Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 404: 78

$$\int \sin m x \cos n x d x=-\frac{\cos (m-n) x}{2(m-n)}-\frac{\cos (m+n) x}{2(m+n)}+C,\ \ \ \ m\neq n$$

Since $$ \sin m x \cos n x=\frac{1}{2}(\sin (m-n) x+\sin (m+n) x)$$ Then \begin{align*} \int \sin m x \cos n x d x&=\frac{1}{2}\int (\sin (m-n) x+\sin (m+n) x) d x\\ &=\frac{1}{2}\left[-\frac{\cos (m-n) x}{(m-n)}-\frac{\cos (m+n) x}{(m+n)}\right]+C \end{align*}