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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
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