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Simple Integration by substitution

  • #1

Homework Statement


Find by letting [tex]U^2=(4 + x^2) [/tex]the following [tex] \int_0^2\frac{x}{\sqrt{4 + x^2}}dx[/tex]?
I can solve it by letting [tex]\mbox{x=2} tan(\theta)[/tex], But I want to be able to do it by substitution.

The Attempt at a Solution


[tex] \frac{du}{dx}=\frac{d\sqrt{(4+x^2)}}{dx}=\frac{x}{\sqrt{4+x^2}}\mbox{, therefore du}=\frac{x}{u^\frac{1}{2}}\times dx\\[/tex] Therefore the integral is [tex] \int_{x=0}^{x=2}\frac{1}{u^\frac{1}{2}}du=[/tex]0.26757, it should be [tex] 2(\sqrt{2}-1)[/tex]. Can you tell me where I went wrong. Thanks for the help.
 
Last edited:

Answers and Replies

  • #2
Dick
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[[tex] \frac{du}{dx}=\frac{d\sqrt{(4+x^2)}}{dx}=\frac{x}{\sqrt{4+x^2}}\mbox{, therefore du}=\frac{x}{u^\frac{1}{2}}\times dx\\[/tex]
What happened to the x in the above expression for du? Did you just drop it?
 
  • #3
There is a mistake in the equation, it should be [tex]\int_0^2\frac{x}{\sqrt{4+x^2}}dx[/tex]. And the answer I then get =.5351. And thanks for the quick reply.
 
  • #4
Dick
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Ah, ha. Then you want to evaluate the final integral between the u limits, not the x limits.
 
  • #5
Dick
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[tex] \frac{du}{dx}=\frac{d\sqrt{(4+x^2)}}{dx}=\frac{x}{\sqrt{4+x^2}}\mbox{, therefore du}=\frac{x}{u^\frac{1}{2}}\times dx\\[/tex]
And something is going awry here. A sqrt(u) would be the 4th root of 4+x^2. You are being sloppy.
 
  • #6
I know there is something awry in the calculation, maybe you can find what I did wrong. Since [tex]u^2=4+x^2 [/tex] that implies [tex] u=\sqrt{4+x^2}[/tex] and [tex] \frac{du}{dx}[/tex] expression follows.
 
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  • #7
Dick
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This shows du=(x*dx)/u. Not u^(1/2).
 
  • #8
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First try substituting u=4+x²⇒du=2xdx⇒(1/2)du=xdx for the substitution part
that should be a lot easier . you are still heading the right direction
(1/2)∫(1/(u^{1/2}))du= √u
I think you can take it from here
 
  • #9
Dick
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He can also do it with his substitution. He just needs to be more careful of exactly what that is. :smile:
 
  • #10
Thanks for the help lads I at last got it out using my substitution.
 

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