substitution technique for integral

[mathaintmybag]

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

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



2. Relevant equations



3. The attempt at a solution

[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
ex 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 dΘ


And then from there it's pretty simple... Is that right?

[mathaintmybag]

I haven't learnt 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 (http://www.sosmath.com/tables/integral/integ12/integ12.html).

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.