Question

Find the integrals of the function: $sin\,3x\,cos\,4x$

Answer 1

The correct answer is: $= \frac{-cos7x}{14}+\frac{cosx}{2}+C$
It is known that, $sin A\, cos B = \frac{1}{2}[sin(A+B) + sin(A-B)]$
$∴ ∫sin\,3x\,cos\,4x\, dx = \frac{1}{2} ∫{sin(3x+4x)+sin(3x-4x)} dx$
$=\frac{1}{2} ∫{sin7x+sin(-x)} dx$
$=\frac{1}{2} ∫sin7x dx-\frac{1}{2} ∫sinx dx$
$=\frac{1}{2}(\frac{-cos7x}{7})-\frac{1}{2}(-cosx)+C$
$= \frac{-cos7x}{14}+\frac{cosx}{2}+C$