Question

Integrate:
$\int {\sin }^{4}xdx$

Answer 1

$I=\int si{n}^{4}xdx=\int \frac{(1-cos2x{)}^2}{4}dx$

$[\because \frac{1-cos2x}{2}=sinx]$

$=\frac{1}{4}\int (1-2cos2x+co{s}^{2}2x)dx$

$=\frac{1}{4}\int 1-2cos2x+\frac{(1+cos4x)}{2}dx$

$=\frac{1}{4}\int (\frac{3}{2}-2cos2x+\frac{cos4x}{2})dx$

$=\frac{1}{4}(\frac{3}{2}x-\frac{2sin2x}{2}+\frac{sin4x}{2})+c$

$I=\frac{3}{8}x-\frac{1}{4}sin2x+\frac{1}{32}sin4x+c$