My Math Forum Substitution Required

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 September 15th, 2016, 08:55 PM #1 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra Substitution Required I'm trying to find a non-trigonometric substitution with which to compute the integral $$\int (x^2+1)^{-\frac32}\,\mathrm dx$$ Any thoughts?
 September 16th, 2016, 01:03 AM #2 Senior Member   Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 put x=sh(t). hyperbolic sinus. use ch^2-sh^2=1. Thanks from duc
 September 16th, 2016, 04:39 AM #3 Senior Member   Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 The result is $\frac{x}{\sqrt{x^2+1}}+C$ Last edited by abdallahhammam; September 16th, 2016 at 04:44 AM.
September 16th, 2016, 08:56 AM   #4
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Quote:
 Originally Posted by abdallahhammam The result is $\frac{x}{\sqrt{x^2+1}}+C$
The derivative doesn't equal the integrand.

 September 16th, 2016, 10:40 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra Hyperbolic substitution is also not allowed. I already knew the answer, it's the substitution that I needed. I've found it now, so this becomes a challenge problem: solve the integral using a substitution that requires only the four basic operations and the square root.
 September 16th, 2016, 02:07 PM #6 Senior Member   Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 my result is correct. $(1+x^2)^{\frac{-3}{2}}=\frac{x^2+1-x^2}{(1+x^2)\sqrt{1+x^2}}$ the derivative of x/sqrt(1+x^2) is (1+x^2)^(-3/2). you could also try the substitution x=1/t. Last edited by abdallahhammam; September 16th, 2016 at 02:15 PM.
 September 16th, 2016, 02:45 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra Does it work?
 September 16th, 2016, 02:52 PM #8 Senior Member   Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 yes
 September 16th, 2016, 03:35 PM #9 Senior Member     Joined: Sep 2015 From: USA Posts: 2,638 Thanks: 1474 I have to confess I didn't see how $x=\dfrac 1 t$ aided the evaluation of this integral. Perhaps you can show your workings.
 September 16th, 2016, 03:43 PM #10 Senior Member   Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 it becomes $-\int (1+t^2)^{\frac{-3}{2}} t dt$ which is easier.

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