Integrate sin9x cos3x dx ?

1 Answer
Øko
Apr 4, 2018

#I=-1/24(cos(12x)+2cos(6x))+C#

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#

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