How do you integrate $[(e^{2 x}) sin x] d x$ $[(e^(2x))sinx]dx$ ?

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How do you integrate $[(e^{2 x}) sin x] d x$ $[(e^(2x))sinx]dx$ ?

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Answer 1 · 2015-09-12T13:32:27

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

Explanation:

After the second integration by parts, you'll have

$\int e^{2 x} sin x d x = - e^{2 x} cos x + 2 e^{2 x} sin x - 4 \int e^{2 x} sin x d x$ $int e^(2x)sinx dx = -e^(2x)cosx + 2e^(2x)sinx - 4 int e^(2x)sinx dx$

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

$I = - e^{2 x} cos x + 2 e^{2 x} sin x - 4 I$ $I = -e^(2x)cosx + 2e^(2x)sinx - 4 I$

So $I = \frac{1}{5} [- e^{2 x} cos x + 2 e^{2 x} sin x] + C$ $I = 1/5[-e^(2x)cosx + 2e^(2x)sinx] +C$

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

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How do you integrate $[(e^{2 x}) sin x] d x$ $[(e^(2x))sinx]dx$ ?

1 Answer

Explanation:

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How do you integrate $[(e^{2 x}) sin x] d x$ $[(e^(2x))sinx]dx$ ?