Trig Integral, Product to Sum

Question

I was trying to solve the integral $\displaystyle\int\sin(2x)\sin(4x)\text{d}x$. Here's my attempt:

$$\int\frac{1}{2}(\cos(2x-4x)-\cos(2x+4x))\text{d}x$$ $$\frac{1}{2}\int\cos(-2x)\text{d}x-\frac{1}{2}\int\cos(6x)\text{d}x$$ $$\text{Let u = 2x and v = 6x}$$ $$\frac{1}{4}\int\cos(u)\text{d}u-\frac{1}{12}\int\cos(v)\text{d}v$$ $$\frac{1}{4}\sin(u)-\frac{1}{12}\sin(v)$$ $$\frac{1}{4}\sin(2x)-\frac{1}{12}\sin(6x)+C$$

But the answer is:

$$\frac{\sin^3(2x)}{3} + C$$

Where did I go wrong? Thanks.

Answer 1

They're equivalent. You'd get their answer with $u=\sin 2x$ since $\sin 4x=2\sin 2x\cos 2x$.

Answer 2

You didn't go wrong. You can check it's the same function, usuing the trig formula (once taught in high school): $$\sin 3x=3\sin x-4\sin^3x, $$ so $\sin 6x=\dotsm$