[tex]\displaystyle \int\limits {-\frac{16}{\sin x} } \, dx = \left|\begin{array}{ccc}t = \text{tg} \dfrac{x}{2} \ \ \ \ \ \ \ \ \\x = 2 \, \text{arctg} \, t \ \ \\ dx = \dfrac{2}{1 + t^{2}}dt \end{array}\right| = \int\limits {-\dfrac{16}{\sin (2 \, \text{arctg} \, t)} } \, \dfrac{2}{1 + t^{2}}dt =[/tex]
[tex]= -16\displaystyle \int\limits {\dfrac{2(1 + t^{2})}{2t(1 + t^{2})} } \, dt = -16\int \frac{dt}{t} = -16\ln |t| + C = -16 \ln \left|\text{tg} \dfrac{x}{2} \right| + C[/tex]