Solving an Improper Integral Homework Equation

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SUMMARY

The discussion focuses on solving the improper integral \(\int \frac{dx}{\sqrt{x^2-4}}\) using trigonometric substitution. The user initially applied the substitution \(\cot\theta = \frac{4}{\sqrt{x^2-4}}\) and \(-4\sin\theta = dx\), leading to an incorrect result of \(-\frac{\sqrt{x^2-4}}{x}\). The correct solution, as provided in the reference material, is \(\ln|x+\sqrt{x^2-4}|\). The discrepancy arises from errors in the substitution process, specifically the incorrect triangle leg and the expression for \(dx\).

PREREQUISITES
  • Understanding of improper integrals
  • Familiarity with trigonometric substitution techniques
  • Knowledge of differentiation and integration rules
  • Ability to manipulate logarithmic expressions
NEXT STEPS
  • Review trigonometric substitution methods for integrals
  • Study the derivation of \(\ln|x+\sqrt{x^2-4}|\) for this integral
  • Practice solving similar improper integrals
  • Learn about the geometric interpretation of trigonometric substitutions
USEFUL FOR

Students studying calculus, particularly those tackling improper integrals and trigonometric substitution techniques.

Bashyboy
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Homework Statement


[itex]\int \frac{dx}{\sqrt{x^2-4}}[/itex]


Homework Equations





The Attempt at a Solution



I tried trig-substitution, by realizing that [itex]cot\theta = \frac{4}{\sqrt{x^2-4}}[/itex]

and that [itex]-4sin\theta = dx[/itex]

My answer, though, found after the substitution and integration, is very different from the books: mine is [itex]- \frac{\sqrt{x^2-4}}{x}[/itex], theirs is [itex]ln|x+\sqrt{x^2-4}|[/itex]

How do you account for this variation?
 
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Bashyboy said:

Homework Statement


[itex]\int \frac{dx}{\sqrt{x^2-4}}[/itex]


Homework Equations





The Attempt at a Solution



I tried trig-substitution, by realizing that [itex]cot\theta = \frac{4}{\sqrt{x^2-4}}[/itex]

and that [itex]-4sin\theta = dx[/itex]
You have mistakes in your substitution. One of the legs in your triangle should be 2, not 4. Also, your equation for dx is incorrect.
Bashyboy said:
My answer, though, found after the substitution and integration, is very different from the books: mine is [itex]- \frac{\sqrt{x^2-4}}{x}[/itex], theirs is [itex]ln|x+\sqrt{x^2-4}|[/itex]

How do you account for this variation?
 

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