Answer
$\frac{e^x}{2}(\sin x+ \cos x)+C$
Work Step by Step
Integrating by parts, we have
$I= \int e^{x}\cos xdx= $$e^{x}\sin x-\int e^{x}\sin xdx$$=e^{x}\sin x-I_{1}$
$I_{1}=\int e^{x}\sin xdx$
$= e^{x}(-\cos x)+\int e^{x}\cos xdx$
Substituting the value of $I_{1}$, we get
$I= e^{x}\sin x+ e^{x}\cos x- I$
or $2I= e^{x}(\sin x+ \cos x)$
or $I= \frac{e^x}{2}(\sin x+ \cos x)$
