I am struggling with the following integral:
$$\int{3\sin^2x\cos x \;dx}$$
I have so far tried to solve this using every tool at my disposal, I have set $t = \sin x$ but I get an even harder quantity to integrate:
$$\int{\frac{t^2-t^4}{\sqrt{1-t^2}} dt}$$
so I have dismissed this way of proceeding. Then I tried to rewrite $\sin^2x = 1-\cos^2x $ but as a result I get stuck at:
$$\int{3dx}-3\int \cos^3xdx$$
which I don't know how to cope with because integrating by parts I end up with an endless chain of integrals. Hope you could shed some light on the way to properly solve this, really feel a bit lost.
Update: sorry I have made a mistake in copying the exercise over the site.
Best Answer
$$\sin^2(x) \cos^2(x) = \dfrac{\sin^2(2x)}4 = \dfrac{1-\cos(4x)}8$$
EDIT
(Question was changed) Setting $\sin(x) = t$. We get that $\cos(x) dx = dt$. Hence, we have $$I = \int 3 \sin^2(x) \cos(x) dx = \int 3 t^2 dt$$