anonymous
  • anonymous
Integrate by parts \[\int\limits_{}^{} \arccos(x)dx = (\arccos(x)x^{2}/2) - \int\limits_{}^{} x^{2}/\sqrt{1-x^{2}} dx\] I don't know what do do now :l
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
crud dx should be in the neumorator fyi for the last integral
anonymous
  • anonymous
Don't forget 1. It should be xInvCos - ... You can multiply by 1, so you use integration by parts with InvCos and 1
anonymous
  • anonymous
oh yeah it should be positive but I still dont see what I can do next :l

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experimentX
  • experimentX
suppose x = sin u => dx = cos u du
anonymous
  • anonymous
You can multiply anything by 1 and get the same number, so you are using 1 which integrates to x. \[x \cos^{-1} x - \int\limits_{?}^{?}(derivative of InvCos)*(x)\]
anonymous
  • anonymous
brian it should be multiplied by x^(2)/2, if you recal the integration by parts intergral of f(x)g'(x) = f(x)g(x) - integral of f'(x)g(x)
anonymous
  • anonymous
experiment what logically tells you that it should be x = sin(x)
anonymous
  • anonymous
The equation is \[\int\limits_{?}^{?}u dv = u*v - \int\limits_{?}^{?} v du\]
anonymous
  • anonymous
https://www.youtube.com/watch?v=ouYZiIh8Ctc
anonymous
  • anonymous
I know how to apply integration by parts it isn't the issue
anonymous
  • anonymous
He gives the same equation. You have to treat dv as 1 and InvCos as u. Int 1= x so... xInvCos - Int of (Derivative of InvCos * x) The derivative of InvCos = \[-1/ \sqrt{1-u^{2}}\]
anonymous
  • anonymous
yes I Know
anonymous
  • anonymous
and g'(x) = x, G(x) = x^(2)/2
anonymous
  • anonymous
ugh
anonymous
  • anonymous
It should be positive above on the right and x, not x^2 on top.
anonymous
  • anonymous
sorry you are wrong brianshot3. watch the youtube video, I messed up the negative sign on the integral I get that though
anonymous
  • anonymous
\[\int\limits_{?}^{?}\cos^{-1} xdx= x \cos^{-1} x + \int\limits_{?}^{?}x/\sqrt{1-x ^{2}}\]
anonymous
  • anonymous
no just no you have to take the antiderivtive of x watch the youtube video
anonymous
  • anonymous
learn how to integrate by parts then try to help me
anonymous
  • anonymous
You can't take the antiderivative of x because x isn't in the equation to start with. You only have InvCosx. I know how to do this, I'm in multivariable calculus.
anonymous
  • anonymous
oh ok sorry Now I understand
anonymous
  • anonymous
Sorry for being rude it was uncalled for just frustrated and stressed out :(
anonymous
  • anonymous
It's fine
anonymous
  • anonymous
oh with x/(1-x)^(1/2) I can simply sub u into the equation and use the identity dx/(a^(2)-x^(2)) = arcsin(x/a), a>0
anonymous
  • anonymous
thanks for showing me that it helps me so much :)
anonymous
  • anonymous
So, u = x du = dx integral of dx/(1+u^(2)) = arcsin(x) + c
anonymous
  • anonymous
The \[\int\limits_{?}^{?}du/\sqrt{1-u ^{2}} = ArcSin + C\] It looks like you mixed up the equations for the Inv functions. The one you have is ArcTan + C
anonymous
  • anonymous
how did I mess up the inv functions? arcsin(x) integral of derivative https://www.wolframalpha.com/input/?i=integrate+1%2F%281^%282%29+-+x^%282%29%29^%281%2F2%29 derivative https://www.wolframalpha.com/input/?i=d%2Fdx+arccos%28x%29 arctan(x) integral of derivative https://www.wolframalpha.com/input/?i=integrate+1%2F%28x^%282%29+%2B+1%29 derivative https://www.wolframalpha.com/input/?i=d%2Fdx+arctan%28x%29 The rules I apply seem fine, but if I'm missing something please point it out :)
anonymous
  • anonymous
You wrote dx/(1+u^(2)) = arcsin(x) + c. That doesn't match up with the functions you just gave. No Sqrt and it should be 1- u^2
anonymous
  • anonymous
so the answer isn't arccos(x)x + arcsin(x) + c what am I doing wrong?

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