Substitution technique for integral

[mathaintmybag]

Homework Statement

I need to use a substitution technique and then the table of integration to integrate the following:
∫ x⁵ √(x⁴ – 4) dx

and i am given a hint which is x⁵ = (x³)(x²)

I would assume that
u = x⁴ – 4, then du = 4x³
and that at some point a² = 4 and a = 2

However, the x⁵ = (x³)(x²) is confusing the heck out of me. Need help!

 

[Mark44]

What do a^2 and a have to do with anything?

The hint looks to me like it's guiding you to use integration by parts. Is that a technique that you have learned yet? I wouldn't advise it unless you have exhausted the possibilities for ordinary substitutions, however.

 

[LCKurtz]

What, so you have a limited table of integrals that doesn't include this and you are suggested a substitution to give you a form you can find in your table?

I think the point of the x⁵ = x³x² is to use the x³ as you have suggested with your u = x⁴ - 4. You can substitute for the extra x² like this:

x⁴ = u+4, x² = sqrt(u+4)

This will give you a pure u integral that may be in your table.

 

[mathaintmybag]

The reason i used a^2 = 4 and a = 2 is because in the book they used a number of different examples where they changed the number to a letter.

We have touched on integration by parts but everything we did learned involved using
e^x so I am not really sure how to use it here.

 

[thepatient]

It can be done with trig substitution I believe. Setting x^4/4 to (x^2/2)^2 and substituting for cos(theta).

∫ x5 √(x4 – 4) dx

∫ x5 2√(x4/4 – 1) dx

∫ x5 2√((x2/2)^2 – 1) dx

d/dx cosΘ = d/dx x^2/2]
[2cosΘ] = x^2]
[4cos^2(Θ)] = x^4]

-sinΘ dΘ = 2x dx∫ x5 2√((x2/2)^2 – 1) dx

∫ x^4(2x)√((x2/2)^2 – 1) dx
∫4cos^2(Θ) (-sinΘ)sinΘ dΘAnd then from there it's pretty simple... Is that right?

 

[mathaintmybag]

I haven't learned how to use trig substitutions yet so that will not help. Thanks for tip anyways.

 

[Bohrok]

x⁵√(x⁴ - 4) = x²∙x³√(x⁴ - 4)
and you can use integration by parts on that.

 

[mathaintmybag]

Would this be correct?
∫ x⁵ √(x⁴ – 4) dx
= ∫ ((x²*x²*x √(x⁴ – 4)) dx


Let u = x², then du/2 = xdx
= ∫ 1/2 (u²√u² – 4) du

Let a = 4 and a² = 2
= ∫(u²√u² – a²) du
= [x(u² – a²)^3/2 / 4] + [a²x√(u² – a²) / 8] – (a⁴/8) ln(x + √(u² – a²))

 

[Mark44]

It looks like you are using formula #10 here.

If so, you final result should not have u or a in it. You also need the constant of integration, which is needed in all indefinite integrals.