Answer
\[-\frac{1}{6}\cos 3x-\frac{1}{26}\cos 13x+C\]
Work Step by Step
Let \[I=\int\sin 8x\:\cos 5x\:dx\]
\[I=\frac{1}{2}\int 2\sin 8x\:\cos 5x\:dx\]
\[\left[2\sin A\cos B=\sin (A+B)+\sin (A-B)\right]\]
\[I=\frac{1}{2}\int \left[\sin 13x\:+\sin 3x\right]\:dx\]
\[I=\frac{1}{2}\left[\frac{-1}{13}\cos 13x-\frac{1}{3}\cos 3x\right]+C\]
\[I=-\frac{1}{6}\cos 3x - \frac{1}{26}\cos 13x+C \]
Hence \[I=-\frac{1}{6}\cos 3x -\frac{1}{26}\cos 13x+C \].
