How do I find the integral $int(cos(x)/e^x)dx$ ?

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Answer 1 · 2014-09-21T14:53:25

$int{cosx}/e^xdx={e^{-x}}/2(sinx-cosx)+C$

Let us look at some details.

Let

$I=int{cos x}/{e^x}dx$

by rewritng $1/e^x$ as $e^{-x}$ ,

$=int e^{-x}cos x dx$

by Integration by Parts

Let $u=e^{-x}$ and $dv=cosx dx$
$Rightarrow du=-e^{-x}dx$ and $v=sin x$

$=e^{-x}sinx+int e^{-x}sinx dx$

by another Integration by Parts,

Let $u=e^{-x}$ and $dv=sinx dx$
$Rightarrow du=-e^{-x}dx$ and $v=-cosx$

$=e^{-x}sinx-e^{-x}cosx-I$

So, we have

$I=e^{-x}(sinx-cosx)-I$

by adding $I$ ,

$2I=e^{-x}(sinx-cosx)$

by dividing by 2,

$I={e^{-x}}/2(sinx-cosx)+C$

How do I find the integral int(cos(x)/e^x)dxint(cos(x)/e^x)dx ?