Euler-Lagrange equation (equation of motion) solution with hairy Lagrangian [closed]

I'm going through Zwiebach Chapter 6 on relativistic strings to try to solve a similar problem. I got all the way to my equation of motion

\begin{eqnarray*} \delta S & = & [ p' \delta \theta]_{z 0}^{z 1} + \int_{z 0}^{z 1} d z \left( p - \frac{\partial ( p')}{\partial z} \right) \delta \theta\\ & & \\ \Longrightarrow p - \frac{\partial ( p')}{\partial z} & = & 0\\ & & \\ {where} : & & \\ p & = & \frac{\partial L ( z, \theta, \theta')}{\partial \theta}\\ p' & = & \frac{\partial L ( z, \theta, \theta')}{\partial \theta'} \end{eqnarray*} The Lagrangian I have is $\begin{array}{lll} L ( z, \theta, \theta') & = & \frac{\cos^3 \theta}{z^5} \sqrt{1 + z^2 \theta'^2} \end{array}$, and I know that this equation is the Euler Lagrange equation.

\begin{eqnarray*} \frac{\partial L}{\partial \theta} - \frac{\partial}{\partial z} \left( \frac{\partial L}{\partial \theta'} \right) & = & 0\\ & & \\ {where} : & & \\ \theta' ( z) & = & \frac{\partial \theta ( z)}{\partial z}\\ L ( z, \theta, \theta') & = & \frac{\cos^3 \theta}{z^5} \sqrt{1 + z^2 \theta'^2} \end{eqnarray*} Just putting the expression for $L$ into the equation gives me a mess \begin{array}{lll} \frac{- 3 \sin \theta \cos^2 \theta}{z^5} \sqrt{1 + z^2 \theta'^2} - \frac{\sqrt{2} \theta'' ( 2 \theta'^2 z^2 + 3) \cos^3 \theta}{z^4 ( \theta'^2 z^2 + 2)^{3 / 2}} & = & 0 \end{array} I don't think I'm supposed to brute force this. I know the solution to be $\theta = \arcsin ( z)$, but I can't see how to get it. I'm thinking that there is something I'm missing about the equation of motion.

closed as off-topic by ACuriousMind♦, Rob Jeffries, JamalS, Kyle Kanos, Wolphram jonnyDec 16 '14 at 15:08

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• oh and here is the link to the same question on physicsforums physicsforums.com/threads/… – user52205 Dec 16 '14 at 5:17
• This question seems to have very little physical content; is it purely a question about the manipulations to solve your particular equation, or do you think there's something physical that's stopping you from getting it? – Danu Dec 16 '14 at 12:53
• Which exercise in Zwiebach's book is this? Please provide context. – Qmechanic Dec 18 '14 at 23:02