\[\begin{array}{l}
 T10. \to \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}  = \arcsin x + C =  - \arccos x + C \\ 
 {\rm{Dokaz}}\,{\rm{(Proof):}} \\ 
 \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}  = \left| \begin{array}{l}
 x = \sin t \Rightarrow t = \arcsin x \\ 
 dx = \cos tdt \\ 
 \end{array} \right| = \int {\frac{{\cos tdt}}{{\sqrt {1 - {{\sin }^2}t} }}}  = \left| \begin{array}{l}
 {\sin ^2}t + {\cos ^2}t = 1 \\ 
 {\cos ^2}t = 1 - {\sin ^2}t \\ 
 \end{array} \right| = \int {\frac{{\cos tdt}}{{\cos t}}dt}  = \int {dt}  = t = \arcsin x \\ 
 Ili\,(Or) \\ 
 \int {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}  = \left| \begin{array}{l}
 x = \cos t \Rightarrow t = \arccos x \\ 
 dx =  - \sin tdt \\ 
 \end{array} \right| = \int {\frac{{ - \sin tdt}}{{\sqrt {1 - {{\cos }^2}t} }}}  = \left| \begin{array}{l}
 {\sin ^2}t + {\cos ^2}t = 1 \\ 
 {\sin ^2}t = 1 - {\cos ^2}t \\ 
 \end{array} \right| = \int {\frac{{ - \sin tdt}}{{\sin t}}dt}  =  - \int {dt}  =  - t =  - \arccos x \\ 
 \end{array}\]