\[\begin{array}{l}
 T14. \to \int {\frac{{dx}}{{{{\cosh }^2}x}}}  = {\mathop{\rm tgh}\nolimits} x + C \\ 
 {\rm{Dokaz}}\,{\rm{(Proof):}} \\ 
 \int {\frac{{dx}}{{{{\cosh }^2}x}}}  = \left| \begin{array}{l}
 {\mathop{\rm tgh}\nolimits} x = t \\ 
 x = {\mathop{\rm arctgh}\nolimits} t \\ 
 dx = \frac{1}{{1 - {t^2}}}dt \\ 
 \cos x = \frac{1}{{\sqrt {1 - {t^2}} }} \\ 
 \end{array} \right| = \int {\frac{{\frac{1}{{1 - {t^2}}}}}{{{{\left( {\frac{1}{{\sqrt {1 - {t^2}} }}} \right)}^2}}}dt}  = \int {\frac{{\frac{1}{{1 - {t^2}}}}}{{\frac{1}{{1 - {t^2}}}}}dt}  = \int {dt}  = t = {\mathop{\rm tgh}\nolimits} x \\ 
 \end{array}\]