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

[crybllrd]

Homework Statement

$\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}$

Homework Equations

The Attempt at a Solution

$\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}$$tan^{-1}x|\stackrel{tan(8)}{tan(1)}$$tan^{-1}(tan(8))-tan^{-1}(tan(1))$$8-1=7$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?

 

[SammyS]

Homework Statement

$\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}$

The Attempt at a Solution

$\int\stackrel{tan(8)}{tan(1)}\frac{dx}{x^{2}+1}$

$tan^{-1}x|\stackrel{tan(8)}{tan(1)}$

$tan^{-1}(tan(8))-tan^{-1}(tan(1))$

$8-1=7$

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 $\displaystyle \ -\frac{\pi}{2}<\theta<\frac{\pi}{2}\,.$ The number, 8, is not in that interval.

 

[crybllrd]

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