Integrate x^2/Sqrt[1 - x^2] - Solve 0=0

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The integral of the function \( \frac{x^2}{\sqrt{1-x^2}} \) can be solved using both substitution and integration by parts methods. The substitution \( x = \sin u \) simplifies the integral significantly, while integration by parts requires careful selection of \( u \) and \( dv \). Specifically, choosing \( u = x \) and \( dv = \frac{x}{\sqrt{1-x^2}} \, dx \) is essential for applying the integration by parts formula correctly. The discussion highlights the importance of proper notation and understanding the components of integration techniques.

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\!\(∫x^2/Sqrt[1 - x^2] \[DifferentialD]x\)

I need to find the integral of
(x^2)/ Sqrt(1-(x^2))
if the above doesn't work properly
integration by parts results in 0=0 how do i do this?
 
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You mean

\int \frac{x^{2}}{\sqrt{1-x^{2}}} \ dx

How about the substitution x=\sin u [/tex] and then a nice trigonometrical identity involving a double angle...?<br /> <br /> Daniel.
 
ok, i got it using x= sinU
can you give me a hint how to do it using integration by parts?
 
Integration by parts uses

\int u \ dv = uv - \int v \ du

choose u = x and

v = \frac{x}{\sqrt{1-x^2}}

Edit: That should be dv = (x / \sqrt{1-x^2}) \ dx!
 
Last edited:
Not really,Data.U needn't specify "u" & "v",but the factors in the LHS,"u" & "dv"...

So

u=x \ \mbox{and} \ dv=\frac{x}{\sqrt{1-x^{2}}} \ dx

Daniel.

P.S.Data,u see the difference,right...?:rolleyes:
 
indeed, silly me :-p~
 
Last edited:

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