$$I = \int {\sqrt {\sec 2x - 1} dx=\int {\sqrt {{{1 - \cos 2x} \over {\cos 2x}}} dx} } $$

$$ = \int {\sqrt {{{2{{\sin }^2}x} \over {2{{\cos }^2}x - 1}}} dx} $$

Let $$\sqrt {2\cos } x = t - \sqrt 2 \sin xdx = dt$$

$$I = \int { - {{dt} \over {\sqrt {{t^2} - 1} }} = - \ln \left| {t + \sqrt {{t^2} - {a^2}} } \right| + c} $$

$$ = - \ln \left| {\sqrt 2 \cos x + \sqrt {2{{\cos }^2}} x - 1} \right| + c$$

$$ = - {1 \over 2}\ln \left| {2{{\cos }^2}x + \cos 2x + 2\sqrt 2 \sqrt {\cos 2x.{{\cos }^2}x} } \right| + c$$

$$ = - {1 \over 2}\ln \left| {2\cos 2x + 1 + 2\sqrt {\cos 2x(1 + \cos 2x)} } \right| + c$$

$$ = - {1 \over 2}\ln \left| {\cos 2x + {1 \over 2} + \sqrt {\cos 2x(1 + \cos 2x)} } \right| + c$$

$$\therefore$$ $$\alpha = {{ - 1} \over 2},\beta = {1 \over 2}$$

$$\therefore$$ $$\beta - \alpha = 1$$