Question

∫ sin5x cos3x dx

Answer 1

$\int$ sin5xcos3xdx$\begin{array}{l} =\int \frac{2}{2} \sin (5 x) \cos (3 x) d x \\ =\frac{1}{2} \int 2 \sin (5 x) \cos (3 x) d x \\ =\frac{1}{2} \int(\sin (5 x+3 x)+\sin (5 x-3 x)) d x \\ =\frac{1}{2} \int(\sin (8 x)+\sin (2 x)) d x \\ =\frac{1}{2}\left(\int \sin (8 x) d x+\int \sin (2 x) d x\right) \\ =\frac{1}{2}\left(\frac{-\cos (8 x)}{8}+\frac{-\cos (2 x)}{2}\right)+C \\ =\frac{-\cos (8 x)}{16}-\frac{\cos (2 x)}{4}+C \end{array}$