Find the indefinite integral.
\[\int\limits\frac{ \arcsin2x }{ \sqrt{1-4x^2} }dx\]
isn't the derivative of the numerator the denominator?
Is it? >_<
got stuck sorry
make a u substitution \[u=\arcsin(2x), du =\frac{2}{\sqrt{1-4x^2}}\] and you are mostly done
It's okay! Ironically, I was just going to seperate that integral and and go from there. :x
always a bad idea
>_< Really?
I don't know, I guess I don't really get it. :(
do you get the answer to this?
you have a function times its derivative, so the integral is easy
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...
you mean the derivative of arcsine
Or would it be as simple as just.... setting dx = du...
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
you got that or you want to go slow and do it step by step?
Whichever way is less complicated. :/
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
Oh.... yeah. Had to pull up the rule sheet. :)
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 \[\frac{1}{2}\int udu\]
anti derivative is \[\frac{1}{4}u^2\] sustitite back and get \[\frac{1}{4}(\arcsin(2x))^2\]
how we doing @fate?
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?
i spose
i find this stuff less interesting that eating plain rice crackers and drinking luke warm water
Lol.
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