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November 28, 2014

November 28, 2014

Posted by **Heather** on Thursday, October 4, 2012 at 1:28pm.

- Calculus 2 -
**Steve**, Thursday, October 4, 2012 at 2:58pmusing a rather than x

_{0}for ease of typing,

∫sinh(p ln(x/a)) dx

recall that sinh(u) = (e^u - e^-u)/2

so, since

e^(p ln(x/a)) = (e^(ln(x/a)))^p = (x/a)^p

and, since e^-u = 1/e^u,

e^(-p ln(x/a)) = (a/x)^p

sinh(p ln(x/a)) = 1/2 (x^p/a^p - a^p/x^p)

integrate that to get

1/2 (x^(p+1)/((p+1)(a^p)) + a^p/(p-1)x^(p-1)))

(p-1)x^(2p+1) + (p+1)xa^(2p)

--------------------------------------

2(p^2-1) a^p x^p

you can play around with that if you want. If you get really ambitious, you might even end up with

letting u = p*ln(x/a),

x(p cosh(u) - sinh(u))/(p^2-1)

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