Integral of a hyperbolic function

[Karol]

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]

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]

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]

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

 

[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.

 

[Karol]

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]

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? $\int \frac{f'}{f}$

 

[LCKurtz]

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]

@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 ...