Answer
$$\int_{0}^{\pi}\sin6x\cos3xdx=\frac{4}{9}$$
Work Step by Step
$$\int_{0}^{\pi}\sin6x\cos3xdx=\int_{0}^{\pi}(2\sin3x\cos3x)\cos3xdx$$
$$=2\int_{0}^{\pi}\sin3x\cos ^{2}3xdx=-\frac{2}{3}\int_{0}^{\pi}\cos^{2}3x\,d(\cos3x)$$
$$=-\frac{2}{9}\left [\cos^{3}3x\right ]_{0}^{\pi}=\frac{4}{9}$$
