Answer
$\frac{1}{2}$
Work Step by Step
$u$ = $e^{-x}$
$u'$ = $-e^{-x}$
$v'$ = $cosx$
$v$ = $sinx$
$\int{e^{-x}cosxdx}$ = $e^{-x}sinx$ - $\int{-e^{-x}sinxdx}$ = $e^{-x}sinx$ + $\int{e^{-x}sinxdx}$
use integration by parts again
$u$ = $e^{-x}$
$u'$ = $-e^{-x}$
$v'$ = $sinx$
$v$ = $-cosx$
$\int{e^{-x}cosxdx}$ = $e^{-x}sinx+[-e^{-x}cosx-\int{-e^{-x}cosxdx}]$
$\int{e^{-x}cosxdx}$ = $\frac{1}{2}e^{-x}(sinx-cosx)+C$
$\int_0^{\infty}{e^{-x}cosxdx}$ = $\lim\limits_{R \to {\infty}}$$(\frac{sinR-cosR}{2e^{R}}+\frac{1}{2})$ = $\frac{1}{2}$
