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

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#### Explanation

Explain in detail...

#### Explanation:

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Jim H Share
Sep 12, 2015

Integrate by parts twice using $u = {e}^{2 x}$ both times.

#### Explanation:

After the second integration by parts, you'll have

$\int {e}^{2 x} \sin x \mathrm{dx} = - {e}^{2 x} \cos x + 2 {e}^{2 x} \sin x - 4 \int {e}^{2 x} \sin x \mathrm{dx}$

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

$I = - {e}^{2 x} \cos x + 2 {e}^{2 x} \sin x - 4 I$

So $I = \frac{1}{5} \left[- {e}^{2 x} \cos x + 2 {e}^{2 x} \sin x\right] + C$

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

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