What is the Integration by Parts Method for Solving Integrals?

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Homework Statement



\int\frac{x^3}{\sqrt{1-x^2}}dx

I have to use integration by parts on the above integral.

Homework Equations




The Attempt at a Solution



u=x^3
du=3x^2dx

dv=\frac{1}{\sqrt{1-x^2}}dx
v=arcsin (x)

=x^3arcsin (x)-3\int\ x^2arcsin (x)dx

u=arcsin (x)
du=\frac{1}{\sqrt{1-x^2}}dx

dv=x^2dx
v=\frac{1}{3}x^3

\int\frac{x^3}{\sqrt{1-x^2}}dx==x^3arcsin (x)-3[x^2 arcsin(x)-\frac{1}{3} \int\ \frac{x^3}{\sqrt{1-x^2}} dx

\int\frac{x^3}{\sqrt{1-x^2}}dx=x^3arcsin (x)-3x^2 arcsin(x) + \int\ \frac{x^3}{\sqrt{1-x^2}} dx

Here I was hoping I could move the integral over but, given the signs, that isn't going to work. Any tips on what course I should take instead?
 
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That isn't going to work because your 2nd application of by parts essentially undoes your first application. Can you think of choosing u and v in another way?
 
Choosing u = x2 works for integration by parts. This leaves you with a somewhat complicated dv, but one that you can integrate to get v, using an ordinary substitution.
 
So setting u=x^2 leaves you with dv=arcsin(x)dx. I can see what v would equal by looking at a table of integrals, but I have no idea how I would go about integrating arcsin(x) on my own. I was thinking maybe using Trig. Substitution but I wouldn't really know how to go about it.
 
No, u = x2, and dv = xdx/sqrt(1 - x2). I'm talking about starting from the original problem.
 
Sorry, I made a mistake.:(
 
Last edited:
Alright I got it, thanks for your help.
 
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