Question

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

Answer 1

$\int \sin 2x\cos 3xdx=\frac{1}{2}\int 2\sin 2x\cos 3xdx$

$=\frac{1}{2}\int (\sin 5x-\sin x)dx$

$(\because \sin (A+B)-\sin (A-B)=2\sin B\cos A)$

Here $A=3x,B=2x$

$=\frac{1}{2}(\int \sin 5xdx-\int \sin xdx)$

$=\frac{1}{2}(\frac{1}{5}\int \sin \left(5x\right)d\left(\sqrt{x}\right)-\int \sin xdx)$

$=\frac{1}{2}(\frac{1}{5}([-\cos \left(5x\right)]+C_1)-\left([-\cos x]\right)+C_2)$

$=\frac{1}{2}(\cos x-\frac{\cos 5x}{5})+(\frac{C_1}{10}+\frac{C_2}{2})$

$\int \sin 2x\cos 3xdx=\frac{1}{2}(\cos x-\frac{\cos 5x}{5})+k$ where, $k=\frac{C_1+5C_2}{10}$