I want to minimize the arclength of a function $u(x) \geq 0$ for $x\in [-1,1]$ that is contrained by $u(-1)=0=u(1)$ and $\int_{-1}^{1} u(x) \, dx = A$, where $0<A<\pi/2.$ I have reduced the differential equation implied by the Euler-Lagrange theorem to $$ -u''(x)=z(1+u'(x)^2)^{3/2} $$ where $z$ is an unknown constant whose existence is implied by the integral constraint. I'm quite unsure how to solve this differential equation. The presence of the $1+u'(x)^2$ and the $\pi/2$ seem to indicate a tangent substitution but my setup doesn't seem to work. If we set $u'(x) = \tan(\theta)$ then taking the derivative w.r.t. $x$ produces $$ u''(x) = \sec(\theta)^2 \frac{d \theta}{ d x} \Longrightarrow u''(x) \, dx = \sec(\theta)^2 \, d\theta $$
But when I try to use this I just go in a circle. Any other ideas of how to fix my approach or find another way to solve the differential equation are appreciated.
You already have that $u''=-z(1+u'^2)^{3/2} $ and that $u'=tan(\theta),\ u''=sec^2(\theta)\frac{d\theta}{dx} $, so just go ahead and make the substitution to get $sec^2(\theta)\frac{d\theta}{dx}=-z(1+tan^2(\theta))^{3/2} $. Use some trig identities and you can simplify it down to $cos(\theta)d\theta=-zdx $, integrate to get $sin(\theta)=c-zx $ and substitute back. $u'=tan(\theta) $ so $\theta=tan^{-1}(u') $ and $sin(tan^{-1}(u'))=\frac{u'}{\sqrt{1+u'^2}}=c-zx $. Solve for $u'$ to get $u'=\sqrt{\frac{(c-zx)^2}{1-(c-zx)^2}} $ and integrate both sides to get $u=\frac{c-zx}{z}\sqrt{\frac{1-(c-zx)^2}{(c-zx)^2}} $.