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.