# Simple integral

1. May 3, 2008

### Howsertal

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

Is there a simple way to evaluate

S x^5 / [rt(x^2-ax)] dx ?

That is, the indefinite integral of (x^5) / [square root of (x^2-ax)].

3. The attempt at a solution

My idea was to rewrite it as x^(9/2) / [rt(x-a)] and then do the substitution u = rt(x-a).
Then you get

S 2(u^2+a)^(9/2) du

But this too is difficult to evaluate.

Last edited: May 3, 2008
2. May 3, 2008

### tiny-tim

Welcome to PF!

Hi Howsertal! Welcome to PF!

Hint: it's always worth trying completing the square … that is, putting (x² - ax) in the form ((x-b)² - c).

3. May 3, 2008

### Howsertal

Hmm okay, so that puts it in the form

∫ x^5 / √[(x-(a/2))² - (a²/4)] dx

One possible substitution seems to be u = x-(a/2) but that puts it in the form

∫(u+(a/2))^5 / [u²-(a/2)²] du = ∫ (u+(a/2))^(9/2)) / [u-(a/2)] du

which doesnt seem alot better than what we started with. A trig substitution of the form
[x-(a/2)] = (a/2)secψ in the first integral above would yield

∫ [(((a/2)secψ) + (a/2))^5)/tanψ] secψtanψ d ψ

= ∫ (((a/2)secψ +(a/2))^5 secψ d ψ

which doesn't seem to yield easily either =/. Those are the only obvious substitutions I can think of. Is there a trick to this or am I just not seeing something? Thanks a lot for your help (btw having the list of symbols in your sig is a great idea!)

Last edited: May 3, 2008
4. May 3, 2008

### HallsofIvy

Staff Emeritus
secant= 1/cosine. And odd powers of trig functions are relative easy to integrate.

5. May 3, 2008

### tiny-tim

Hi Howsertal!
Looks ok to me …

Hint: what is (d/dψ)(sec^{n}ψ tanψ)?

(yes … I tried to get them to put the symbols on the Reply page … but this is almost as good … and I only use the signature if I need it! )

6. May 3, 2008

### Howsertal

Do you mean that we can get

∫ (((a/2)secψ +(a/2))^5 secψ d ψ

into an expression just involving odd powers of trig functions?

7. May 3, 2008

### Howsertal

Tiny-tim,

That derivative came out to
[sec^{n+2}ψ](1+n) - nsec^{n}ψ

which seems like it could be helpful but I'm not sure how. Sorry if I seem really dense.

Last edited: May 3, 2008
8. May 3, 2008

### tiny-tim

Hi Howsertal!

Well, you want to integrate (secψ +(a/2))^5 secψ, which is powers of secψ up to the sixth, and you now know that:

∫((n+1)sec^{n+2}ψ - nsec^{n}ψ) dψ = sec^{n}ψ tanψ

9. May 3, 2008

### Howsertal

Ah i see, thank you very much. So that's a nifty way of getting expressions for all the integrals of powers of secant recursively.

Last edited: May 3, 2008