In solving the integral
$$\int\sin^5(2x)\cos^3(2x)~dx,$$
I misread it as
$$\int\sin^5(2x)\cos^3(\color{red}{3x})~dx.$$
After a while of failing to solve this, I realized my mistake. How would I go about solving the second integral?
On first glance I thought it could be integration by parts, but since there are no $\cos(2x)$ or $\sin(3x)$ terms that seems like the wrong approach. I thought of using the $\sin(x)\cos(y) = \dots$ and $\sin(x)\sin(y) = \dots$ identities, but that didn't pan out either.
Best Answer
Hint:
Use double and triple angle trigonometric identities and reduce them to the same angle:
$$\cos(3x)=4\cos^3(x)-3\cos(x)$$
$$\sin(2x)=2\sin(x)\cos(x)$$
Expanding them will lead you to multiple integrals involving different exponents of the $\sin(x)$ and $\cos(x)$ (their product) terms which becomes similar to your original problem which you have already solved.
Can you try it now?