How do you find the integral of $e^{7 x} \cdot sin (2 x) d x$ $e^(7x)*sin(2x)dx$ ?

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How do you find the integral of $e^{7 x} \cdot sin (2 x) d x$ $e^(7x)*sin(2x)dx$ ?

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Answer 1 · 2018-04-11T15:13:18

$\int e^{7 x} sin 2 x d x = \frac{e^{7 x} (7 sin 2 x - 2 cos 2 x)}{53} + C$ $int e^(7x) sin 2x dx = (e^(7x)(7sin 2x - 2cos2x))/53 +C$

Explanation:

Integrate by parts:

$\int e^{7 x} sin 2 x d x = \int sin 2 x d (\frac{e^{7 x}}{7})$ $int e^(7x) sin 2x dx = int sin 2x d(e^(7x)/7)$

$\int e^{7 x} sin 2 x d x = \frac{e^{7 x} sin 2 x}{7} - \frac{2}{7} \int e^{7 x} cos 2 x d x$ $int e^(7x) sin 2x dx = (e^(7x)sin 2x)/7 - 2/7 int e^(7x)cos 2x dx$

and then again:

$\int e^{7 x} sin 2 x d x = \frac{e^{7 x} sin 2 x}{7} - \frac{2}{7} \int cos 2 x d (\frac{e^{7 x}}{7})$ $int e^(7x) sin 2x dx = (e^(7x)sin 2x)/7 - 2/7 int cos 2x d(e^(7x)/7)$

$\int e^{7 x} sin 2 x d x = \frac{e^{7 x} sin 2 x}{7} - \frac{2 e^{7 x} cos 2 x}{49} - \frac{4}{49} \int e^{7 x} sin 2 x d x$ $int e^(7x) sin 2x dx = (e^(7x)sin 2x)/7 - (2e^(7x)cos2x)/49 - 4/49 int e^(7x) sin 2x dx$

The same integral now appears on both sides of the equation and we can solve for it:

$\frac{53}{49} \int e^{7 x} sin 2 x d x = \frac{e^{7 x} sin 2 x}{7} - \frac{2 e^{7 x} cos 2 x}{49} + C$ $53/49 int e^(7x) sin 2x dx = (e^(7x)sin 2x)/7 - (2e^(7x)cos2x)/49 +C$

$\int e^{7 x} sin 2 x d x = \frac{e^{7 x} (7 sin 2 x - 2 cos 2 x)}{53} + C$ $int e^(7x) sin 2x dx = (e^(7x)(7sin 2x - 2cos2x))/53 +C$

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How do you find the integral of $e^{7 x} \cdot sin (2 x) d x$ $e^(7x)*sin(2x)dx$ ?

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Explanation:

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