Question

integral problem

Answer 1

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

Answer 2

I know that the 1/sqrt(1-x^2) is part of the integral for arcsin but I am not sure how to remove the arcsin using substitution

Answer 3

Alright, so this is more simple than you might think! You probably know about this. So, using u-substitution with u being arcsin(x) and du being 1/sqrt(1-x^2)dx, what do you get?

Answer 4

the integral of u?

Answer 5

so 1/2u^2+c

Answer 6

\[\int\limits_{0}^{1/2}\frac{\arcsin(x)}{\sqrt{1-x^2}} =\int\limits_{0}^{1/2} \arcsin(x)\frac{1}{\sqrt{1-x^2}}dx=\int\limits_{0}^{1/2}u \ du\]
(Limits of integration not changed, but you can if you want, I prefer converting back.)

Yep, correct! However...

Are you sure about that +C?

Answer 7

Yeah this is where Zarkon knit-picks on my integration haha <.<

Answer 8

Hah. I get points off for it from my profs, just letting you know so it doesn't happen to you, too.

Answer 9

I would say let \(\sf u = sin^{-1}(x)\)

Answer 10

It makes the problem look prettier, and easier to take the integral of.

Answer 11

That's what he did, didn't he?

Or are you just talking about the notation of arcsin(x) vs sin^{-1}(x)?

Answer 12

Oh did he do that? I wasn't paying much attention.

Answer 13

recon14193 Best Response Medals 0
so 1/2u^2+c
8 minutes ago


^ YESSS YOU GOT IT.

Answer 14

so I have \[\frac{ 1 }{ 2 }(\frac{ \pi }{ 6 })^2\]

Answer 15

which gives me \[\frac{ \pi ^{2} }{ 72 }\]

Answer 16

thank you for all your help

Answer 17

http://i.imgur.com/Ay29K.jpg

Answer 18

\[\sf x= 0 \implies u=0 ~ , ~ x=\frac{1}{2} \implies .... whatever ~ this~ u =...\]

Answer 19

Easier to change limits at first before integration, at least for me anyway. It keeps your problem looking consistent.