Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 404: 60

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Given $$ \int_{0}^{\pi}\sin x\sin 3xdx$$ Use $$ \sin (a x) \sin (b x)=\frac{1}{2} \cos ((a-b) x)-\frac{1}{2} \cos ((a+b) x) $$ Then \begin{align*} \int_{0}^{\pi}\sin x\sin 3xdx&=\frac{1}{2}\int_{0}^{\pi}(\cos 2x -\cos 4x)dx\\ &=\frac{-1}{2}\left(\frac{1}{2}\sin x-\frac{1}{7}\sin4 x\right)\bigg|_{0}^{\pi} \\ &=0 \end{align*}