How do you find the antiderivative of $e^{2 x} \cdot sin (3 x) d x$ $e^(2x)*sin(3x) dx$ ?

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How do you find the antiderivative of $e^{2 x} \cdot sin (3 x) d x$ $e^(2x)*sin(3x) dx$ ?

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Answer 1 · 2016-08-27T16:49:26

$\frac{e^{2 x}}{13} (2 sin (3 x) - 3 cos (3 x)) + C$ $e^(2x)/13(2sin(3x)-3cos(3x))+C$

Explanation:

This kind of integral can be easily handled by inmersion in the complex domain. Solving instead

$\int e^{2 x} cos (3 x) d x + i \int e^{2 x} \cdot sin (3 x) d x = \int e^{2 x} e^{i 3 x} d x$ $int e^(2x)cos(3x)dx + i int e^(2x)*sin(3x) dx =int e^(2x) e^(i3x)dx$

(We are using de Moivre's identity $e^{i ϕ} = cos (ϕ) + i sin (ϕ))$ $e^(i phi) = cos(phi)+i sin(phi))$

but

$\int e^{2 x} e^{i 3 x} d x = \int e^{2 x + i 3 x} d x = \int e^{(2 + 3 i) x} d x$ $int e^(2x) e^(i3x)dx = int e^(2x+i3x)dx = int e^((2+3i)x)dx$

which gives

$\int e^{(2 + 3 i) x} d x = \frac{e^{(2 + 3 i) x}}{2 + 3 i} = \frac{e^{(2 + 3 i) x} (2 - 3 i)}{13} = \frac{2 - 3 i}{13} e^{2 x} e^{i 3 x} = \frac{2 - 3 i}{13} e^{2 x} (cos (3 x) + i sin (3 x))$ $int e^((2+3i)x)dx = e^((2+3i)x)/(2+3i)=(e^((2+3i)x)(2-3i))/13 = (2-3i)/13 e^(2x) e^(i3x)=(2-3i)/13e^(2x)(cos(3x)+i sin(3x))$ with imaginary part given by

$\frac{e^{2 x}}{13} (2 sin (3 x) - 3 cos (3 x))$ $e^(2x)/13(2sin(3x)-3cos(3x))$

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How do you find the antiderivative of $e^{2 x} \cdot sin (3 x) d x$ $e^(2x)*sin(3x) dx$ ?

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Explanation:

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