Evaluate the integral:
$$\int \sin^4 2x \cos 2x\, dx$$
What I currently did is
$$\int \left(\frac{1-\cos 2x}{2}\right)^2 \cos2x\, dx$$
Honestly I'm not sure what to do, I was absent from class that day. If I try substitution I get $du$ with $\sin(2x)$ and you cant mix variables.. Any help would be appreciated.
Best Answer
$$\frac{1}{2}\int \left(\sin^4 2x\right)\left(\cos 2x\right)d(2x)=\frac{1}{2}\int \left(\sin^4 2x\right)d(\sin 2x)$$
Now let $u=\sin 2x$. The integral becomes
$$\frac{1}{2}\int u^4\ du=\frac{1}{2}\left(\frac{u^5}{5}+C_1\right)=\frac{(\sin 2x)^5}{10}+C_2$$