Hi,

I'd be happy if someone could check my work on the integral $\displaystyle sin^6(x)dx$

$\displaystyle sin^6(x) = (sin^2(x))^3$

$\displaystyle sin^2(x) = \frac{1}{2}(1-cos(2x))$

$\displaystyle \frac{1}{8}\int(1-cos(2x))^3$

let u = cos(2x);

$\displaystyle (1-u)^2\cdot(1-u) = 1-3u+3u^2-u^3$

$\displaystyle = \frac{1}{8}\int~1-3cos(2x)+3cos^2(2x)-cos^3(2x)$

=$\displaystyle 3~\int cos^2(2x) = \frac{1}{2}(1+cos(2x)) = \frac{1}{2}x+\frac{1}{4}sin(2x)$