How do you integrate $\int cos 2 x \cdot e^{- x}$ $int cos2x*e^-x$ by integration by parts method?

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Answer 1 · 2016-11-24T09:02:25

The answer is $= \frac{e^{- x} (2 sin 2 x - cos 2 x)}{5} + C$ $=(e^(-x)(2sin2x-cos2x))/5+C$

$intu'v=uv-intuv'$

$v=cos2x$ $=>$ , $v'=-2sin2x$

$u'=e^(-x)$ , $=>$ , $u=-e^(-x)$

$inte^(-x)*cos2xdx=-e^(-x)*cos2x-2inte^(-x)sin2xdx$

We apply once more the integration by parts

$v=sin2x$ , $=>$ , $v'=2cos2x$

$u'=e^(-x)$ , $=>$ $u=-e^(-x)$

so, $2inte^(-x)sin2xdx=-2e^(-x)sin2x+4inte^(-x)cos2xdx$

So,

$inte^(-x)*cos2xdx=-e^(-x)*cos2x-(-2e^(-x)sin2x+4inte^(-x)cos2xdx)$

$inte^(-x)*cos2xdx=$

$-e^(-x)*cos2x+2e^(-x)sin2x-4inte^(-x)cos2xdx$

$5inte^(-x)*cos2xdx=-e^(-x)*cos2x+2e^(-x)sin2x$

$inte^(-x)*cos2xdx=(e^(-x)(2sin2x-cos2x))/5+C$

How do you integrate int cos2x*e^-x∫cos2x⋅e−xint cos2x*e^-x by integration by parts method?