"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
 A: 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 $$\begin{align}\int \left(u^2-u^4\right)\,\left(-\frac12 du\right)&=\frac12 \left(\frac15 u^5-\frac13 u^3\right)+C\\\\&=\frac{1}{10}\cos^5 (2x)-\frac16 \cos^3(2x)+C\end{align}$$

A: $$
I = \int sin^3 (2x) cos^2(2x) dx
$$ 
Substituting
$$
t = 2x
\\dt = 2 dx
\\dx = \frac{dt}{2}
$$
So,
$$
I = \frac{1}{2}\int sin^3(t) cos^2(t) dt
$$
I'm now going to factor out 
$$
sin^3 t = sin(t) (1 - cos^2 t)
$$
Because the factor of $sin(t)$ will simplify the integral.
$$
I = \frac{1}{2}\int sin(t)\; (1 - cos^2 (t)) \;cos^2(t) dt
\\I = \frac{1}{2}\int sin(t) cos^2 (t) dt - \frac{1}{2}\int sin(t) cos^4(t) dt
\\I = \frac{1}{2}\int cos^2(t) (-dcos(t)) - \frac{1}{2}\int cos^4(t) (-dcos(t))
\\I = -\frac{1}{2}\int cos^2(t) (dcos(t)) + \frac{1}{2}\int cos^4(t) (-dcos(t))
\\I = -\frac{1}{6} cos^3(t) + c1 + \frac{1}{10} cos^5(t) + c2
\\I = \frac{1}{10} cos^5(2x) + -\frac{1}{6} cos^3(2x) + c
$$
