\[\begin{array}{l}
 T9. \to \int {\frac{{dx}}{{{{\sin }^2}x}}}  =  - {\mathop{\rm ctg}\nolimits} x + C \\ 
 {\rm{Dokaz}}\,{\rm{(Proof):}} \\ 
 \int {\frac{{dx}}{{{{\sin }^2}x}}}  = \left| \begin{array}{l}
 {\mathop{\rm tg}\nolimits} x = t \\ 
 x = {\mathop{\rm arctg}\nolimits} t \\ 
 dx = \frac{1}{{1 + {t^2}}}dt \\ 
 \sin x = \frac{t}{{\sqrt {1 + {t^2}} }} \\ 
 \end{array} \right| = \int {\frac{{\frac{1}{{1 + {t^2}}}}}{{{{\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right)}^2}}}dt}  = \int {\frac{{\frac{1}{{1 + {t^2}}}}}{{\frac{{{t^2}}}{{1 + {t^2}}}}}dt}  = \int {\frac{{dt}}{{{t^2}}}}  = \frac{{{t^{ - 1}}}}{{ - 1}} =  - \frac{1}{t} =  - \frac{1}{{{\mathop{\rm tg}\nolimits} x}} =  - {\mathop{\rm ctg}\nolimits} x \\ 
 \end{array}\]