Question

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

Answer 1

$\int \left(2x\right).cos\left(3x\right).dx$

Now,

2 sin A . cosB = sin(A+B) + sin(A-B)

=$\frac{1}{2}\int 2sin\left(2x\right).cos\left(3x\right).dx$

$=\frac{1}{2}\int \left[sin\right(5x)+sin(-x\left)\right]dx$

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

$\frac{1}{2}(\frac{-cos\left(5x\right)}{5}-(-cosx\left)\right)$

$=\frac{-cos\left(5x\right)}{10}+\frac{cosx}{2}+C$