π2π4ecotxsin2xdx=?

I used u-sub for this one, and I got 1212e2. However, the correct answer is 1e.

Is my answer wrong or I need to simplify it? If so, how?

Thank you!

1 Answer
Apr 10, 2018

π2π4ecotxsin2xdx=1e

Explanation:

For π2π4ecotxsin2xdx, let u=cotx

then du=csc2xdx and upper and lower limits are 0 and 1 respectively. And

π2π4ecotxsin2xdx

= 01eudu

= [eu]01

= 1e