How do you integrate #[(e^(2x))sinx]dx#?

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Jim H Share
Sep 12, 2015

Answer:

Integrate by parts twice using #u = e^(2x)# both times.

Explanation:

After the second integration by parts, you'll have

#int e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx - 4 int e^(2x)sinx dx#

Note that the last integral is the same as the one we want. Call it #I# for now.

#I = -e^(2x)cosx + 2e^(2x)sinx - 4 I#

So #I = 1/5[-e^(2x)cosx + 2e^(2x)sinx] +C#

You may rewrite / simplify / factor as you see fit.

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