Let us take the required equation

$$y = \sqrt {{{1 + \sin x} \over {\cos x}}} $$ and $$y = \sqrt {{{1 - \sin x} \over {\cos x}}} $$

$$\int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{1 + \sin x} \over {\cos x}}} - \sqrt {{{1 - \sin x} \over {\cos x}}} } \right)dx} $$

$$\because$$ $$\left( {{{1 + \sin x} \over {\cos x}} > {{1 - \sin x} \over {\cos x}} > 0} \right)$$

$$ = \int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{1 + {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \over {{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}}}} - \sqrt {{{1 + {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \over {{{{{(1 - \tan {x \over 2})}^2}} \over {1 - \tan {x \over 2}}}}}} } \right)dx} $$

$$ = \int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{{{\left( {1 + {{\tan }^2}{x \over 2}} \right)}^2}} \over {1 - {{\tan }^2}{x \over 2}}}} - \sqrt {{{{{\left( {1 - \tan {x \over 2}} \right)}^2}} \over {1 - {{\tan }^2}{x \over 2}}}} } \right)dx} $$

$$\int\limits_0^{{\pi \over 4}} {{{1 + \tan {x \over 2} - 1 + \tan {x \over 2}} \over {\sqrt {1 - {{\tan }^2}{x \over 2}} }}dx} $$

$$ = \int\limits_0^{{\pi \over 4}} {{{2\tan {x \over 2}} \over {\sqrt {1 - {{\tan }^2}{x \over 2}} }}dx} $$

Put $$\tan {x \over 2} = t$$

$$dx = {{2\,dt} \over {1 + {t^2}}}$$

$$A = \int\limits_0^{\sqrt 2 - 1} {{{4t} \over {(1 + {t^2})\sqrt {1 - {t^2}} }}dt} $$