Evaluate : `int_0^(2pi)e^xcos(pi/4+x/2)dx`

Explanation

Take, `frac{x}{2}+frac{pi}{4}=theta so that x=2 theta-pi / 2 and dx=2 dtheta `

So,
`
I=2 int_{pi / 4}^{5 pi / 4} {e}^{2 theta-pi / 2} cos theta d theta `
`
=2 {e}^{-pi / 2} {I}_{1}`


Here `I_{1}=int_{pi / 4}^{5 pi / 4} e^{2 theta} cos theta d theta`
`

.=frac{1}{2} {e}^{2 theta} cos theta]_{pi / 4}^{5 pi / 4}+frac{1}{2} int_{pi / 4}^{5 pi / 4} {e}^{2 theta} sin theta d theta`
`
...