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 Calculus Calculus Math Forum

 February 19th, 2008, 09:35 AM #1 Newbie   Joined: Feb 2008 Posts: 3 Thanks: 0 integrating squareroot[ (z/x) - 1 ] dx I'm having trouble integrating the form that's in the subject squareroot[ (z/x) - 1 ] dx from 0 to infinity anyone willing to share some insight on where to start? is there some identity for this form? February 19th, 2008, 10:41 AM #2 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 First of all, the integral does not converge on 0 to infinity. This isn't so hard to see since the function it self isn't defined for x > z. Still, let's try to find the integral it self, not concerning about the boundaries (I'll do the calculation replacing infinity with z). First, use integration by parts. Using: v=sqrt(z/x-1) u'=1 this gives you: I=x sqrt(z/x-1)+z/2 int[1/(x sqrt(z/x-1))dx] now we have another integral that can be solved using substitution: t=sqrt(z/x-1) This gives you: I=x sqrt(z/x-1)-z arctan(sqrt(z/x-1))+C Now, let us do the boundaries: lim(I(x),x->z) = 0 lim(I(x),x->0) = -z arctan(infinity)=-z pi/2 so: I form 0 to z = z pi/2 for all z in R Tags integrating, squareroot, z or x Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post tuetue Calculus 14 February 10th, 2014 10:38 AM BenFRayfield Number Theory 1 September 30th, 2013 04:58 AM warnexus Calculus 4 March 3rd, 2013 12:47 PM MathematicallyObtuse Calculus 10 November 1st, 2010 04:04 PM ElMarsh Algebra 8 July 21st, 2009 04:52 PM

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