integral of $\cos^3(x)\cos(2x)$

Question

I have a midterm in a few hours and cannot solve this problem no matter how many different ways I try.

I need to find the indefinite integral of $\cos^3(x)\cos(2x)$

I started off by using IBP but I ended up coming full circle to the original statement. Then I tried to see if I could find a trig identity but I'm not sure which one to go about using, or what to try instead.

Answer 1

Try $\cos^3 x = (1-\sin^2x)\cos x$ and $\cos 2x = 1-2\sin^2 x$. Then substitute $u=\sin x$ in the integral.

Answer 2

Simple: $$\cos 2x = 1 - 2 \sin^2 x,$$ and we also have $$\cos^3 x = \cos x \,(1 - \sin^2 x).$$ Therefore, $$\int \cos^3 x \cos 2x \, dx = \int (1 - 2 \sin^2 x)(1 - \sin^2 x) \cos x \, dx,$$ and with a single substitution $u = \sin x$, $du = \cos x \, dx$, we obtain $$\int (1 - 2u^2)(1-u^2) \, du,$$ and the rest is trivial.