Integral of a hyperbolic function

  • Thread starter Karol
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  • #1
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Homework Statement


$$\int \tanh=?$$

Homework Equations


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

The Attempt at a Solution


$$\int \tanh=\int \frac{\sinh}{\sqrt{1+\sinh^2}}$$
That's as far as i can go
 

Answers and Replies

  • #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 ...
 
  • #3
PeroK
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The Attempt at a Solution


$$\int \tanh=\int \frac{\sinh}{\sqrt{1+\sinh^2}}$$
That's as far as i can go

Hint: perhaps you went too far!
 
  • #4
epenguin
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Analogously to cos and sin but even simpler, cosh and sinh are each other's derivatives. :oldwink:
 
  • #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.
 
  • #6
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Do you know how to integrate$$\int \tan x~dx = \int \frac {\sin x}{\cos x}~dx\text{ ?}$$Same idea works.
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.
 
  • #7
PeroK
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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.

What about, in general? ##\int \frac{f'}{f}##
 
  • #8
LCKurtz
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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:
$$\int \tan x~dx = \int \frac {\sin x}{\cos x}~dx=\int\frac{e^x}{e^x+e^{-x}}-\int\frac{e^{-x}}{e^x+e^{-x}}$$
No, I certainly didn't mean that. Sines and cosines can't be expressed as real exponentials. Try the "obvious" u-substitution.
 
  • #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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