QJEEM23Ma6618: JEE Main 2023 Mathematics - Class 12 JEEM23Ma6618 Solution: If constant, then is equal to ____________.

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

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

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

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

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

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

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

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

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

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