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
Best Answer
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.