$\displaystyle \int sin^2(2x)cos^3(2x) dx$

$\displaystyle \int sin^2(2x)cos^2(2x)cos(2x) dx$

$\displaystyle \int sin^2(2x)(1-sin^2(2x))cos(2x) dx$

let $\displaystyle u = sin(2x)$ $\displaystyle du = 2cos(2x)dx \rightarrow \frac{1}{2}du = cos(2x)dx$

$\displaystyle \frac{1}{2}\int u^2(1-u^2)du = \frac{1}{2} \int u^2 - u^4 du$

$\displaystyle \frac{1}{2}\left[\frac{u^3}{3} - \frac{u^5}{5}\right] + C \rightarrow \frac{1}{2}\left[\frac{5u^3 - 3u^5}{15}\right] + C$

$\displaystyle =\frac{1}{6}sin^3(2x) - \frac{1}{10}sin^5(2x) + C$

I thought I did everything right but wolfram got a different answer than me. Is what I did here correct?