Find $e^{\Phi}$

let $b=\int_{\pi/2}^{\pi}\sin{x} dx$

$b=\int_{\pi/2}^{\pi}\sin{x} dx\\ b=-\cos{\pi} -(-\cos{\pi/2})\\ b=-(-1)-(-0)\\ b=1$

let $a = \int_{0}^{\pi/2} \cos{x} dx$

$a= \sin{\pi/2} - \sin{0}\\ a= 1-0\\ a=1$

use: $\displaystyle\int_{a}^{b} f(x) dx =0 \iff a=b\\$

then: $e^\Phi = e^0 = 1$

Both of the above definite integral bounds equal unity:

$\int_{\pi/2}^{\pi} \sin(x) dx = -\cos(x)|_{\pi/2}^{\pi} = 1$

$\int_{0}^{\pi/2} \cos(x) dx = \sin(x)|_{0}^{\pi/2} = 1$

which leads to $\Phi = 0 \Rightarrow e^{\Phi} = \boxed{1}.$