Answer
$\int$ sin 6$\theta$sin2$\theta$d$\theta$ = $\frac{1}{2}$$\int$ cos(6$\theta$-2$\theta$) - cos(6$\theta$+2$\theta$) d$\theta$
=$\int$$\frac{1}{2}$cos4$\theta$ -$\frac{1}{2}$cos8$\theta$ d$\theta$
=$\frac{1}{8}$4$\theta$- $\frac{1}{16}$sin8$\theta$ + C
= $$\frac{1}{8}4\theta- \frac{1}{16}sin8\theta+ C$$
Work Step by Step
Use property: sin A sin B =$\frac{1}{2}$[cos(A-B) - cos(A+B)]
