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Integral of a hyperbolic function

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    $$\int \tanh=?$$

    2. Relevant equations
    $$\cosh^2-\sinh^2=1$$
    $$(\tanh)'={\rm sech}^2=\frac{1}{\cosh^2},~~(\coth)'=-{\rm csch}^2=-\frac{1}{\sinh^2}$$
    $$({\rm sech})'=\left( \frac{1}{\cosh} \right)'=-{\rm sech}\cdot\tanh=-\frac{\sinh}{{\rm cosh}^2}$$
    $$({\rm csch})'=\left( \frac{1}{\sinh}\right)'=-{\rm csch}\cdot\coth=-\frac{\rm cosh}{\sinh^2}$$

    3. The attempt at a solution
    $$\int \tanh=\int \frac{\sinh}{\sqrt{1+\sinh^2}}$$
    That's as far as i can go
     
  2. jcsd
  3. Oct 24, 2016 #2

    Ssnow

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    Hi, you can start with the definition ##\tanh(x)=\frac{\sinh{x}}{\cosh{x}}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}## , if you split the fraction you have two integral simpler ...
     
  4. Oct 24, 2016 #3

    PeroK

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    Hint: perhaps you went too far!
     
  5. Oct 24, 2016 #4

    epenguin

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    Analogously to cos and sin but even simpler, cosh and sinh are each other's derivatives. :oldwink:
     
  6. Oct 24, 2016 #5

    LCKurtz

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    Another hint: Do you know how to integrate$$\int \tan x~dx = \int \frac {\sin x}{\cos x}~dx\text{ ?}$$Same idea works.
    [Edit:]epenguin must type faster than me.
     
  7. Oct 24, 2016 #6
    No, i don't know to integrate that either, i am ashamed to say. if it were to derive ##\frac{\sinh}{\cosh}## then yes. maybe you mean:
    $$\int \tan x~dx = \int \frac {\sin x}{\cos x}~dx=\int\frac{e^x}{e^x+e^{-x}}-\int\frac{e^{-x}}{e^x+e^{-x}}$$
    But i don't know that. i don't want to use integral tables.
     
  8. Oct 25, 2016 #7

    PeroK

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    What about, in general? ##\int \frac{f'}{f}##
     
  9. Oct 25, 2016 #8

    LCKurtz

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    No, I certainly didn't mean that. Sines and cosines can't be expressed as real exponentials. Try the "obvious" u-substitution.
     
  10. Oct 26, 2016 #9

    Ssnow

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    @Karol remember that ##sin{x}## and ##cos{x}## cannot be wrote in terms of real exponential function ##e^{x}##, only ##\sinh{x}## and ##cosh{x}## ... in any way the two integrals you wrote with ##e^{x}## and ##e^{-x}## are correct for the ##\tanh{x}## and performing a strategic substitution you can solve the problem ...
     
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