Integration by substitution

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Hi all,

I've been studying calculus out of Tom Apostol's book "Calculus". I'm having troube with the following problem in the section on integration by substitution:

Integrate [tex]\int(x^2+1)^{-3/2}\,dx[/tex].

I tried the substitution [tex]u=x^2+1[/tex] but it didn't seem to work. I can't see anything else that may help. Any hints or solution would be greatly appreciated!
 

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  • #2
HallsofIvy
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There are two fairly standard substitutions that can by used for something of the form [itex]x^1+ 1[/itex] inside a square root. Since [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex], dividing through by [itex]cos^2(\theta)[/itex] gives [itex]sin^2(\theta)/cos^2(\theta)+ 1= 1/cos^2(\theta)[/itex] so that [itex]tan^2(\theta)+ 1= sec^2(\theta)[/itex] which suggests the substitution [itex]x= tan(\theta)[/itex]. If arildno have gotten here first, he would have suggested x= sinh(u). That works, and may be simpler, because [itex]cosh^2(u)- sinh^2(u)= 1[/itex] so that [itex]sinh^2(u)+ 1= cosh^2(u)[/itex].
 
  • #3
arildno
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There are two fairly standard substitutions that can by used for something of the form [itex]x^1+ 1[/itex] inside a square root. Since [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex], dividing through by [itex]cos^2(\theta)[/itex] gives [itex]sin^2(\theta)/cos^2(\theta)+ 1= 1/cos^2(\theta)[/itex] so that [itex]tan^2(\theta)+ 1= sec^2(\theta)[/itex] which suggests the substitution [itex]x= tan(\theta)[/itex]. If arildno have gotten here first, he would have suggested x= sinh(u). That works, and may be simpler, because [itex]cosh^2(u)- sinh^2(u)= 1[/itex] so that [itex]sinh^2(u)+ 1= cosh^2(u)[/itex].
I would have, indeed..:smile:

The result is then, of course:
[tex]I=Tanh(u)=Tanh(Sinh^{-1}(x))=\frac{x}{\sqrt{1+x^{2}}}[/tex]
 
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  • #4
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Hint: Use sec^2(x) - tan^2(x)= 1 and then use another subst involving derivative of tan x.

That should help.
 
  • #5
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Thanks for the help, all! I used HallsofIvy's suggestion which worked marvelously.

Maybe arildno's would have been simpler, however I've only vaguely heard of the hyperbolic trig functions and don't really know them.
 
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I'm also having trouble with the following integral: [tex]\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx[/tex]. Any ideas?
 
  • #7
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I'm also having trouble with the following integral: [tex]\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx[/tex]. Any ideas?
First of all, post this as a seperate q. You can use the hint I gave above and then just keep simplifying it, keep simplifying it, and after a while you'll arrive at an expression where it should be clear that you have to make another subst. involving sec(u) + 1

That hint should help.
 
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  • #8
Gib Z
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I'm also having trouble with the following integral: [tex]\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx[/tex]. Any ideas?
It works out nicely and exactly analogously with both a circular trig substitution and a hyperbolic trig substitution. Firstly the trig substitution, then a linear one.
 
  • #9
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[tex]\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx[/tex]
It works out nicely and exactly analogously with both a circular trig substitution and a hyperbolic trig substitution. Firstly the trig substitution, then a linear one.
I think I have a better way.. [please see attachment] {sorry if my handwriting ain't clear..}
 

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  • #10
Mute
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You made a mistake doing your final integral in the attachment. The correct result is

[tex]\int \frac{du}{\sqrt{1+u}} = \sqrt{1 + u} =\sqrt{1 + \sqrt{x^2+1}}[/tex]

(up to a constant of integration, of course).


You can then check by differentiation that this is correct.
 
  • #11
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You made a mistake doing your final integral in the attachment. The correct result is

[tex]\int \frac{du}{\sqrt{1+u}} = \sqrt{1 + u} =\sqrt{1 + \sqrt{x^2+1}}[/tex]

(up to a constant of integration, of course).


You can then check by differentiation that this is correct.
oops.. I mistakenly took the power of [itex]1 + u[/itex] to be [itex]\frac{1}{2}[/itex] instead of [itex]\frac{-1}{2}[/itex]..
 
  • #12
Gib Z
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Circular Trig method

[tex]\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx[/tex]

Let x= tan u. Then dx = sec^2 u du.

[tex]\int \frac{\tan u \sec^2 u}{\sqrt{\sec^2 u + \sec^3 u}} du = \int \frac{ (\sec u)'}{\sqrt{1+ \sec u}} du[/tex].

Now let t= 1+ sec u.
[tex]\int \frac{1}{ \sqrt{t} } dt = 2\sqrt{t} + C = 2\sqrt{1+ \sec u} + C = 2\sqrt{1+ \sqrt{x^2+1}} + C[/tex].

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

Hyperbolic Trig method

[tex]\int \frac{x}{\sqrt{x^2+1+\sqrt{(x^2+1)^3}}}\,dx[/tex]

Let x = sinh u, dx = cosh u du

[tex]\int \frac{ \sinh u \cosh u }{ \sqrt{\cosh^2 u + \cosh^3 u}} du = \int \frac{(\cosh u)'}{\sqrt{1+ \cosh u}} du[/tex].

Let t= 1+ cosh u.

[tex]\int \frac{1}{\sqrt{t}} dt = 2\sqrt{t} + C = 2\sqrt{ 1+ \cosh u} + C = 2\sqrt{1+ \sqrt{1+x^2}} + C[/tex].


PS. You both missed a factor of 2 when evaluating the integral.
 
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  • #13
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PS. You both missed a factor of 2 when evaluating the integral.
i hate myself.. and u :p
 
  • #14
Mute
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PS. You both missed a factor of 2 when evaluating the integral.
Eh, factors of 2 between friends aren't important. ; )
 
  • #15
Gib Z
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