Answer
$\frac{2}{9}Sin^3(3x)+C$
Work Step by Step
$\int Sin(3x)Sin(6x) dx$
Using $Sin(2\theta)=2Sin(\theta)Cos(\theta)$
We get $\int Sin(3x)[2Sin(3x)Cos(3x)] dx = 2\int Sin^2(3x)Cos(3x) dx=\frac{2}{3}\int Sin^2(3x)3Cos(3x) dx$
By using $\int f^n(x)f'(x) dx = \frac{f^{n+1}}{n+1} + C$
We get: $\frac{2}{3}\frac{Sin^3(3x)}{3} + C = \frac{2}{9}Sin^3(3x) + C.$
