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.

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