Integrate sin9x cos3x dx ?
1 Answer
Apr 4, 2018
Explanation:
We want to solve
I=∫sin(9x)cos(3x)dx
Use the product identity
sin(a)cos(b)=12(sin(a+b)+sin(a−b))
Thus
I=12∫sin(9x+3x)+sin(9x−3x)dx
I=12∫sin(12x)+sin(6x)dx
I=12(−112cos(12x)−16cos(6x))+C
I=−124(cos(12x)+2cos(6x))+C