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

Question

2457 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-31T11:45:40

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

Remembering that

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

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

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