Answer
\[\frac{e^x}{2}\left(\cos x+\sin x\right)+C\]
Work Step by Step
Let \[I=\int e^x \cos x\: dx\;\;\;\ldots (1)\]
Using integration by parts
\[I=\cos x\int e^x \: dx-\int\left((\cos x)'\int e^x \:dx\right)dx\]
\[I=(\cos x )\: e^x+\int(\sin x) \: e^x \:dx\]
Again using integration by parts
\[I=e^x \cos x+\sin x\int e^x dx-\int \left((\sin x)'\int e^x\:dx\right)dx\]
\[I=e^x \cos x+e^x \sin x-\int (\cos x)\: e^x\:dx+C\]
Where $C$ is constant of integration
\[I=e^x (\cos x+\sin x)-I+C\]
\[I=\frac{e^x}{2}(\cos x+\sin x)\]
Hence \[I=\frac{e^x}{2}(\cos x+\sin x)\].
