Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

hmmm

Answer 7

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

Answer 8

still have to take "anti derivative" right?

Answer 9

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

Answer 10

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?

Answer 11

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

Answer 12

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

Answer 13

good to be busy, me as well.

Answer 14

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?

Answer 15

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

Answer 16

you have*

Answer 17

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?

Answer 18

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.

Answer 19

ok yeah but where do i go from here?

Answer 20

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.