Integrate $ \cos^3x \sin^2 x \ \text{d}x$ $$\int \cos^3x \sin^2 x \ \text{d}x$$
I've simplified it to $\int \cos^3x-\cos^5x \,\mathrm{d}x$ using Pythagorean identities. It this the right way to do it? How do I integrate it from here?

Hint: Start with rewriting as
$$
\int \color{blue}{\cos^2(x)}\color{red}{\sin}^2(x)\color{green}{\cos(x)dx}
$$
Use that $\color{blue}{\cos^2(x) = 1 - \sin^2(x)}$. Now let $\color{red}{u = \sin(x)}$ so that $\color{green}{du = \cos(x)dx}$ (as Dario suggests in his comment above).
