Integrate 1/3(x^2(50 - x^2)^(3/2))

[Stochastic13]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I tried by parts with no luck :(

 

[bob1182006]

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

 

[Stochastic13]

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

 

[bob1182006]

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

 

[SammyS]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I tried by parts with no luck :(

Do you mean: [itex]\displaystyle \int{\frac{1}{3x^2(50-x^{2})^{3/2}}}\,dx\ ?[/itex]

What did you try for parts?

A trig. substitution may work better.

 

[Stochastic13]

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

 

[Stochastic13]

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

How would you do it with trig sub?

 

[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.

 

[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.
[tex]\int \frac{1}{3x^2x^3(50x^{-2}-1)^{3/2}} dx[/tex]
u = 50x^(-2) - 1

 

[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.

 

[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.

 

[Stochastic13]

Thanks I'll give it a shot.

 

[Stochastic13]

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

 

[SammyS]

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

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

 

[Stochastic13]

Ok I'll try that as well, thanks.