# Integral of a hyperbolic function

## 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

Ssnow
Gold Member
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 ...

PeroK
Homework Helper
Gold Member
2020 Award
1

## 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!

epenguin
Homework Helper
Gold Member
Analogously to cos and sin but even simpler, cosh and sinh are each other's derivatives.

LCKurtz
Homework Helper
Gold Member
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.

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.

PeroK
Homework Helper
Gold Member
2020 Award
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}##

Ssnow
LCKurtz
Homework Helper
Gold Member
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.

Ssnow
Gold Member
@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 ...