Integrate $\sin(3x)\cos(3x)$

Question

Integrate $\sin(3x)\cos(3x)$

I looked at various answers on different sites but still do not understand how to use the u-substitution method in this question or the double angle rule.

Answer 1

$\int\sin3x\cos3x \mathop{\mathrm{d}x}=\int\frac{1}{2}\sin6x \mathop{\mathrm{d}x}=-\frac{1}{12}\cos6x+C$

Answer 2

Here's one way to do this : \begin{align*}\int \sin(3x)\cos(3x)\mathop{\mathrm{d}x} & = \frac{1}{3}\int 3\sin(3x)\cos(3x)\mathop{\mathrm{d}x} \\ & = \frac{1}{3} \int \sin(3x)(\sin(3x))' \mathop{\mathrm{d}x} \\ & = \frac{1}{6} \int (\sin(3x)^2)'\mathop{\mathrm{d}x} \\ & = \frac{1}{6}\sin(3x)^2 +K\end{align*}

This is in fact the same as doing a substitution $u=\sin(3x)$.

Note also that though it may seem different than almagest's solution, the two are equivalent. Indeed, $$\cos(6x)^2=1-2\sin(3x)^2,$$ so that $$-\frac{1}{12}\cos(6x)+C=-\frac{1}{12}(1-2\sin(3x)^2)+C=\frac{1}{6}\sin(3x)^2+C-\frac{1}{12},$$ which is the same as my answer if you choose $K=C-\frac{1}{12}$.