Mathematics 16 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (mathmagician):

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

OpenStudy (mathmagician):

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

OpenStudy (anonymous):

hmmm

OpenStudy (anonymous):

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

OpenStudy (anonymous):

still have to take "anti derivative" right?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

good to be busy, me as well.

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you have*

OpenStudy (anonymous):

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?

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

ok yeah but where do i go from here?

OpenStudy (anonymous):

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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