Improper Integral

  • Thread starter Bashyboy
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  • #1
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?
 

Answers and Replies

  • #2
36,332
8,293

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.
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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