00:01
We want to solve this initial value problem.
00:03
So what that really means is we want to find some function v, that when we take the derivative of it with respect to t, we will get a derivative of 8 over 1 plus t squared plus secan squared t.
00:16
And for our function, when we plug in 0, we get an output of 1.
00:20
So we really just have the 0 .01 on this.
00:24
So let's go ahead and integrate this first.
00:27
So we can get a v to pop up.
00:28
So i'm going to integrate each side with respect to.
00:31
T so we're going to get v is equal to so we can use the linearity of the integral to rewrite as eight times the integral of one plus one over t square plus the integral of secant square t d t now if we were to go ahead and look at our anti -derivative chart this first one is going to be arc tan or inverse tangent so tangent inverse of t plus and secant squared, well, that's going to be tangent.
01:07
Because we know if we take the derivative of tangent, we get second squared...