which is a standard integral.
Now if you wanna prove such result, make the substitution
I need help integrating from 0 to 1, the integral of (41) all over (square root of 1-x^2). I think that this integral converges, but I'm not sure if I'm evaluating the integral right, I started by putting the 41 in front of the integral and maybe using trig substitution?! Ah please help me, I am out of it, with math!
Okay! I used x=sintheta and so I got (1) over the square root of (1-(sintheta squared) squared. That looks kind of weird, but then I got rid of the square root and got (1) over (1-sin(theta)^2). That is the same as (1) over cos(theta)^2, which is the same thing as secant squared, and then integrating that gets you tangent I think. So, I'm that far. I think I have to use the triangle, to change theta back to x. I think that is opposite over adjacent, which is (x) over (the square root of 1-x^2). I'm not sure where to go from there.
Hmm, okay So I have the integral of (1) over the square root of 1-sintheta squared and cos(theta) dtheta. I simplified that to be (1) over the square root cosine squared, cosine theta d theta. So I simplified that to be (1) over cosine theta, cosine theta d theta. I canceled out a cosine theta, and therefore got 1 dtheta. Can that be right? Integrating that, I would get just theta?