Recent content by 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.

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. :frown:

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?

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 doesn't seem a lot better than what we started with. A trig...

Homework Statement 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)].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...