Question

$\int \sin x\sin 2x\sin 3xdx$

Answer 1

$$\begin{gathered} \int \sin x·\sin 2x·\sin 3xdx \\ =\frac{1}{2}\int \left(2\sin 2x·\sin x\right)\sin 3xdx \\ =\frac{1}{2}\int \left[\cos \left(2x-x\right)-\cos \left(2x+x\right)\right]\sin 3xdx\left[\because 2\sin A\sin B=\cos (A-B)-\cos (A+B)\right] \\ \Rightarrow =\frac{1}{2}\int \left[\cos x-\cos 3x\right]\sin 3xdx \\ =\frac{1}{2}\int \sin 3x·\cos xdx-\frac{1}{2}\int \sin 3x·\cos 3xdx \\ =\frac{1}{4}\int 2\sin 3x·\cos xdx-\frac{1}{4}\int 2\sin 3x·\cos 3xdx \\ =\frac{1}{4}\int \left[\sin 4x+\sin 2x\right]dx-\frac{1}{4}\int \sin 6xdx\left[\because 2\sin A\cos B=\sin \left(A+B\right)+\sin \left(A-B\right)\right] \\ =\frac{1}{4}\left[\frac{-\cos 4x}{4}-\frac{\cos 2x}{2}\right]-\frac{1}{4}\left[-\frac{\cos 6x}{6}\right]+C \\ =-\frac{\cos 4x}{16}-\frac{\cos 2x}{8}+\frac{\cos 6x}{24}+C \end{gathered}$$