How do you integrate $int e^xsin4xdx$ using integration by parts?

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Answer 1 · 2015-11-03T14:01:04

$inte^xsin(4x)dx=frac{e^xsin(4x)-4e^xcos(4x)}{17}+c$ ,
where $c$ is the integration constant.

$inte^xsin(4x)dx=-1/4inte^xfrac{d}{dx}(cos(4x))dx$

$=-1/4[e^xcos(4x)-intfrac{d}{dx}(e^x)cos(4x)dx]$

$=1/4inte^xcos(4x)dx-1/4e^xcos(4x)$

$=1/16inte^xfrac{d}{dx}(sin(4x))dx-1/4e^xcos(4x)$

$=1/16[e^xsin(4x)-intfrac{d}{dx}(e^x)sin(4x)dx]-1/4e^xcos(4x)$

$=1/16e^xsin(4x)-frac{1}{16}inte^xsin(4x)dx-1/4e^xcos(4x)$

$frac{17}{16}inte^xsin(4x)dx=1/16e^xsin(4x)-1/4e^xcos(4x)+c_1$ ,
where $c_1$ is the constant of integration.

$inte^xsin(4x)dx=frac{e^xsin(4x)-4e^xcos(4x)}{17}+c_2$ ,
where $c_2=frac{16c_1}{17}$ .

How do you integrate int e^xsin4xdxint e^xsin4xdx using integration by parts?