In the following, I don't understand how they put $\tan{\phi} = \sinh{\frac{\psi}{\sqrt{2}}}$.
Is there a relationship?
This relationship between trigonometric and hyperbolic function
calculuscomplex numbersmultivariable-calculus
Related Solutions
Do you mean $\frac {dy}{dx}$ is reminiscent of the triangle definition of tangent $\tan \theta=\frac{\text{opposite}}{\text{adjacent}}$?
The derivative of a function at a point can be interpreted as the slope of the tangent line to that point on the graph of the function. This is distinct from the function tangent, which can be geometrically interpreted as the length of a special tangent to a unit circle (see below) given a certain angle.
You could connect them in a roundabout way - if you take the tangent line to a function's graph at a certain point, then extent it to intersect the x-axis, the tangent of the angle it forms with the x-axis (measured counterclockwise from the x-axis) will be the derivative of the function at that point.
There are in fact two sets of formulas for $F(\sin^{-1}u,m)$ with $u>1$, depending on whether $u>\frac1{\sqrt m}$ or not. Both are very simple and don't have the problem of the tangent blowing up at $\pi/2$.
If $1\le u\le\frac1{\sqrt m}$, Byrd and Friedman 115.02 gives $$F(\sin^{-1}u,m)=K(m)+iF\left(\sin^{-1}\frac{\sqrt{u^2-1}}{u\sqrt{1-m}},1-m\right)$$ If $\frac1{\sqrt m}\le u$, Byrd and Friedman 115.03 gives $$F(\sin^{-1}u,m)=F\left(\sin^{-1}\frac1{u\sqrt m},m\right)+iK(1-m)$$
Best Answer
This isn't asserting a relationship. It's simply setting up a substitution that trades circular functions for hyperbolics in order to escape the square root. It's really no different than having $\sqrt{4-x^2}$ and setting $x=2\sin\theta$.
In the exercise, first note that the expression (ignoring $d\theta$) can be rewritten as $$a^2\left((1+\tan^2\phi)+\tan^2\phi\right)^{1/2} \sec^2\phi\, d\phi = a^2\left(1+2\tan^2\phi\right)^{1/2} \sec^2\phi\,d\phi$$ The author is suggesting that the form $1+2\tan^2\phi$ calls for the substitution $$\tan\phi = \frac{\sinh\psi}{\sqrt{2}} \qquad\qquad \sec^2\phi\,d\phi=\frac{\cosh\psi}{\sqrt{2}}\,d\psi$$ (note that the entire $\sinh\psi$ is divided by $\sqrt{2}$, not just $\psi$) so that we have $$\begin{align} a^2\left(1+2\cdot\frac{\sinh^2\psi}{2}\right)^{1/2} \frac{\cosh\psi\,d\psi}{\sqrt{2}} &= \frac1{\sqrt{2}} a^2\left(1+\sinh^2\psi\right)^{1/2} \cosh\psi\,d\psi \\[4pt] &= \frac1{\sqrt{2}}a^2\left(\cosh^2\psi\right)^{1/2} \cosh\psi\,d\psi \\[4pt] &= \frac1{\sqrt{2}}a^2\cosh^2\psi\,d\psi \tag1 \end{align}$$
Alternatively, one could make a standard circular substitution thusly: $$\tan\phi = \frac{\tan\omega}{\sqrt{2}} \qquad\qquad \sec^2\phi\,d\phi = \frac{\sec^2\omega}{\sqrt{2}}d\omega$$ to give $$\begin{align} a^2\left(1+2\cdot\frac{\tan^2\omega}{2}\right)^{1/2} \frac{\sec^2\omega d\omega}{\sqrt{2}} &= \frac1{\sqrt{2}}a^2\left(1+\tan^2\omega\right)^{1/2} \sec^2\omega\,d\omega \\[4pt] &= \frac1{\sqrt{2}}a^2\left(\sec^2\omega\right)^{1/2} \sec^2\omega\,d\omega \\[4pt] &= \frac1{\sqrt{2}}a^2\sec^3\omega\,d\omega \tag2 \end{align}$$ Expression $(1)$ has fairly straightforward integral, which in turn has a fairly straightforward back-substitution into $\phi$-form. Expression $(2)$ is perhaps a little messier on both counts. (The author may have anticipated this, or the author may have had some other reason to favor the hyperbolic substitution.) The end result in both cases is the same.
By the way, a "third" way to proceed is with this substitution: $$\tan\phi = \frac{u}{\sqrt{2}} \qquad\qquad \sec^2\phi\,d\phi=\frac{du}{\sqrt{2}}$$ This gives $$\begin{align} a^2\left(1+2\cdot\frac{u^2}{2}\right)^{1/2} \frac{du}{\sqrt{2}} &= \frac1{\sqrt{2}}a^2\left(1+u^2\right)^{1/2}du \tag3 \end{align}$$ From here, substitution $u=\sinh\psi$ reverts to $(1)$, and $u=\tan\omega$ reverts to $(2)$, so this isn't necessarily a different approach. On the other hand, integration by parts and back-substituting are pretty straightforward from here, too.