The integral of 0 to π/2 cos x e^sinx dx = a - b. Where a =____? And b =___?

Answer 1

∫ cos(x) e^sin(x) dx = a - b (from 0 to π/2)

u = sin(x) and du = cos(x)

∫ e^u du = e^u = e^sin(x)

e^sin(x)| evaluated from (0 to π/2) =

e^sin(π/2) - e^sin(0) = e¹ - e⁰ = e - 1

So a = e and b = 1.

Feel free to reach out to me if you have questions.

Answer 2

∫cos(e)^sin(x) dx=. 1. I am going to use u substitution u=sin(x). so du=cos (x)dx.

2. I am going to substitute sin(x) and cos(x)dx. as follow. ∫e^u du=

3.Remember that integration of e^u. = e^u note: sin(0)=0 , sin (∏/2) =1

4.e^u from 0 to ∏/2 =. e^sin(x). from 0 to ∏/2 = e^sin (∏/2) - e^sin(0)= e- 1

so the integration of ∫cos(x)(e^sin(x)dx= e-1. or 1.71828. graphically using ti-84,