Question

How do you integrate `int e^(sinpix)cospix` from `[0,pi/2]`?

Answer 1

`1/pi(e^(sin(pi^2/2))-1)`

Explanation:

Remembering that

`d/(dx)e^(sin(pix))=pi cos(pix)e^(sin(pix))` we have

`int cos(pix)e^(sin(pix))dx=1/pi(e^(sin(pi^2/2))-1)`