Express $\operatorname{sech}^{-1}(x)$ in terms of logarithms

Question

I'm trying to express the following $\operatorname{sech}^{-1}(x)$ in terms of logarithms, and would warmly appreciate feedback towards my approach. The solution should be :

$$\ln\left(\dfrac{1+\sqrt{(1-x^2)}}{x}\right)$$ when $x > 0.$ However, I cannot seem to get this. My working out:

$y = \operatorname{sech}^{-1}(x)\implies \operatorname{sech}(y) = x$

$$-\operatorname{sech}(y)\operatorname{tan}(y) = x; $$

Once I got to here, I decided to transform the $\operatorname{sech}$ and $\tan$ into their hyperbolic identities.

$$-\dfrac{\operatorname{sinh}(y)}{\operatorname{cosh}^2(y)}=\dfrac{e^y-e^{-y}}{2}(\dfrac{e^y+e^{-y}}{2})^2 = x$$

Though I've tried reworking this into a quadratic form $\dfrac{b \space \pm \space \sqrt{b^2-4ac}}{2a}$, however, I couldn't manage to get the right form. I would greatly appreciate some help on the next steps towards this.

Answer 1

Once you have $\operatorname{sech}(y) = x$ it implies $$\dfrac{e^y+e^{-y}}2=\dfrac1x$$ which simplifies to the quadratic $$xe^{2y}-2e^y+x=0$$ of $e^y.$ Now solve for $y$ as you discribed in the last sentence.

Answer 2

By definition $\operatorname{sech}(x)$ is equal to $$y = \frac{2}{e^x+e^{-x}}.$$ What we now want to do is solve for $x$. First, divide both sides by $2$ and then take the multiplicative inverse of both sides to get $$\frac{2}{y}=e^x+e^{-x}$$ Now substitute $u=e^x.$ We will get a new and easy equation. $$u+\frac{1}{u}=\frac{2}{y}.$$ You should be able to solve this equation for $u$ $$u=\pm\sqrt{y^{-2}-1}+\frac{1}{y}.$$ If you are done, don't forget to substitute back $e^x=u$.

Answer 3

Forming a quadratic as outlined in the above answers is a fine approach, and one I recommend above this one. Here's a different one, which relies on knowing beforehand that $$\cosh^{-1}x=\ln(x+\sqrt{x^2-1})$$ Let $y=\operatorname{sech} ^{-1} x$. Then $$\begin{align} \operatorname{sech}y=x&\iff \cosh y=\frac{1}{x}\\ &\iff y=\operatorname{sech}^{-1}x=\cosh^{-1}\frac{1}{x}=\ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right) \end{align}$$ Simplifying yields the correct answer. I hope that was helpful. If you have any questions please don't hesitate to ask :)

Answer 4

Note

\begin{align} \text{sech}\left(\ln\frac{1+\sqrt{1-x^2}}x \right) =& \frac2{ \exp\left(\ln\frac{1+\sqrt{1-x^2}}x \right)+\exp\left(-\ln \frac{1+\sqrt{1-x^2}}x\right)}\\ = &\frac2{ \frac{1+\sqrt{1-x^2}}x + \frac x{1+\sqrt{1-x^2}}} =\frac2{ \frac{1+\sqrt{1-x^2}}x + \frac{1-\sqrt{1-x^2}}x } =x \end{align}

Thus

$$\text{sech}^{-1}x =\ln\frac{1+\sqrt{1-x^2}}x $$