Find the integrals of the function: $cos^4 2x$

Answer 1

cos4 2x = (cos2 2x)2

$=(\frac{1+cos4x}{2})^2$

=$\frac14$[1+cos2 4x+2cos 4x]

$= (\frac{1}{4}) [1+(\frac{1+cos8x}{2})+2cos4x]$

$= (\frac{1}{4}) [1+\frac{1}{2}+\frac{cos8x}{2}+2cos4x]$

$= (\frac{1}{4}) [ \frac{3}{2}+\frac{cos8x}{2}+2cos4x]$

∴ ∫cos4 2x dx = ∫$(\frac{3}{8}+\frac{cos8x}{8}+\frac{cos4x}{2})dx$

$=\frac{3}{8}x+\frac{sin8x}{64}+\frac{sin4x}{8}+C$