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

Question

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

anonymous · Answer

crud dx should be in the neumorator  fyi for the last integral

anonymous · Answer

Don't forget 1. It should be xInvCos - ... You can multiply by 1, so you use integration by parts with InvCos and 1

anonymous · Answer

oh yeah it should be positive  but I still dont see what I can do next :l

experimentx · Answer

suppose x = sin u => dx = cos u du

anonymous · Answer

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 · Answer

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 · Answer

experiment what logically tells you that it should be x = sin(x)

anonymous · Answer

The equation is $$\int\limits_{?}^{?}u dv = u*v - \int\limits_{?}^{?} v du$$

anonymous · Answer

https://www.youtube.com/watch?v=ouYZiIh8Ctc

anonymous · Answer

I know how to apply integration by parts it isn't the issue

anonymous · Answer

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 · Answer

yes I Know

anonymous · Answer

and g'(x) = x, G(x) = x^(2)/2

anonymous · Answer

ugh

anonymous · Answer

It should be positive above on the right and x, not x^2 on top.

anonymous · Answer

sorry you are wrong brianshot3. watch the youtube video, I messed up the negative sign on the integral I get that though

anonymous · Answer

$$\int\limits_{?}^{?}\cos^{-1} xdx= x \cos^{-1} x  + \int\limits_{?}^{?}x/\sqrt{1-x ^{2}}$$

anonymous · Answer

no just no you have to take the antiderivtive of x watch the youtube video

anonymous · Answer

learn how to integrate by parts then try to help me

anonymous · Answer

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 · Answer

oh ok sorry Now I understand

anonymous · Answer

Sorry for being rude it was uncalled for just frustrated and stressed out :(

anonymous · Answer

It's fine

anonymous · Answer

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 · Answer

thanks for showing me that it helps me so much :)

anonymous · Answer

So,

u = x
du = dx

integral of dx/(1+u^(2))  = arcsin(x) + c

anonymous · Answer

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 · Answer

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 · Answer

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 · Answer

so the answer isn't

arccos(x)x + arcsin(x) + c

what am I doing wrong?