integration - 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.

Answer

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
$$
\begin{align}
\int{\rm sech}\: x\:dx&=\int({\rm sech}\: x)^2 \times \frac{1}{\sqrt{({\rm sech}\: x)^2 }}\:dx\\\\
&=\int(\tanh x)'\times \frac{1}{\sqrt{1-(\tanh x)^2}}\:dx\\\\
&=\arcsin (\tanh x)+C.
\end{align}
$$