$\cfrac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }},x \in \left[ {0,1} \right]$

: Let $I = \int {\cfrac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }}dx}$

$= \int {\cfrac{{{{\sin }^{ - 1}}\sqrt x - \left( {\cfrac{\pi }{2} - {{\sin }^{ - 1}}\sqrt x } \right)}}{{\cfrac{\pi }{2}}}} dx$

$= \cfrac{2}{\pi }\int {\left[ {2{{\sin }^{ - 1}}\sqrt x - \cfrac{\pi }{2}} \right]dx} = \cfrac{4}{\pi }\int {{{\sin }^{ - 1}}\sqrt x dx} - \int {\left( 1 \right)dx}$

…(i)
Let ${I_1} = \int {{{\sin }^{ - 1}}\sqrt x dx}$

Put $\sqrt x = t$ $\Rightarrow$ $x = {t^2}$ $\Rightarrow$ $dx = 2t\,dt$

$\therefore$ ${I_1} = 2\int {{{\sin }^{ - 1}}t \cdot t} dt$
$= 2\left[ {{{\sin }^{ - 1}}t\int {tdt} - \int {\left( {\cfrac{d}{{dt}}\left( {{{\sin }^{ - 1}}t} \right)\int t dt} \right)} dt} \right]$

$= 2\left[ {{{\sin }^{ - 1}}t \cdot \cfrac{{{t^2}}}{2} - \int {\cfrac{1}{{\sqrt {1 - {t^2}} }}\cfrac{{{t^2}}}{2}} dt} \right] + {C_1}$

$= {t^2}{\sin ^{ - 1}}t + \int {\cfrac{{1 - {t^2} - 1}}{{\sqrt {1 - {t^2}} }}dt + {C_1}}$

$= {t^2}{\sin ^{ - 1}}t + \int {\sqrt {1 - {t^2}} dt} - \int {\cfrac{1}{{\sqrt {1 - {t^2}} }}dt} + {C_1}$

$= {t^2}{\sin ^{ - 1}}t + \left( {\cfrac{t}{2}\sqrt {1 - {t^2}} + \cfrac{1}{2}{{\sin }^{ - 1}}t} \right) - {\sin ^{ - 1}}t + {C_1}$

$= x{\sin ^{ - 1}}\sqrt x + \cfrac{{\sqrt x \sqrt {1 - x} }}{2} - \cfrac{1}{2}{\sin ^{ - 1}}\sqrt x + {C_1}$

$= \left( {x - \cfrac{1}{2}} \right){\sin ^{ - 1}}\sqrt x + \cfrac{{\sqrt x \sqrt {1 - x} }}{2} + {C_1}$

$\therefore$ Form (i)

we have
$I = \cfrac{4}{\pi }\left( {\cfrac{{2x - 1}}{2}} \right){\sin ^{ - 1}}\sqrt x + \cfrac{{2\sqrt x \sqrt {1 - x} }}{\pi } - x + C$

$\left[ {C = \cfrac{4}{\pi }{C_1}} \right]$

$= \cfrac{{2\left( {2x - 1} \right)}}{\pi }{\sin ^{ - 1}}\sqrt x + \cfrac{{2\sqrt x \sqrt {1 - x} }}{\pi } - x + C$