How to calculate the integral of $ \int\frac{1}{\cosh^{2}x}dx $

Question

Ofcourse one can notice that

$ \left(\frac{\sinh\left(x\right)}{\cosh\left(x\right)}\right)'=\frac{\cosh^{2}\left(x\right)-\sinh^{2}\left(x\right)}{\cosh^{2}\left(x\right)}=\frac{1}{\cosh^{2}\left(x\right)} $

But im looking for a straight way to prove that

$ \int\frac{1}{\cosh^{2}x}dx=\frac{\sinh\left(x\right)}{\cosh\left(x\right)}+ C $

Here's what I tried:

$ \int\frac{1}{\cosh^{2}x}=\int\frac{1}{1+\sinh^{2}x}dx=\frac{\arctan\left(\sinh x\right)}{\cosh x} + C$

Now im not sure how to continue. Thanks in advance

Answer 1

Based on your comment to another answer, you want to show that $\displaystyle \int \mathrm{sech}^2 x dx = \tanh x + c$.

I find the easiest way is to use complex numbers. Start with the circular trigonometric version $\displaystyle \int \sec^2 x dx = \tan x + c$, which I assume you can assume or you know how to prove. Then transform to the hyperbolic trigonometric version with the substitution $x=iy$:

Note that $\cos iy = \cosh y \implies \sec iy = \mathrm{sech}\ y$ and $\tan iy = i\tanh y$.

$\displaystyle \int \sec^2 iy d(iy) = \tan iy + c$

$\displaystyle i\int \mathrm{sech}^2 y dy = i\tanh y + c$

$\displaystyle \int \mathrm{sech}^2 y dy = \tanh y + z$

where $z$ is an arbitrary complex constant. If you're dealing with a purely real domain, $z$ will be real as well.

Answer 2

There are multiple ways of evaluating this integral. You can let $t=\text{tanh}{x}$ so $\text{cosh}^2{x}=\frac{1}{1-t^2}$ and $dx=\frac{dt}{1-t^2}$: $$\int \frac{\frac{dt}{1-t^2}}{\frac{1}{1-t^2}}= \int \; dt = t + C = \text{tanh}{x}+C$$

You can also rewrite $\text{cosh}^2{x}$ as $\frac{e^x+e^{-x}}{2}$ and let $u=e^x$ to evaluate this integral.

Answer 3

Switch to circular trig functions if you are more comfortable with them, and then back again.

$$ x\rightarrow iy \;; \; y \rightarrow x/i \; ; $$

$$ \int \text{sech}^2{x} \; dx\rightarrow \int \text{sech}^2{iy}\; idy \rightarrow \int \text{sec}^2{y} \;idy $$ $$= i\; \tan y = i\tan \dfrac{x}{i} = i\tan \dfrac{ix}{i^2} = \dfrac{i^2}{-1}\cdot\tanh x $$ $$ = \tanh x $$

Answer 4

In order to do integrals involving hyperbolic functions, you will need to know the basic derivatives: $$ \frac{d}{dx} \sinh x = \cosh x \\ \frac{d}{dx} \cosh x = \sinh x \\ \frac{d}{dx} \tanh x = \operatorname{sech}^2 x $$ You will not get very far without knowing at least these.

Looking at the third line, you already know $$ \int\frac{dx}{\cosh^2 x} . $$

Answer 5

With $u=e^{2x}$ this is$$\int\frac{2du}{(u+1)^2}=\frac{-2}{u+1}+C=\tanh x+C-1.$$