Question

Find the integrals of the functions.
$\int sin3xcos4x$ dx.

Answer 1

$\int si{n}^{2}3xcos4xdx=\frac{1}{2}\int 2sin3xcos4xdx=\frac{1}{2}\int (sin7x+sin(3x-4x\left)\right)dx[\because 2sinAandB=sin(A+B)+sin(A-B\left)\right]=\frac{1}{2}\int (sin7x-sinx)dx[\therefore sin-\theta )=-sin\theta ]=\frac{1}{2}(\frac{-cos7x}{7}+cosx)+C(\therefore \int sinaxdx=\frac{-cosax}{a})=\frac{-cos7x}{14}+\frac{cosx}{2}+C$