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Mathematics 15 Online
OpenStudy (moonlitfate):

Find the indefinite integral.

OpenStudy (moonlitfate):

\[\int\limits\frac{ \arcsin2x }{ \sqrt{1-4x^2} }dx\]

OpenStudy (anonymous):

isn't the derivative of the numerator the denominator?

OpenStudy (moonlitfate):

Is it? >_<

OpenStudy (anonymous):

got stuck sorry

OpenStudy (anonymous):

make a u substitution \[u=\arcsin(2x), du =\frac{2}{\sqrt{1-4x^2}}\] and you are mostly done

OpenStudy (moonlitfate):

It's okay! Ironically, I was just going to seperate that integral and and go from there. :x

OpenStudy (anonymous):

always a bad idea

OpenStudy (moonlitfate):

>_< Really?

OpenStudy (moonlitfate):

I don't know, I guess I don't really get it. :(

OpenStudy (anonymous):

do you get the answer to this?

OpenStudy (anonymous):

you have a function times its derivative, so the integral is easy

OpenStudy (moonlitfate):

No, I mean... the only thing I had noticed so far was that it looks like the the integral for arcsin. I mean the bottom does, so I recognize that...

OpenStudy (anonymous):

you mean the derivative of arcsine

OpenStudy (moonlitfate):

Or would it be as simple as just.... setting dx = du...

OpenStudy (anonymous):

you should get \[\frac{1}{4}\left(\arcsin(2x)\right)^2\] instantly in your head and to see that it is correct, take the derivative and get what you started with

OpenStudy (anonymous):

you got that or you want to go slow and do it step by step?

OpenStudy (moonlitfate):

Whichever way is less complicated. :/

OpenStudy (anonymous):

ok you recognize that the derivative of \[\arcsin(2x)\] is \[\frac{2}{\sqrt{1-4x^2}}\] right? you need to basically know that at the start, otherwise you cannot do the problem, so i will assume that you know it

OpenStudy (moonlitfate):

Oh.... yeah. Had to pull up the rule sheet. :)

OpenStudy (anonymous):

once you know that you write the following \[u=\arcsin(2x), du =\frac{2}{\sqrt{1-4x^2}}dx,\frac{1}{2}du=\frac{1}{\sqrt{1-4x^2}}dx\] and your integral becomes the simple \[\frac{1}{2}\int udu\]

OpenStudy (anonymous):

anti derivative is \[\frac{1}{4}u^2\] sustitite back and get \[\frac{1}{4}(\arcsin(2x))^2\]

OpenStudy (anonymous):

how we doing @fate?

OpenStudy (moonlitfate):

I kind of understand. I'm sure as I see more problems and try to work through them, this will make more sense. :) Practice makes perfect, right?

OpenStudy (anonymous):

i spose

OpenStudy (anonymous):

i find this stuff less interesting that eating plain rice crackers and drinking luke warm water

OpenStudy (moonlitfate):

Lol.

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