Integral Substitution

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hi guys
im debating whether this question requires trignometric substitution or just normal substitution.

∫ √9-2(x-1)²

Im leaning towards normal substitution, with u = x-1, but im not sure
Any ideas
Thanx heaps
 

Answers and Replies

  • #2
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Is it

[tex] \int \sqrt{9-(2x-1)^2} [/tex]?

If so a normal one wont do, youll need a trig one.
 
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  • #3
dextercioby
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Try to bring it to the form

[tex] C\int\sqrt{1-(something)^2} \ dx [/tex]

,where C is a constant.

Daniel.
 
  • #4
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2 is on the outside of the (x+1)
 
  • #5
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(x-1) sorry
 
  • #6
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Well if you substitute u = x-1, you get the integral...

[tex]\int \sqrt{9 - 2u^2}du[/tex]

I think trig substitution is where this is leading.
 
  • #7
GCT
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yeah from what I remember, trig substitution was the way to go.

[tex] \sqrt(2) \int \sqrt{(9/2-u^{2})} ~du [/tex]
 
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  • #8
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hi guys
im still stuck on this question, it has really got me stumped
i know that for the form

√a² - x².dx

x=asinθ, dx = acosθ.dθ

but i just cant seem to put it together
 
  • #9
dextercioby
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Why not.U've got an |a| coming out of the sqrt and u'll get a square

[tex] \int \sqrt{a^{2}-x^{2}} \ dx =a|a|\int \cos^{2}\theta \ d\theta [/tex]

Then use the double angle formula.

Daniel.
 
  • #10
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the 2 outside the (x-1)^2 is confusing me
Im not sure what my x and a squared is?
 
  • #11
dextercioby
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Look at post #6.If you make the substitution

[tex] \sqrt{2}u=t [/tex],

u can put that integral under the form

[tex]\frac{1}{\sqrt{2}} \int \sqrt{3^{2}-t^{2}} \ dt [/tex]

And now do that substitution involving "sin".

Daniel.
 
  • #12
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so, sqrt 2u = sin theta
 
  • #13
dextercioby
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Nope.[itex] u\sqrt{2}=3\sin\theta [/itex]

Daniel.
 

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