How can I prove that $ \int \text{sech}(x) ~ \mathrm{d}{x} = {\sin^{-1}}(\tanh(x)) + c $?

Question

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 1

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

Answer 2

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 $\tanh'(x)=\text{sech}^2(x)$.

Answer 3

here is one way to do this. $$\begin{align}\int {dx}\, {\text{sech}\, x} &= \int\frac{2e^x\, dx}{e^{2x} + 1} \\ &= 2\int \frac{du}{1+u^2}, u = e^x\\ &=2\tan^{-1}(e^x)+c\\ &=\sin^{-1}\left(\frac{2e^x}{1+e^{2x}} \right) + C, \text{ used }\sin 2t = 2\sin t \cos t\\ &=\sin^{-1}\left(\text{sech}\, x \right) + C \end{align}$$

Answer 4

$ \operatorname{sech}^2 x + \tanh^2 x =1 $ So, let $ u = \tanh x $ ...

Answer 5

Substituting $u=\tanh x$, you will arrive at the standard integral for $\arcsin (u)$.

Answer 6

$$ \int \text{sech}x\, dx=\int \dfrac{\text{sech}^2 x}{\sqrt{\text{sech}^2x}}dx ,$$ 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.