00:01
All right, hello, everybody.
00:02
So in this problem, we're asked to find the indefinite integral or the anti -derivative of this function here, one over square root of 1 plus x squared.
00:12
Unfortunately, there are multiple ways to solve this, but unfortunately, we don't have an easy way to solve it, like just a simple substitution, or we just know the integral just by looking at it.
00:25
And so we're going to have to use something special called a trig substitution to be able to do this.
00:29
Because if you notice, in the bottom i have 1 plus 8.
00:31
X squared that is in a form that uh kind of helps me know that i could use a trig substitution and use a pythagrin identity to rewrite this integral into something that's more um recognizable and that i can actually take the anti derivative of pretty easily and so for this problem we're going to use a trick substitution that x equals tangent of theta and so that's going to give us later we're going to need this triangle remember all the trig functions come from our right triangle is tangent of theta um equals opposite over adjacent side.
01:03
And so that's telling me that the opposite over adjacent side of theta in this right triangle would be x over any number is over one.
01:11
And so that'd be opposite side is x over one is my adjacent side.
01:15
And so that's going to be helpful later because we're going to use the pythagrin theorem, x squared plus one squared equals my hypotenuse squared, which means that my hypotenuse equals the square root of x squared plus one.
01:30
And so i'm going to go ahead and write that to that this hypothesis is going to be the square of x squared plus one, which will help us later.
01:39
All right? so i want to find the, i want to use my u -substitution x equals 10 theta, which means that my d -x, my derivative, is going to be secant -square to theta, d -theta, just like that, because i took the derivative of tangent, which is c -squared.
01:57
So now i can rewrite this integral in terms of theta, it would be one over the square.
02:03
Root of 1 plus tangent squared theta times secant squared of theta d theta.
02:12
That is a square d theta, just like that.
02:19
Why we did this is because like i said, there's a pythagin identity that says one plus tangent squared of theta equals secant square to theta.
02:28
And so now i get to replace even more, the denominator in the square root, one plus tangent squared actually becomes secant squared times secant squared d theta.
02:42
Let's like that.
02:43
Well, the squared is something squared is itself, and so that's going to be one over secant times secant square theta d theta...