Answer
\[{ = \frac{{{e^x}\sin x}}{2} + \frac{{{e^x}\cos x}}{2} + C}\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{e^x}\cos xdx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = {e^x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,du = {e^x}dx \hfill \\
dv = \cos x\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,v = \sin x \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
\hfill \\
= {e^x}\sin x - \int_{}^{} {{e^x}\sin xdx} \hfill \\
{\text{integrate}}\,\,{\text{by}}\,\,{\text{parts}}\,\,{\text{again}} \hfill \\
\hfill \\
u = {e^x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,du = {e^x}dx \hfill \\
du = \sin xdx\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,v = - \cos x \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
\hfill \\
= {e^x}\sin x - \,\,\left[ { - {e^x}\cos x - \int_{}^{} {\,\left( { - \cos x} \right){e^x}dx} } \right] \hfill \\
\hfill \\
= {e^x}\sin x + {e^x}\cos x - \int_{}^{} {{e^x}\cos xdx} \hfill \\
\hfill \\
solving\,\,for\,\,\,\int_{}^{} {{e^x}\cos xdx} \hfill \\
\hfill \\
= 2\int_{}^{} {{e^x}\cos xdx} = {e^x}\sin x + {e^x}\cos x + C \hfill \\
\hfill \\
\int_{}^{} {{e^x}\cos xdx = \frac{{{e^x}\sin x}}{2} + \frac{{{e^x}\cos x}}{2} + C} \hfill \\
\end{gathered} \]
