Verify the hyperbolic integration formula Verify the integration formula:
$$\int \operatorname{sech} x \, dx =\sin^{-1}(\tanh x)+C$$
I know that I have to manipulate the equation, but I am not sure how. Here is what I have done so far:
\begin{align}
& \int \operatorname{sech} x \, dx \\[6pt]
= {} & \int{\operatorname{sech}^3x\over \operatorname{sech}^2 x} dx \\[6pt]
= {} & \int{\operatorname{sech}^3x\over 1-\operatorname{tanh}^2x} \,dx
\end{align}
$$u= \operatorname{tanh} x, \quad du= \operatorname{sech}^2x \, dx$$
I am not even sure if I am even doing anything right up to this point. If someone could steer me in the right direction that would be great. 
 A: Use the definition of the hyperbolic secant:
$$\int\text{sech}(x)\ \text{d}x = \int\frac{1}{\cosh(x)}\ \text{d}x = \int\frac{2}{e^x + e^{-x}}\ \text{d}x$$
Let's collect an $e^x$ term
$$\int\frac{2e^{-x}}{1 + e^{-2x}}\ \text{d}x$$
We can now use the Geometric Series for the term $\frac{1}{1 + e^{-2x}}$
$$\int 2e^{-x}\sum_{k = 0}^{+\infty}(-e^{-2x})^k = \sum_{k = 0}^{+\infty}2(-1)^k\int e^{-x}e^{-2kx}\ \text{d}x = \sum_{k = 0}^{+\infty}2(-1)^k\int e^{-x(1 + 2k)}\ \text{d}x$$
Integration is trivial and you end up with
$$\sum_{k = 0}^{+\infty}2(-1)^k \frac{-1}{1 + 2k}e^{-x(1+2k)} = \sum_{k = 0}^{+\infty} \frac{2\cdot (-1)^{k+1}}{1+2k}e^{-x(1+2k)}$$
This is a well known series and it's
$$\sum_{k = 0}^{+\infty} \frac{2\cdot (-1)^{k+1}}{1+2k}e^{-x(1+2k)} = -2\arctan(e^{-x}) = -2\operatorname{arccot}(e^x)$$
A: $\newcommand{\sech}{\operatorname{sech}}\newcommand{\tanh}{\operatorname{tanh}}$To verify, Take the derivative of both sides of the equation:
$$\int \sech(x)\,dx=\arcsin(\tanh(x)) + c$$
$$\sech(x)=\frac{\mathrm{d} }{\mathrm{d} x}[\arcsin(\tanh(x)) + c]$$
$\frac{\mathrm{d} }{\mathrm{d} x}[\arcsin(u)]=\frac{u'}{\sqrt{1-u^2}}$ and $\frac{\mathrm{d} }{\mathrm{d} x}[\tanh(x)]=\sech^2(x)$
$$\sech(x)=\frac{\sech^2(x)}{\sqrt{1-\tanh^2(x)}}$$
$1-\tanh^2(x)=\sech^2(x)$
$$\sech(x)=\frac{\sech^2(x)}{\sqrt{\sech^2(x)}}$$
$$\sech(x)=\sech(x)$$
A: A slightly unconventional procedure is to force hyperbolic functions to the circular 
ones by $ x\rightarrow ix $ on either side of the equation.
$$\text{LHS}= \int \operatorname{sech} ix \, d ix = i \int \sec x \, dx = i \log ( \sec x + \tan x) $$
$$\text{RHS}= \sin^{-1}(\operatorname{tanh} ix)+C = i \operatorname{sinh}^{-1}(\tan x)  = i \log( \sec x + \tan x) $$
