Answer
$\dfrac{e^x}{5}[2 \sin 2x+\cos 2x]+C$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
$\int e^x \cos 2x dx=(\dfrac{1}{2})e^x \sin 2x - (\dfrac{1}{2})\int e^x \sin 2x dx=(\dfrac{4}{2})e^x \sin 2x - (\dfrac{4}{2})\int e^x \sin 2x dx=(2)e^x \sin 2x - (2)\int e^x \sin 2x dx$...(1)
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
Also, $\int e^x \cos 2x dx=e^x \cos 2x+2\int e^x \sin 2x dx$...(2)
Now, add equations (1) and (2), we get
$(5) \int e^x \cos 2x dx=2e^x \sin 2x+e^x \cos 2x+C$
Hence, $\int e^x \cos 2x dx=\dfrac{e^x}{5}[2 \sin 2x+\cos 2x]+C$
