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.