Definite Integral from 0 to 1/sqrt(2) arcsin(x) OVER sqrt(1-x^2)

\[\int\limits_{0}^{1/\sqrt{2}} \arcsin(x)\div \sqrt{1-x ^{2}}\]

This is trivial. You need to use a substitution \(u=\sin^{-1}(x) \implies du=\frac{dx}{\sqrt{1-x^2}}\). That's: \(\int_0^{\frac{1}{\sqrt{2}}}\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}dx=\int_0^\frac{\pi}{4}udu=\frac{1}{2}u^2|_0^\frac{\pi}{4}=\frac{\pi^2}{32}.\)

straight up u- sub since if \[u=\sin^{-1}(x)\] \[du=\frac{du}{\sqrt{1-x^2}}\]

\[=\int\limits_{0}^{1/\sqrt(2)}\arcsin(x)d(\arcsin(x)=\arcsin(0)-\arcsin(\sqrt(2)/2)=\pi/4\]

because \[1/\sqrt(1-x^2)\] is derrivative of arcsin(x)

hmmm

Yeah, the result should be then \(\frac{1}{2}\arcsin^2(x).\)

still have to take "anti derivative" right?

I don't understand what you mean! How have you been satellite?!

good. i mean that you are right, you have to write either \[\frac{u^2}{2}\] or \[\frac{1}{2}\arcsin^2(x)\] before you can evaluate. how are you? where you been?

The idea that mathmagician is trying I think is that \(\int f^n(x) f'(x)dx=\frac{f^{n+1}(x)}{n+1}.\)

I'm good! I've been a little bit busy.

good to be busy, me as well.

ok....hold up everybody!!! Anwar....on your first statement, how did you change the interval on the integral? It goes from 1/sqrt2 to pi/4.....is that because that is the angle where sin IS 1/sqrt2? or another reason?

Yeah. We substitute \(u=\sin^{-1}(x)\), so we need to find the limits for u now. How to do that? It's simple, just plug the values you for \(x\). That means the lower limit is \(u=\sin^{-1}(0)=0\), and the upper limit will be \(u=\sin^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}.\)

you have*

ok so i get that if you u-sub you get \[\int\limits u du\]...but i don't get how that helps...isn't it still anti-product rule that we can't do? should I do integration by parts?

It's not necessarily that you get this form. You get this only if you have a function multiplied by its derivative, which I can say the simplest case like the one we have here.

ok yeah but where do i go from here?

OH. i didn't realize the du wasn't really part of it...just like the dx in normal ones. I think I got it now. thanks.

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