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

Question

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

mathmagician · Answer

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

mathmagician · Answer

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

anonymous · Answer

hmmm

anonymous · Answer

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

anonymous · Answer

still have to take "anti derivative" right?

anonymous · Answer

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

anonymous · Answer

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?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

good to be busy, me as well.

anonymous · Answer

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?

anonymous · Answer

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 $u=\sin^{-1}(\frac{1}{\sqrt{2}})=\frac{\pi}{4}.$

anonymous · Answer

you have*

anonymous · Answer

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?

anonymous · Answer

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.

anonymous · Answer

ok yeah but where do i go from here?

anonymous · Answer

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.