I have tried to integrate the following indefinite integral but I'm not sure if I get the right answer. Please tell me if I'm wrong and if so, please indicate what went wrong.
$$ \int\sqrt{\cos2x}\sin^32x\,dx $$ $$ \int\sqrt{\cos2x}(\sin^22x)(\sin2x)\,dx $$ $$ \int\sqrt{\cos2x}(1-\cos^22x)(\sin2x)\,dx $$ $$ \frac {-1}2 \int\sqrt{u}(1-u^2)\,du $$ $$ \frac {-1}2 \int(u^{\frac 12}-u^{\frac 52})\,du $$ $$ \frac {-1}2 (\frac {2u^{\frac 32}}3-\frac {2u^{\frac 72}}7)\,+C $$ $$ \frac {u^{\frac 72}}7-\frac {u^{\frac 32}}3\,+C $$ $$ \frac {\sqrt{\cos^72x}}7-\frac {\sqrt{\cos^32x}}3\,+C $$
Using the fact that $\cos^3 2x = \frac{\cos 6x+3 \cos 2x}{4}$ you get the answer given by Wolfram Alpha.