# How to integrate?

1. Jul 13, 2011

### Stochastic13

1. The problem statement, all variables and given/known data

How do you integrate: 1/3(x^2(50 - x^2)^(3/2))

2. Relevant equations

3. The attempt at a solution

I tried by parts with no luck :(

2. Jul 13, 2011

### bob1182006

Start out by taking out x^2 to get
$$\frac{1}{x^2x^3(50x^{-2}-1)^{3/2}}$$

3. Jul 13, 2011

### Stochastic13

Sorry, it's kinda hard to read, I still don't follow.

4. Jul 13, 2011

### bob1182006

Take out (x^2) from the denominator's root.
$$\int \frac{1}{3x^2(50-x^2)^{3/2}}dx = \int \frac{1}{3x^2x^3(50x^{-2}-1)^{3/2}} dx$$

5. Jul 13, 2011

### SammyS

Staff Emeritus
Do you mean: $\displaystyle \int{\frac{1}{3x^2(50-x^{2})^{3/2}}}\,dx\ ?$

What did you try for parts?

A trig. substitution may work better.

6. Jul 13, 2011

### Stochastic13

Oh, I see :) Very Nice, Thanks :)

7. Jul 13, 2011

### Stochastic13

For parts I tried u=x, dv=x(50-x^2)^(3/2)

How would you do it with trig sub?

8. Jul 13, 2011

### Stochastic13

actually I'm stuck after factoring out x^2 because I get: 1/3 * (50/(x^2) - 1)^(3/2), but I don't know how to proceed.

9. Jul 14, 2011

### bob1182006

The x's don't cancel outside of the 3/2 root.
The second step would be to do a substitution of everything inside the 3/2 root.
$$\int \frac{1}{3x^2x^3(50x^{-2}-1)^{3/2}} dx$$
u = 50x^(-2) - 1

10. Jul 14, 2011

### Stochastic13

The original equation is: (1/3)(x^2(50 - x^2)^(3/2)) when I take out x^2 I get: (1/3)(x^2*x^3(50/x^2 -1)^(1/3)) and if u= 50/x^2 -1 then du = -100/x^3 which doesn't help

P.S. Latex code that you are using doesn't work and it makes it real hard to understand what you mean, if you could just use the keyboard I would get a better idea of what you mean.

11. Jul 14, 2011

### bob1182006

Ah, sorry I thought it was all in the denominator.
For that one all I think of is a trig substitution, which isn't my best area.

But looking at it I'd think you want to try x = 5 sqrt(2) sin(u) so that you can get rid of the root into a cosine.

12. Jul 14, 2011

### Stochastic13

Thanks I'll give it a shot.

13. Jul 14, 2011

### Stochastic13

Yeah, it's a pretty nasty problem, I don't know why they would include it in our book without explanation.

14. Jul 14, 2011

### SammyS

Staff Emeritus
So, it really is: $\displaystyle \int{\frac{1}{3}x^2(50-x^{2})^{3/2}}\,dx\ ?$

For parts, I would use u = (1/3)(50-x^{2})^{3/2} & dv = x2dx.

15. Jul 14, 2011

### Stochastic13

Ok I'll try that as well, thanks.