Trying an alternative to Trig Substitutions

  • #1
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I wanted to try an alternative method to the proverbial technique used in trig substitution. Is this method a dead-end or is there hope for it?

[tex] \int \frac{\sqrt{x^2-3}}{x} dx[/tex]
Using trig substitution
[tex]c^2=a^2+b^2[/tex]
[tex]a = \sqrt{c^2-b^2}[/tex]
[tex]∴ c = x, b = \sqrt{3}[/tex]
Assigning these values to a triangle, where a is adjacent to theta and b is opposite to theta we get..
[tex] \int \frac{\sqrt{x^2-3}}{x} dx = \int \frac{a}{x} dx = \int cos \theta dx[/tex]

And I've hit a wall. I can't seem to be able to adequately convert dx to dθ.
 

Answers and Replies

  • #2
DryRun
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You could try the substitution: [tex]x=\sqrt 3 \cosh \theta[/tex]
 
  • #3
SammyS
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I wanted to try an alternative method to the proverbial technique used in trig substitution. Is this method a dead-end or is there hope for it?

[tex] \int \frac{\sqrt{x^2-3}}{x} dx[/tex]
Using trig substitution
[tex]c^2=a^2+b^2[/tex]
[tex]a = \sqrt{c^2-b^2}[/tex]
[tex]∴ c = x, b = \sqrt{3}[/tex]
Assigning these values to a triangle, where a is adjacent to theta and b is opposite to theta we get..
[tex] \int \frac{\sqrt{x^2-3}}{x} dx = \int \frac{a}{x} dx = \int cos \theta dx[/tex]

And I've hit a wall. I can't seem to be able to adequately convert dx to dθ.
Your substitution appears to be [itex]\displaystyle \cos(\theta)=\frac{\sqrt{x^2-3}}{x}\ .[/itex]

In that case, how is dx related to dθ ?
 
  • #4
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You could try the substitution: [tex]x=\sqrt 3 \cosh \theta[/tex]

I just did, it starts becoming ugly. And your still plagued by the dx so it hasn't made a difference.

Edit: I think I over-complicated things, I'm reviewing my method.

Edit: Either way, I feel a bit uncomfortable with the approach because I don't know how

[tex] x = \sqrt{3} cosh \theta [/tex]
 
Last edited:
  • #5
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Your substitution appears to be [itex]\displaystyle \cos(\theta)=\frac{\sqrt{x^2-3}}{x}\ .[/itex]

In that case, how is dx related to dθ ?

Well, it doesn't seem that I can know until I can explicitly define the function in terms of x.
 
  • #6
DryRun
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It works out by the substitution that i suggested. What is the final answer given in your book/notes?
 
  • #7
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It works out by the substitution that i suggested. What is the final answer given in your book/notes?

[tex] \sqrt{3} \int (sec^2 \theta - 1) d\theta [/tex]
[tex] \sqrt{3} (tan \theta - \theta) [/tex]
 
  • #8
dextercioby
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The substitution reccomended in post #2 should solve the problem.
 
  • #9
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The substitution reccomended in post #2 should solve the problem.

I can use any of the proverbial substitutions given in the book but that is besides the point. I was trying to do it in a different way and just need to know if my method is a dead end or not.

And at any rate, I'm not too familiar with the hyperbolic functions and how he derived that substitution equivalent.
 
  • #10
HallsofIvy
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Why are the substitutions in your book "proverbial"? I don't recally any proverbs related to Calculus!

In any case, your method is not a "dead end" but it is exactly the trig substitution you apparently are trying to avoid. Taking x to be the hypotenuse of your right triangle and the "near side" to be [itex]\sqrt{x^3- 3}[/itex], the "opposite side" has length 3 so [itex]sin(\theta)= 3/x[/itex], [itex]x= 3/sin(\theta)= 3csc(\theta)[/itex]. From that, [itex]dx= 3 d(csc(\theta))= -3 csc(\theta)cot(\theta) d\theta[/itex].
 
  • #11
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Why are the substitutions in your book "proverbial"? I don't recally any proverbs related to Calculus!

Proverbial as in widely known and common.

In any case, your method is not a "dead end" but it is exactly the trig substitution you apparently are trying to avoid. Taking x to be the hypotenuse of your right triangle and the "near side" to be [itex]\sqrt{x^3- 3}[/itex], the "opposite side" has length 3 so [itex]sin(\theta)= 3/x[/itex], [itex]x= 3/sin(\theta)= 3csc(\theta)[/itex]. From that, [itex]dx= 3 d(csc(\theta))= -3 csc(\theta)cot(\theta) d\theta[/itex].

Hmm, I see now. ^.^ Thanks.
 

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