00:01
First thing i notice here, i have a lot of e to the x's.
00:05
So i'm going to let u be e to the x.
00:08
And then du is going to be e to the x dx.
00:12
So i'm going to rewrite this equation as sine inverse e to the x.
00:21
And then move this e to the x over here with the dx.
00:26
And then i'm going to write this as 1 minus e to the x to the 2, because that's e to the 2x.
00:35
Now i'm going to make the substitution.
00:37
So that's sine inverse of u.
00:40
Over here i have du over the square root of 1 minus u squared.
00:49
Now it still looks hard, but then all of a sudden i remember that if i take the derivative of the sine inverse of u, i get 1 over 1 minus u squared...