00:01
Section 7 .2, problem 37, we're dealing with integrals here that can be evaluated with integration by parts.
00:09
So that integration by parts formula tells us if i can look at an integran, which is a lot just like looking at a puzzle, and if i can fit those puzzle pieces together where it looks like two functions, u and dv, then i can integrate this by part using this integration by parts formula.
00:29
And my hope is the second integral is easier to work with.
00:33
So what i like to do first, let's just do the anti -derivative.
00:37
Let's worry about the limits of integration later.
00:40
So how would i tackle the arc sine derivative? so in this case, let u equal the arc sign of y, then du is going to be 1 over the square root of 1 minus y squared, and then dv is going to equal d y which tells me v is equal to y so this anti -derivative is going to be uv so that's y arc sine of y minus the integral of v -d -u so that's going to be y over one minus y squared and so now in order to work on this integral let's just make a substitution.
01:36
So let's let z equal 1 minus y squared.
01:43
Then dz is going to be minus 2y, d .y.
01:48
So to get a minus 2y, dy here, i can just modify this integral as such.
01:57
And so this turns into y, the arc sine of y, plus one half, and then this is the integral of 1 over the square root of z d z...