The anti-derivative `int e^x cos x dx` has to be determined.

Use integration by parts, which gives `int u dv = u*v - int v du`

Starting with `int e^x*cos x dx`

Let `dv = cos x dx => v = sin x`

`u = e^x => du = e^x dx`

This gives `int e^x cos x dx = e^x*sin x - int e^x*sin x dx`

Now take `int e^x*sin x dx`

Let `dv = sin x dx => v = -cos x`

`u = e^x => du = e^x dx`

This gives `int e^x*sin x dx = -e^x*cos x + int e^x* cos x dx`

`int e^x cos x dx`

=> `e^x*sin x - int e^x*sin x dx`

Substitute `-e^x*cos x + int e^x* cos x dx` for `int e^x sin x dx`

=> `e^x*sin x - (-e^x*cos x + int e^x* cos x dx )`

`int e^x cos x dx = e^x*sin x - (-e^x*cos x + int e^x* cos x dx )`

=> `2*int e^x cos x dx = e^x*sin x + e^x*sin x`

=>` int e^x*cos x dx = (e^x*sin x + e^x*sin x)/2`

**This gives **`int e^x*cos x dx = (e^x*sin x + e^x*sin x)/2`