Question

Evaluate:

$\int si{n}^{4}xco{s}^{4}xdx$

Answer 1

$I=\int {\sin }^{4}x.{\cos }^{4}x.dx$

We know that

${\sin }^{2}x=\frac{1-\cos 2x}{2},{\cos }^{2}x=\frac{1+\cos 2x}{2}$

$\int {\left(\frac{1-\cos 2x}{2}\right)}^2.{\left(\frac{1+\cos 2x}{2}\right)}^{2}dx$

$=\frac{1}{16}\int (1-{\cos }^{2}2x{)}^{2}dx$

$=\frac{1}{16}\int (1+{\cos }^{4}2x-2{\cos }^{2}2x)dx$

$=\frac{1}{16}\int 1+\frac{(1+\cos 4x{)}^2}{4}-2\left(\frac{1+\cos 4x}{2}\right)dx$

$=\frac{1}{16}\int 1+\frac{1+{\cos }^{2}4x+2\cos 4x}{2}-1-\cos 4xdx$

$=\frac{1}{16}\int \frac{1+\frac{1+\cos 8x}{2}+2\cos 4x-2\cos 4x}{2}dx$

$=\frac{1}{32}\int 1+\frac{1+\cos 8x}{2}dx$

$=\frac{1}{64}\int 3+\cos 8xdx$

$=\frac{1}{64}[3x+\frac{\sin 8x}{8}]+c$