Integral of a hyperbolic function

1. Oct 24, 2016

Karol

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. Oct 24, 2016

Ssnow

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. Oct 24, 2016

PeroK

Hint: perhaps you went too far!

4. Oct 24, 2016

epenguin

Analogously to cos and sin but even simpler, cosh and sinh are each other's derivatives.

5. Oct 24, 2016

LCKurtz

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. Oct 24, 2016

Karol

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. Oct 25, 2016

PeroK

What about, in general? $\int \frac{f'}{f}$

8. Oct 25, 2016

LCKurtz

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

9. Oct 26, 2016

Ssnow

@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 ...