How can I prove that $ \int \text{sech}(x) ~ \mathrm{d}{x} = {\sin^{-1}}(\tanh(x)) + c $? How can I prove that
$$
\int \text{sech}(x) ~ \mathrm{d}{x} = {\sin^{-1}}(\tanh(x)) + c?
$$
I don’t know how to prove this identity. Any help?
I tried to multiply by $ \dfrac{\cosh(x)}{\cosh(x)} $, and everything is okay, but at last I didn’t get the same answer.
 A: Since you know the result, why not just differentiate it:
$$
\begin{align}
(\arcsin (\tanh x))'&=(\tanh x)'\times \frac{1}{\sqrt{1-(\tanh x)^2}}\\\\
&=({\rm sech}\: x)^2 \times \frac{1}{\sqrt{({\rm sech}\: x)^2 }}\\\\
&={\rm sech}\: x.
\end{align}
$$ 

I don’t know how to prove this identity. Any help?

This proves that
$$
\int{\rm sech}\: x\:dx = \arcsin (\tanh x)
$$ up to a constant.

Edit: you might want to write
$$
\begin{align}
\int{\rm sech}\: x\:dx&=\int({\rm sech}\: x)^2 \times \frac{1}{\sqrt{({\rm sech}\: x)^2 }}\:dx\\\\
&=\int(\tanh x)'\times \frac{1}{\sqrt{1-(\tanh x)^2}}\:dx\\\\
&=\arcsin (\tanh x)+C.
\end{align}
$$
A: Your first attempt is indeed correct, since 
$$
\int \text{sech}(x) dx = \int \frac{\cosh(x)}{\cosh^2(x)} dx = \int \frac{\cosh(x)}{1+\sinh^2(x)} dx = \tan^{-1}(\sinh(x)) + C
$$
which is not the answer you have. However note that this also can be solve as follows 
$$
\int \text{sech}(x) dx = \int \frac{\text{sech}^2(x)}{\sqrt{1-\tanh^2(x)}} dx = \sin^{-1}(\tanh(x)) + C
$$
since $\tanh'(x)=\text{sech}^2(x)$.
A: here is one way to do this. $$\begin{align}\int {dx}\, {\text{sech}\,  x} &= \int\frac{2e^x\, dx}{e^{2x} + 1} \\
&= 2\int \frac{du}{1+u^2}, u = e^x\\
&=2\tan^{-1}(e^x)+c\\ 
&=\sin^{-1}\left(\frac{2e^x}{1+e^{2x}} \right) + C, \text{ used }\sin 2t = 2\sin t \cos t\\
&=\sin^{-1}\left(\text{sech}\, x \right) + C
\end{align}$$
A: $ \operatorname{sech}^2 x +  \tanh^2 x =1 $
So, let $ u = \tanh x $ ...
A: Substituting $u=\tanh x$, you will arrive at the standard integral for $\arcsin (u)$.
A: $$ \int \text{sech}x\, dx=\int \dfrac{\text{sech}^2 x}{\sqrt{\text{sech}^2x}}dx ,$$ and since $$\text{sech} ^2 x =1- \tanh^2 x,$$ letting $u=\tanh x$, $du= \text{sech} ^2 x \, dx$ and solving the integral $$\int \dfrac{du}{\sqrt{1-u^2}},$$ you get the desired result.
