Definite Integral $ 4\pi\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}(t)+\sinh^{2}(t)} dt $ Consider the integral
$$4\pi\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}.$$
This definite integral arose while computing the surface area of a hyperboloid.
The hyperboloid is parameterized by
$$
\begin{align*}
{x}&=(\cosh({t}))(\cos(\theta))\text{;}\\
{y}&=(\cosh({t}))(\sin(\theta))\text{;}\\
{z}&=\sinh({t})\\
\end{align*}
$$
for
$$
\begin{align*}
0&\leq\theta\leq{2}\pi\text{;}\\
-1&\leq{t}\leq{1}\text{.}\\
\end{align*}
$$
Recall that $\cosh^{2}({t})-\sinh^{2}({t})=1$.
The position vector $\vec{r}$ is given by
$$
\vec{r}({x}({t},\theta),{y}({t},\theta),{z}({t},\theta))=(\cosh({t}))(\cos(\theta))\hat{\imath}+(\cosh({t}))(\sin(\theta))\hat{\jmath}+(\sinh({t}))\hat{k}\text{,}
$$
and its partial derivatives are
$$
\begin{align*}
\frac{\partial\vec{r}}{\partial{t}}&=(\sinh({t}))(\cos(\theta))\hat{\imath}+(\sinh({t}))(\sin(\theta))\hat{\jmath}+(\cosh({t}))\hat{k}\text{;}\\
\frac{\partial\vec{r}}{\partial\theta}&=(\cosh({t}))(-\sin(\theta))\hat{\imath}+(\cosh({t}))(\cos(\theta))\hat{\jmath}+0\hat{k}\\
\end{align*}
$$
with respect to $t$ and $\theta$.
The area of the parallelogram formed by the two partial derivatives is given by the magnitude of their cross product.
$$
\begin{align*}
\frac{\partial\vec{r}}{\partial{t}}\times\frac{\partial\vec{r}}{\partial\theta}&=\begin{vmatrix} \hat{\imath}&\hat{\jmath}&\hat{k}\\(\sinh({t}))(\cos(\theta))&(\sinh({t}))(\sin(\theta))&\cosh({t})\\(\cosh({t}))(-\sin(\theta))&(\cosh({t}))(\cos(\theta))&0\end{vmatrix}\\
&=-\cosh^{2}({t})\cos(\theta)\hat{\imath}-\cosh^{2}({t})\sin(\theta)\hat{\jmath}+\sinh({t})\cosh({t})\hat{k}\text{;}\\
\left\|\frac{\partial\vec{r}}{\partial{t}}\times\frac{\partial\vec{r}}{\partial\theta} \right\|&=\cosh({t})\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}\text{.}\\
\end{align*}
$$
Integrating over the surface yields
$$\int_{0}^{2\pi}\int_{-1}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}{d}\theta=2\pi\int_{-1}^{1}\cosh({t})\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}\text{.}$$
By symmetry, the integral is equivalent to
$$4\pi\int_{0}^{1}\cosh({t})\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}\text{.}$$
According to Wolfram Alpha, the surface area of a hyperboloid is
$$4\pi\int_{0}^{1}\cosh({t})\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}=\pi\left(\sqrt{2}\sinh^{-1}(\sqrt{2}\sinh(1))+2\sinh(1)\sqrt{\cosh(2)}\right)\text{.}$$
How does one solve the integral $4\pi\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}$.
 A: Here comes the first steps.
You use the hyperbolic one,
$$
\cosh^2t-\sinh^2t=1,
$$
which gives you
$$
4\pi\int_0^1\cosh t\sqrt{1+2\sinh^2t}\,dt
$$
Now, let $u=\sinh t$, and you will get $du=\cosh t\,dt$, and so
$$
4\pi\int_0^{\sinh 1}\sqrt{1+2u^2}\,du.
$$
Can you proceed from here, finding a primitive of $\sqrt{1+2u^2}$?
A: As mickep suggests, $$ 4\pi\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}\,{dt}$$ can be converted to $$ 4\pi\int_0^{\sinh 1}\sqrt{1+2u^2}\,du$$
The next step is to substitute again with $$ u = \tfrac{\sqrt{2}}{2}\sinh v $$ and $$ du = \tfrac{\sqrt{2}}{2}\cosh v\,dv $$ so that the integral becomes $$  4\pi\int_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}\sqrt{\sinh^2 v + 1}\cdot\tfrac{\sqrt{2}}{2}\cosh v\,dv$$ and $$2\pi\sqrt{2}\int_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}\cosh^2 v\,dv $$
which evaluates to $$2\pi\sqrt{2}\left(\tfrac{1}{2}\sinh v \cosh v + \tfrac{1}{2} v\right|_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}$$ $$
\pi\sqrt{2}\left(\sinh v \cosh v + v\right|_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}$$ $$
\pi\sqrt{2}\left(\sinh v \sqrt{\sinh^2 v + 1} + v\right|_0^{\sinh^{-1}\left(\sqrt{2}\sinh 1\right)}$$ $$
\pi\sqrt{2}\left(\sqrt{2}\sinh 1 \sqrt{2\sinh^2 1 + 1} + \sinh^{-1}\left(\sqrt{2}\sinh 1\right)\right) $$ and finally $$\pi\left(2\sinh 1 \sqrt{\cosh 2} + \sqrt{2}\sinh^{-1}\left(\sqrt{2}\sinh 1\right)\right)$$
That last can be changed to $$\pi\sqrt{2}\left(\sqrt{2}\sinh 1 \sqrt{\cosh 2} + \ln\left(\sqrt{2}\sinh 1+\sqrt{\cosh 2}\right)\right)$$ which follows a common pattern of hyperbolic integrals where the result is a product of two values plus the natural logarithm of the sum of those values.
A: Another way 
(with the help of Wolfy):
$\begin{array}\\
I
&=\int_{0}^{1}\cosh(t)\sqrt{\cosh^{2}({t})+\sinh^{2}({t})}{dt}\\
&=\int_{0}^{1}\frac12(e^t+e^{-t})\sqrt{\frac14(e^t+e^{-t})^{2}+\frac14(e^t-e^{-t})^{2}}{dt}\\
&=\frac14\int_{0}^{1}(e^t+e^{-t})\sqrt{(e^{2t}+2+e^{-2t})+(e^{2t}-2+e^{-2t})}{dt}\\
&=\frac14\int_{0}^{1}(e^t+e^{-t})\sqrt{2(e^{2t}+e^{-2t})}{dt}\\
&=\frac{\sqrt{2}}{4}\int_{0}^{1}(e^t+e^{-t})\sqrt{e^{2t}+e^{-2t}}{dt}\\
&=\frac{\sqrt{2}}{4}\int_{0}^{1}(1+e^{-2t})\sqrt{e^{4t}+1}{dt}\\
&=\frac{\sqrt{2}}{4}\int_{1+e^{-2}}^{2}x\sqrt{\frac1{(x-1)^2}+1}\dfrac{dx}{2(x-1)}\\
&=\frac{\sqrt{2}}{8}\int_{1+e^{-2}}^{2}\dfrac{x\sqrt{(x-1)^2+1}}{(x-1)^2}dx\\
&=\frac{\sqrt{2}}{8}\int_{e^{-2}}^{1}\dfrac{(x+1)\sqrt{x^2+1}}{x^2}dx\\
&=\frac{\sqrt{2}}{8}\left(\int_{e^{-2}}^{1}\dfrac{\sqrt{x^2+1}}{x}dx+\int_{e^{-2}}^{1}\dfrac{\sqrt{x^2+1}}{x^2}dx\right)\\
&=\frac{\sqrt{2}}{8}\left( \sqrt{x^2 + 1} - \ln(\sqrt{x^2 + 1} + 1) + \ln(x)+\sinh^{-1}(x) - \dfrac{\sqrt{x^2 + 1}}{x}\right)\big|_{e^{-2}}^{1}
\qquad\text{Wolfy did this}\\
&=\frac{\sqrt{2}}{8}\left( (1-\dfrac1{x})\sqrt{x^2 + 1} - \ln(\sqrt{x^2 + 1} + 1) + \ln(x)+\sinh^{-1}(x) \right)\big|_{e^{-2}}^{1}\\
&=\frac{\sqrt{2}}{8} (f(1)-f(e^{-2}) )\\
f(x)
&= (1-\dfrac1{x})\sqrt{x^2 + 1} - \ln(\sqrt{x^2 + 1} + 1) + \ln(x)+\sinh^{-1}(x) \\
f(1)
&=  - \ln(\sqrt{2} + 1) +\sinh^{-1}(1) \\
f(e^{-2})
&= (1-e^2)\sqrt{e^{-4} + 1} - \ln(\sqrt{e^{-4} + 1} + 1) -2+\sinh^{-1}(e^{-2})\\
I
&=\frac{\sqrt{2}}{8} ( - \ln(\sqrt{2} + 1) +\operatorname{arcsinh}(1)\\
&\quad - ((1-e^2)\sqrt{e^{-4} + 1} - \ln(\sqrt{e^{-4} + 1} + 1) -2+\operatorname{arcsinh}(e^{-2})))\\
\end{array}
$
$x = e^{-2t}+1,
dx = -2e^{-2t}dt
=-2(x-1)dt,
dt
=\dfrac{-dx}{2(x-1)}
$
