minimal surface of revolution integrating Euler Lagrange result In finding minimal surface of revolution after applying to euler lagrange equation $d u \over dx \sqrt{1+(u')^2}$$=0$.
Then $ u \over  \sqrt{1+(u')^2}$$=constant$.
Then solving $\int{ c \over \sqrt{u^2-c^2}}du=x+k$.Then using cosh
$cosh^{-1}{u\over c}=x+k$ and hence $u=c *cosh({x+k \over c})$.  
I don't understand how this integration is performed.  
That is how $\int{ c \over \sqrt{u^2-c^2}}du=x+k$ is computed.Can someone please explain it to me
 A: Starting from $u = c\sqrt{1 + (u')^{2}}$ (your ODE after using Euler-Lagrange, with the radical moved to the right-hand side), solving for $u'$ gives
$$
u^{2} = c^{2}\bigl(1 + (u')^{2}\bigr) = c^{2} + c^{2} (u')^{2},
$$
or
$$
\frac{c\, du}{\sqrt{u^{2} - c^{2}}} = dx.
\tag{1}
$$
The hyperbolic substitution $u = c\cosh t$ gives $du = c\sinh t\, dt$ and
$$
\sqrt{u^{2} - c^{2}} = c\sqrt{\cosh^{2} t - 1} = c\sinh t;
$$
the hyperbolic identity $\cosh^{2} t - 1 = \sinh^{2} t$ replaces the difference of squares under the radical with a perfect square. That is,
$$
\frac{du}{\sqrt{u^{2} - c^{2}}} = \frac{c\sinh t\, dt}{c\sinh t} = dt.
\tag{2}
$$
Consequently, 
\begin{align*}
x + k &= \int dx \\
  &= \int \frac{c\, du}{\sqrt{u^{2} - c^{2}}} && \text{by (1)} \\
  &= \int c\, dt && \text{by (2)} \\
  &= ct
  = c\cosh^{-1}\left(\frac{u}{c}\right).
\end{align*}
(The constant of integration only needs to be added to one side.) Solving for $u$ in terms of $x$ gives
$$
u = c\cosh\left(\frac{x + k}{c}\right).
$$
