00:03
Okay, on this problem, i'm practicing integrating t -trig functions that are a little more complex using substitution.
00:12
So i know the integral of tangent is ln absolute secant, but i have a number inside, an expression inside, so chain rule is going to come into it in this problem.
00:32
So i'm going to let you be the inside of the tangent.
00:36
That makes du e to the negative 2x times the derivative of what's inside times dx.
00:52
So i can replace the tangent e to the negative 2x with a tangent u, and i can replace e to the negative 2x with negative a half of a du.
01:11
I need to divide both sides by the negative 2 to make that work out.
01:15
So the e to the negative 2x dx is a negative 1 half d .u.
01:22
And now i'm ready to integrate.
01:26
The integral of tangent is ln absolute secant...