How do you find the integral of $e^(7x)*sin(2x)dx$ ?

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Answer 1 · 2018-04-11T15:13:18

$int e^(7x) sin 2x dx = (e^(7x)(7sin 2x - 2cos2x))/53 +C$

Integrate by parts:

$int e^(7x) sin 2x dx = int sin 2x d(e^(7x)/7)$

$int e^(7x) sin 2x dx = (e^(7x)sin 2x)/7 - 2/7 int e^(7x)cos 2x dx$

and then again:

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

$int e^(7x) sin 2x dx = (e^(7x)sin 2x)/7 - (2e^(7x)cos2x)/49 - 4/49 int e^(7x) sin 2x dx$

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

$53/49 int e^(7x) sin 2x dx = (e^(7x)sin 2x)/7 - (2e^(7x)cos2x)/49 +C$

$int e^(7x) sin 2x dx = (e^(7x)(7sin 2x - 2cos2x))/53 +C$

How do you find the integral of e^(7x)*sin(2x)dxe^(7x)*sin(2x)dx?