Is Trigonometric Substitution Needed for ∫ √9-2(x-1)²?

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    Integral Substitution
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Discussion Overview

The discussion revolves around the integral ∫ √9-2(x-1)², specifically whether it requires trigonometric substitution or can be solved using normal substitution. Participants explore different substitution methods and the forms needed for trigonometric integration.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using normal substitution with u = x-1, but expresses uncertainty about this approach.
  • Another participant proposes that if the integral is rewritten as ∫ √(9-(2x-1)²), a trigonometric substitution would be necessary.
  • A participant advises transforming the integral into the form C∫√(1-(something)²)dx, indicating a direction towards trigonometric substitution.
  • There is a discussion about the confusion caused by the coefficient of 2 in the expression, with participants trying to clarify the correct substitutions.
  • One participant confirms that substituting u = x-1 leads to the integral ∫ √(9 - 2u²)du, suggesting that trigonometric substitution seems appropriate.
  • Another participant recalls that trigonometric substitution is typically the method to use for integrals of the form √(a² - x²)dx, but struggles to apply it to this specific problem.
  • Participants discuss how to manipulate the integral into a suitable form for trigonometric substitution, with suggestions on using specific substitutions involving sine and constants.

Areas of Agreement / Disagreement

Participants express differing opinions on whether normal substitution is sufficient or if trigonometric substitution is necessary. There is no consensus on the best approach to solve the integral.

Contextual Notes

Participants note the complexity introduced by the coefficient of 2 and the need to clarify the forms of variables involved in the substitutions. There are unresolved steps in transforming the integral into a suitable form for trigonometric substitution.

cogs24
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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 I am not sure
Any ideas
Thanx heaps
 
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Is it

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

If so a normal one won't do, youll need a trig one.
 
Last edited:
Try to bring it to the form

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

,where C is a constant.

Daniel.
 
2 is on the outside of the (x+1)
 
(x-1) sorry
 
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.
 
yeah from what I remember, trig substitution was the way to go.

[tex]\sqrt(2) \int \sqrt{(9/2-u^{2})} ~du[/tex]
 
Last edited:
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 can't seem to put it together
 
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
the 2 outside the (x-1)^2 is confusing me
Im not sure what my x and a squared is?
 
  • #11
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
so, sqrt 2u = sin theta
 
  • #13
Nope.[itex]u\sqrt{2}=3\sin\theta[/itex]

Daniel.
 

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