How do I find the indefinite integral of #sin(3x)cos(3x)#?

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Answer 1 · 2018-08-05T12:35:13

The answer is #=-1/12cos(6x)+C#

We know that

#sin2x=2sinxcosx#

Therefore,

#sin(3x)cos(3x)=1/2sin(6x)#

So,

The integral is

#I=intsin(3x)cos(3x)dx=1/2intsin(6x)dx#

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