Integrating $\int{3\sin^2x\cos x \;dx}$ 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.
 A: $$\sin^2(x) \cos^2(x) = \dfrac{\sin^2(2x)}4 = \dfrac{1-\cos(4x)}8$$
EDIT
(Question was changed) Setting $\sin(x) = t$. We get that $\cos(x) dx = dt$. 
Hence, we have $$I = \int 3 \sin^2(x) \cos(x) dx = \int 3 t^2 dt$$
A: Hint::
$\displaystyle \sin^2x = \frac{1 - \cos 2x}{2}$ and $\displaystyle \cos^2x = \frac{1 + \cos 2x}{2}$
Edit:: Let $\sin x = t $ then $\cos x dx = dt$, then you have $\int 3t^2  dt $
A: If you let $t=\sin x$,  you get $dt = \cos x dx$, so $$\int{3\sin^2x\cos x \;dx}=\int 3t^2dt=t^3+C=\sin^3x+C$$
A: In the integrand, if all arguments of trig functions are "the x", transform all of them into sin's and cos's. Every m and n below matches any integer.


*

*cos2m(x) sin2n(x) where m and n are nonnegative: discussed below.

*cos2n+1(x) f(cos2x, sin x): set y = sin x.

*sin2n+1(x) f(sin2x, cos x): set y = cos x.


If the integrand matches none of these, transform it into tan's and sec's.


*

*sec2n(x) f(tan x, sec2 x): set y = tan x.


If none of the above work, apply the sneaky Weierstrass substitution.
$$ y = \tan \frac x 2 = \frac {\sin x} {\cos x + 1}. $$
Integrate cos2m(x) sin2n(x)
Find the less of m and n. Assume n < m, extract sin(2x)2n with
$$ \cos x \sin x = \frac {\sin 2x} 2.$$
There will be either only cos's or only sin's left. Convert them into double-angles with
\begin{align}
\cos^2 x &= \frac {1 + \cos 2x} 2 \\
\sin^2 x &= \frac {1 - \cos 2x} 2.
\end{align}
