Integral of $\int 2\,\sin^{2}{x}\cos{x}\,dx$ I am asked as a part of a question to integrate $$\int 2\,\sin^{2}{x}\cos{x}\,dx$$
I managed to integrate it using integration by inspection:
$$\begin{align}\text{let } y&=\sin^3 x\\
              \frac{dy}{dx}&=3\,\sin^2{x}\cos{x}\\
              \text{so }\int 2\,\sin^{2}{x}\cos{x}\,dx&=\frac{2}{3}\sin^3x+c\end{align}$$
However, looking at my notebook the teacher did this:
$$\int -\left(\frac{\cos{3x}-\cos{x}}{2}\right)$$
And arrived to this result:
$$-\frac{1}{6}\sin{3x}+\frac{1}{2}\sin{x}+c$$
I'm pretty sure my answer is correct as well, but I'm curious to find out what how did do rewrite the question in a form we can integrate.
 A: $$2\sin^2(x)\cos(x) = (2\sin(x)\cos(x))\sin(x) = \sin(2x)\sin(x) = \frac{\cos(x) - \cos(3x)}{2}$$
The last step comes from :
$$\cos(A - B) - \cos(A + B) = 2\sin(A)\sin(B)$$
A: $$2(\sin x)^2\cos x=(1-\cos 2x)\cos x=\cos x-\cos 2x\cos x=\cos x-\frac{1}{2}(\cos x +\cos 3x)$$
relying on
$$\cos x \cos y  =\frac{1}{2}(\cos(x-y)+\cos(x+y))$$
A: Try these identities:
$$\sin(3x) = 3\sin(x) - 4\sin^3(x)$$
and
$$\cos(3x) = 4\cos^3(x) - 3\cos(x)$$
A: Another natural approach is the substitution $u=\sin x$. 
The path your instructor chose is less simple. We can rewrite $\sin^2 x$ as $1-\cos^2x$, so we are integrating $2\cos x-2\cos^3 x$. Now use the identity $\cos 3x=4\cos^3 x-3\cos x$ to conclude that $2\cos^3 x=\frac{1}{2}\left(\cos 3x+3\cos x\right)$.
Remark: The identity $\cos 3x=4\cos^3 x-3\cos x$ comes up occasionally, for example in a formula for solving certain classes of cubic equations. The same identity comes up when we are proving that the $60^\circ$ angle cannot be trisected by straightedge and compass.
A: Another way using a simple substitution: 
$$I = \int 2\,\sin^{2}{x}\cos{x}\ \ dx$$
Let $u = \sin x, du = \cos x  \ dx$
$$ I = 2\int u^2 \ du$$
$$I = \frac{2}{3} u^3$$
$$I = \frac{2}{3} \sin^3 x + C$$
A: I'd go direct, taking into account that $\,\displaystyle{\int x^2\,dx=\frac{1}{3}x^3 + C}\,$:$$\int 2\sin^2x\cos x\,dx=2\int\sin^2x(d\sin x)=\frac{2}{3}\sin^3x+C$$
This looks simpler to me and less messy than with trigonometric identities.
