Answer
$\dfrac{e^x}{5}[2 \sin 2x+\cos 2x]$
Work Step by Step
Here, $I_1=\int e^x \cos 2x dx=\dfrac{1}{2}e^x \sin 2x - \dfrac{1}{2}\int e^x (\sin 2x) dx$
Now, do multiplication by $4$, we have
$\implies I_1=(\dfrac{4}{2})e^x \sin 2x - (\dfrac{4}{2})\int e^x \sin 2x dx=2(e^x) (\sin 2x) - 2 )\int (e^x) (\sin 2x) dx$
Let us consider that
$I_2=\int (e^x) (\cos 2x) dx=(e^x) (\cos 2x)+2 \int (e^x) (\sin 2x) dx$
Now, $I=I_1+I_2$
or, $I=2(e^x) (\sin 2x)+(e^x)( \cos 2x)$
Thus, $I=\dfrac{e^x}{5}[2 \sin 2x+\cos 2x]$
