Ta có:
$\int\sin^42xdx$
$=\int\dfrac{(1-\cos4x)^2}{4}dx$
$=\int\left(\dfrac{1}{4}-\dfrac{\cos4x}{2}+\dfrac{\cos^24x}{4}\right)dx$
$=\int\left(\dfrac{1}{4}-\dfrac{\cos4x}{2}+\dfrac{1+\cos8x}{8}\right)dx$
$=\dfrac{3x}{8}-\dfrac{\sin4x}{8}+\dfrac{\sin8x}{64}+C$