integral of 1/root(sin) ?
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From: Kenneth Bull on 27 Sep 2009 18:55 Does 1/root(sinx) have an antiderivative?
From: Robert Israel on 27 Sep 2009 19:11 > Does 1/root(sinx) have an antiderivative? I assume you mean the square root. Yes, of course it does (on any interval of real numbers or simply-connected complex domain where it is continuous). If you want an explicit one, for 0 < x < pi/2 try F(x) = sqrt(2)*EllipticF(sqrt(2*sin(x)/(1+sin(x))), 1/sqrt(2)) (using Maple's conventions for elliptic functions: Mathematica's answer will look different). -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: Kenneth Bull on 27 Sep 2009 19:35 I am trying to do definite integral of 1/sqrt(sinx) from 0 to pi/2, do I have to use an improper integral since sinx not defined at 0 ?
From: Kenneth Bull on 27 Sep 2009 19:35 I mean, 1/root(sinx) not defined at 0
From: David W. Cantrell on 27 Sep 2009 20:18
Kenneth Bull <kenneth.bull(a)gmail.com> wrote: > I am trying to do definite integral of 1/sqrt(sinx) from 0 to pi/2, > do I have to use an improper integral since sinx not defined at 0 ? Well, yes, it's an "improper" integral. Its value can be expressed nicely using the Gamma function: 2 sqrt(pi) Gamma(5/4)/Gamma(3/4) which is 2.622057554... As you could deduce from Robert's earlier response, its value can also be expressed in terms of an elliptic integral. David |