1. Mar 18, 2009 #1

   roam

   

   1. I am asked to calculate [tex]\int \frac{dx}{cosh(x)}[/tex]
   
   
   
   2. Relevant equations
   
   
   
   3. The attempt at a solution
   
   I know that [tex]\frac{1}{cosh(x)}[/tex] is equivalent to "sech(x)" which by definition is [tex]\frac{2}{e^x+e^{-x}}[/tex].
   
   I'm confused & I don't know which one I need to use for this question:
   
   [tex]\int sech(x) = 2tan{-1}(tanh(\frac{x}{2}))[/tex]
   
   [tex]\int \frac{2}{e^x + e^{-x}} = 2tan^{-1}(e^x)[/tex]
   
   Any help would be greatly appreciated.
   

    

   Last edited: Mar 18, 2009

   roam, Mar 18, 2009
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3. Mar 18, 2009 #2

   yyat

   

   The two answers you got only differ by a constant of [tex]\frac{\pi}{2}[/tex]. Recall that the indefinite integral is only defined up to a constant, so both solutions are correct.

    

   yyat, Mar 18, 2009

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