<s>[INST] "Trig Substitutions", I tried half- angle and trig indentity in this one, but doesn't work I´m really lost in this one.
$\int \sin^3 (2x) \cos^2 (2x) dx$
I know that the answer is:
$\frac{1}{10}cos^5(2x)-\frac{1}{6}cos^3(2x) + c$
Please help
 [/INST] HINT:
Write $\sin^3(2x)\cos^2(2x)=(1-\cos^2(2x))\cos^2(2x)\sin(2x)$
SPOLIER ALERT:  SCROLL OVER THE SHADED AREA TO REVEAL SOLUTION

We have $$\begin{align}\sin^3(2x)\cos^2(2x)&=\left(1-\cos^2(2x)\right)\cos^2(2x)\sin(2x)\\\\&=\left(\cos^2(2x)-\cos^4(2x)\right)\sin(2x)\end{align}$$Then, $$\int \sin^3(2x)\cos^2(2x)\,dx=\int \left(\cos^2(2x)-\cos^4(2x)\right)\sin(2x)\,dx$$Now, enforce the substitution $u=\cos (2x)$ so that $du=-2\sin(2x)\,dx$.  Then, the integral becomes $$\begin{align}\int \left(u^2-u^4\right)\,\left(-\frac12 du\right)&=\frac12 \left(\frac15 u^5-\frac13 u^3\right)+C\\\\&=\frac{1}{10}\cos^5 (2x)-\frac16 \cos^3(2x)+C\end{align}$$

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