Question

Evaluate the integral from 0 to (Pi/2) of (cosx * e^(sinx)) dx, just stuck

Answer 1

are you familiar with the wallis formula? i think it's applicable here...or if you want you can let u = sin x so du = cosx \[\int_0^{\pi/2} e^u du\] so it becomes just \[[e^u|_0^{\pi/2}\] \[e^{\sin x}|_0^{\pi/2}\]

Answer 2

so you get \[e^{\sin (90)} - e^{\sin (0)}\] \[e^1 - e^0\] \[e - 1\]

Answer 3

you get it right @Njabulod ??

Answer 4

Thanx alot, you made it look very easy for me totally understand it

Answer 5

heh that's what we do here ^_^