Definite Integral with trig

  • Thread starter crybllrd
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  • #1
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Homework Statement



[itex]\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}[/itex]

Homework Equations





The Attempt at a Solution



[itex]\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}[/itex]


[itex]tan^{-1}x|\stackrel{tan(8)}{tan(1)}[/itex]


[itex]tan^{-1}(tan(8))-tan^{-1}(tan(1))[/itex]


[itex]8-1=7[/itex]


I'm fine with this, but my instructor said "Hint: The answer isn't 7."

What did I do wrong here?


As a side note, is there a better way to type a def. integral in LaTeX?
 
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Answers and Replies

  • #2
SammyS
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Homework Statement



[itex]\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}[/itex]

The Attempt at a Solution



[itex]\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}[/itex]

[itex]tan^{-1}x|\stackrel{tan(8)}{tan(1)}[/itex]

[itex]tan^{-1}(tan(8))-tan^{-1}(tan(1))[/itex]

[itex]8-1=7[/itex]

I'm fine with this, but my instructor said "Hint: The answer isn't 7."

What did I do wrong here?

As a side note, is there a better way to type a def. integral in LaTeX?
Look at the value of tan(8 radians). It's negative.

If u < 0, tan-1(u) < 0 .

Also, tan-1(tan(θ)) = θ, only if [itex]\displaystyle \ -\frac{\pi}{2}<\theta<\frac{\pi}{2}\,.[/itex] The number, 8, is not in that interval.
 
  • #3
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Thanks a lot, I got it now:
tan-1(tan(θ)) =8-3pi
tan-1(tan(1)) = 1
 

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