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Ideas as to how I should start to integrate this one?

  • Thread starter LocationX
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  • #1
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[tex]\int (C^2x^2-x^4)^{0.5} dx [/tex]

C is a constant

any ideas as to how I shoud start to integrate this one?
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
859
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Yeah, just factor out x^2, so you have
[tex]\int x\sqrt{C^2-x^2}dx[/tex]
Then just make the substitution [itex]u=x^2[/itex].
 
  • #3
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so then how would I integrate

[tex]\int (C^2-u)^{0.5}[/tex]
 
  • #4
You can make another substitution, e.g. [tex]z = C^2 - u [/tex] or just make your original substitution [tex]u = C^2 - x^2 [/tex], which gives

[tex]du = -2x dx[/tex]

[tex]\frac{du}{-2x} = dx [/tex]

then you just use your normal integration rules.
 
  • #5
arildno
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LeonhardEuler said:
Yeah, just factor out x^2, so you have
[tex]\int x\sqrt{C^2-x^2}dx[/tex]
Then just make the substitution [itex]u=x^2[/itex].
Just a minor correction.
We should have:
[tex]\int\sqrt{C^{2}x^{2}-x^{4}}dx=\int|x|\sqrt{C^{2}-x^{2}}dx[/tex]
 

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