I have an integral here that I'm trying

avalg10o

avalg10o

Answered question

2022-04-01

I have an integral here that I'm trying to figure out.
7sin2xcos4xdx

Answer & Explanation

Jadyn Gentry

Jadyn Gentry

Beginner2022-04-02Added 12 answers

You can use sin(x)cos(x)=12sin(2x) and cos2(x)=1+cos(2x)2 and your integral is
7(12sin(2x))21+cos(2x)2dx
=78sin2(2x)dx+78sin2(2x)cos(2x)dx
The first term is just a sin2 integral, and the second can be dealt with by a u substitution u=sin2(2x)

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