How to evaluate the integral #int e^(2θ)*sin3θ dθ #?

1 Answer
Konstantinos Michailidis
Sep 24, 2015

Refer to explanation

Explanation:

We set #x=theta# hence we have

#int e^(2x)sin3xdx=int 1/2*(e^(2x))'sin3xdx= 1/2*e^(2x)sin3x-3/2*int e^(2x) cos3xdx=1/2*e^(2x)sin3x-3/4*int (e^(2x))'*cos3xdx=1/2*e^(2x)sin3x-3/4*[(e^(2x)cos3x-3*int e^(2x)sin3xdx)]= #
Hence finally

#int e^(2x)sin3xdx=1/13*(e^(2x))*(2sin3x-3cos3x)+c#

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