Integrate $ \cos^3x \sin^2 x \ \text{d}x$

Question

$$\int \cos^3x \sin^2 x \ \text{d}x$$

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?

Answer 1

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).

Answer 2

$$ \int \cos^2 x \sin^2 x\ \underbrace{\Big(\cos x\,dx\Big)}_{\text{HINT}} = \int(1-\sin^2 x)\sin^2 x\ \underbrace{\Big(\cos x\,dx\Big)}_{\text{HINT}} $$

Answer 3

Hint:
As an alternative way, one can show that: $$\cos^3x\sin^2x=\dfrac1{16}\big(2\cos x-\cos(3x)-\cos(5x)\big).$$ Then you can easily integrate it since: $$\int\cos(ax)\,\mathrm dx=\dfrac{\sin(ax)}{a}+\text C.$$