First, consider the identity cos 2x = 2cos^2 x - 1. Use this identity to transform the integral
int (e^x cos^2 x) dx
into the integral
int (e^x (cos 2x + 1))/2 dx = 1/2 int e^x cos 2x dx + 1/2 int e^x dx
Finding int e^x dx is straightforward i.e. int e^x dx = e^x + C
We use integration by parts to find int e^x cos 2x dx. By LIATE, we integrate e^x and differentiate cos 2x:
int (e^x cos 2x) dx = e^x cos 2x - int e^x (–2 sin 2x) dx = e^x cos 2x + 2 int (e^x sin 2x) dx = e^x cos 2x + 2[(e^x sin 2x)-int e^x (2 cos 2x) dx] = e^x cos 2x + 2e^x sin 2x-4 int (e^x cos 2x) dx
Thus,
int (e^x cos 2x) dx = e^x cos 2x + 2e^x sin 2x-4 int (e^x cos 2x) dx
5int (e^x cos 2x) dx = e^x cos 2x + 2e^x sin 2x dx
int (e^x cos 2x) dx = 1/5 e^x cos 2x + 2/5 e^x sin 2x dx
Therefore,
int (e^x (cos 2x + 1))/2 dx = 1/2 (1/5 e^x cos 2x + 2/5 e^x sin 2x) + 1/2 e^x + C = 1/10 e^x cos 2x + 1/5 e^x sin 2x + 1/2 e^x + C
