What is the Integration by Parts Method for Solving Integrals?

[3.141592654]

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?

 

[Mandark]

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?

 

[Mark44]

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.

 

[3.141592654]

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.

 

[Mark44]

No, u = x², and dv = xdx/sqrt(1 - x²). I'm talking about starting from the original problem.

 

[Jophil]

Sorry, I made a mistake.:(

 

[3.141592654]

Alright I got it, thanks for your help.