Beautiful Area Between The Lines And The Curve

Let any point on the hyperbola be $(x, \sqrt{x^{2}-1} )$ .
The area under the line drawn from origin to the point is
$\Delta_{1} = \dfrac{1}{2} \times x \times y = \dfrac{x\sqrt{x^{2}-1}}{2}$

The area under the hyperbola is given by,

$\Delta_{2} = \displaystyle \int_{1}^{x}\sqrt{t^{2}-1} dt$

It can be derived very easily integrating by parts,
$\displaystyle \int \sqrt{t^{2}-a^{2}}dt = \dfrac{t\sqrt{t^{2}-a^{2}} - a^{2}\ln\left( t + \sqrt{t^{2}-a^{2}} \right)}{2}$

Putting a=1 and substituting the limits,

$\therefore \Delta_{2} = \dfrac{x\sqrt{x^{2}-1} - \ln\left( x + \sqrt{x^{2}-1}\right)}{2}$

Required area is,
$\Delta_{req} = \Delta_{1} - \Delta_{2}$
$\therefore \Delta_{req} = \dfrac{\ln \left( x + \sqrt{x^{2}-1}\right)}{2}$
For the given question, substitute $x = \dfrac{e^{2016} + e^{-2016}}{2}$ , and $\Delta_{req}$ comes out to be 1008.