Question

Integrate

$\int {\sin }^{4}x.{\cos }^{4}xdx$

Answer 1

${\sin }^{4}x{\cos }^{4}x=\frac{1}{16}{\sin }^{4}2x=\frac{1}{64}(1-\cos 4x{)}^2=\frac{1}{16}+\frac{1}{16}{\cos }^{2}4x-\frac{1}{32}\cos 4x=\frac{3}{32}+\frac{1}{32}\cos 8x-\frac{1}{32}\cos 4x$

$\int {\sin }^{4}x{\cos }^{4}xdx=\int \frac{3dx}{32}+\frac{1}{32}\int \cos 8xdx-\frac{1}{32}\int \cos 4xdx=\frac{3x}{32}+\frac{1}{256}\sin 8x-\frac{1}{128}\sin 4x+c$