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## Homework Statement

For the following integral, find F and its partial derivatives and plug them into the Euler Lagrange equation

$$F(y,x,x')=y\sqrt{1+x'^2}\\$$

## Homework Equations

Euler Lagrange equation : $$\frac{dF}{dx}-\frac{d}{dy}\frac{dF}{dx'}=0$$

## The Attempt at a Solution

$$A=2\pi\int_{Y1}^{Y2} y\sqrt{1+x'^2}dy$$

$$F(y,x,x')=y\sqrt{1+x'^2}\\$$

$$\frac{dF}{dx}=0\\$$

$$\frac{dF}{dx'}=\frac{yx'}{\sqrt{1+x'^2}}\\$$

$$0-\frac{d}{dy}\frac{dF}{dx'} = -\frac{d}{dy}\frac{yx'}{\sqrt{1+x'^2}}=0$$

The books answer i s this :

$$\frac{yx'}{\sqrt{1+x'^2}}=C\\

x=acosh^-1\frac{y}{a}+b$$

I dont understand how they get a cosh function. The integral according to wolfram alpha is $$\frac{yx*x'}{\sqrt{x'y^2+1}} + c$$

Is there some identity to turn wolframs into the books answer?