Answer
$$\int^\pi_{-\pi}\sin3x\sin3xdx=\pi$$
Work Step by Step
$$A=\int^\pi_{-\pi}\sin3x\sin3xdx$$
Apply the identity of the product of 2 sine functions: $$\sin mx\sin nx=\frac{1}{2}[\cos(m-n)x-\cos(m+n)x]$$
we have $$A=\frac{1}{2}\int^\pi_{-\pi}(\cos0-\cos6x)dx$$ $$A=\frac{1}{2}\int^\pi_{-\pi}(1-\cos6x)dx$$ $$A=\frac{1}{2}\Big(x-\frac{1}{6}\sin6x\Big)\Big]^\pi_{-\pi}$$ $$A=\frac{1}{2}\Big[(\pi-\frac{1}{6}\sin6\pi)-(-\pi-\frac{1}{6}\sin(-6\pi))\Big]$$ $$A=\frac{1}{2}[(\pi-0)-(-\pi-0)]$$ $$A=\frac{1}{2}(2\pi)=\pi$$
