Integral of a tangent function $$ \displaystyle {\int_{0}^{z}} \sqrt {1 + \tan^2(\dfrac{\pi}{4} \dfrac{z}{H} )} dz $$
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$$ gives $$ 
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$$ \dfrac{4H}{\pi} {\sinh^{-1}} ( {\tan \dfrac{\pi}{4} \dfrac{z}{H} } ) $$
Please advise solution
edit:- 
I can get to 
$$\dfrac{4H}{\pi} \displaystyle {\int_{0}^{\dfrac{\pi z}{4H}}} \sec {u} {du}$$
Please help after this step ?
 A: Use the identity:
$$\tan^2 \theta + 1 = \sec^2 \theta$$
Though the answer won't necessarily look like the answer given. Does it have to be in terms of $\sinh^{-1}$?
A: 1.) Define $\frac{\pi/4 }{H}=a $
2.) Substitute $a z'=\arctan(r),\quad dz'=\frac{dr}{a(1+r^2)}$. This gives
$$
I(a)=\frac{1}{a}\int_0^{\tan(a z)}\frac{\sqrt{1+r^2}}{1+r^2}=\frac{1}{a}\int_0^{\tan(a z)}\frac{1}{\sqrt{1+r^2}}
$$
Furthermore $\int\frac{1}{\sqrt{1+r^2}}=\text{arcsinh}(r)+C$
and therefore
$$
I(a)=\frac{1}{a}\text{arcsinh}(\tan(az))
$$
Done!
Sidenote:
This integral is only well be behaved (in a straightforward manner) as long as $az< \pi/2$, because of the divergence of $\tan$
A: There are always more than one way to represent a solution.  
So, let's note a couple of things here.

First, note that the hyperbolic sine function $\sinh x =\frac12 (e^x-e^{-x})$ has inverse function 
$$\sinh^{-1}x=\log\left(x+\sqrt{1+x^2}\right)$$
To see this, let's solve the equation $x=\sinh y$ for $y$.  Thus,
$$\begin{align}
x&=\frac12 (e^y-e^{-y})\\
0&=e^y-e^{-y}-2x\\
0&=(e^y)^2-2xe^y-1
\end{align}$$
whereupon solving the quadratic formula for $e^y$ reveals that
$$e^y=x + \sqrt{x^2+1}$$
Observe that we rejected the "negative" square root solution since $e^y>0$. Finally, taking logarithms on both sides yields the result
$$y=\sinh^{-1}x =\log\left(x+\sqrt{1+x^2}\right)$$

The second thing to note is that the integral of the secant function $\int \sec x dx$ is given by
$$\begin{align}
\int \sec x dx &= \log |\tan x+\sec x|+C\\
&=\log |\tan x+\sqrt{1+\tan^2x}|+C\\
&=\sinh^{-1}(\tan x)+C
\end{align}$$
where in going from the first line to the second, we restricted $x$ such that $|x|<\pi/2$.  So, we see that, as always, there are alternative ways to express the result.  One way here is to use the log function while another way is to use the inverse hyperbolic sine function.
