How Is the Definite Integral of a Trigonometric Function Evaluated Correctly?

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SUMMARY

The definite integral of the function \(\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}\) was incorrectly evaluated as 7. The correct evaluation involves recognizing that \(\tan^{-1}(tan(8))\) does not equal 8 due to the periodic nature of the tangent function, specifically that \(\tan^{-1}(tan(θ)) = θ\) only holds when \(-\frac{\pi}{2} < θ < \frac{\pi}{2}\). Therefore, the correct evaluation is \(\tan^{-1}(tan(8)) = 8 - 3\pi\) and \(\tan^{-1}(tan(1)) = 1\), leading to the correct result of \(8 - 3\pi - 1\).

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  • Understanding of definite integrals
  • Familiarity with trigonometric functions and their properties
  • Knowledge of the inverse tangent function, \(\tan^{-1}(x)\)
  • Proficiency in LaTeX for mathematical typesetting
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  • Study the properties of the inverse tangent function, particularly its range and periodicity
  • Learn about evaluating definite integrals involving trigonometric functions
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Students studying calculus, particularly those focusing on integration techniques, as well as educators seeking to clarify the evaluation of definite integrals involving trigonometric functions.

crybllrd
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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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crybllrd said:

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.
 
Thanks a lot, I got it now:
tan-1(tan(θ)) =8-3pi
tan-1(tan(1)) = 1
 

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