How to integrate Sin(x)e^Cos(x) using substitution.

[donaldduck]

So a question for a test I just had was integrate by substitution:

Sin(x)e^Cos(x).

I did something like this:

Let u=Cos(x)

du=-sin(x) dx

∫sin(x)e^Cos(x) dx = ∫-e^u du

=∫-e^Cos(x) du

= -e ^cos (x) + c

Is that correct??

Thank you.

 

[clamtrox]

=∫-e^Cos(x) du

It's correct but this step is weird. You calculate the integral with respect to u, then substitute back AFTER you've integrated.

 

[donaldduck]

Thanks Clamtrox!

So I meant to write:
∫sin(x)e^Cos(x) dx = ∫-e^u du
=-e^u +c
=-e^cos(x) +c

 

[SammyS]

Thanks Clamtrox!

So I meant to write:
∫sin(x)e^Cos(x) dx = ∫-e^u du
=-e^u +c
=-e^cos(x) +c

Hello donaldduck. Welcome to PF !

That result looks good.

Check the answer by finding the derivative of the result .