"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

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