Integrate sin9x cos3x dx ?
1 Answer
Apr 4, 2018
Explanation:
We want to solve
#I=intsin(9x)cos(3x)dx#
Use the product identity
#color(blue)(sin(a)cos(b)=1/2(sin(a+b)+sin(a-b))#
Thus
#I=1/2intsin(9x+3x)+sin(9x-3x)dx#
#color(white)(I)=1/2intsin(12x)+sin(6x)dx#
#color(white)(I)=1/2(-1/12cos(12x)-1/6cos(6x))+C#
#color(white)(I)=-1/24(cos(12x)+2cos(6x))+C#