Integral 0 to pi e^cosx cos (sinx) dx

Hi,Shreya

Let u = sin(x). Then du = cos(x) dx. So you can now antidifferentiate e^u du. This
is e^u + C = e^sin(x) + C.

Then substitute your range 0 to pi.

e^sin (pi)-e^sin(0)

=0-0

=0