...........................

Решите задачу:

$\displaystyle\int\limits^{\displaystyle\frac{\pi}{3}}_{\displaystyle\frac{\pi}{2}}\frac{dx}{sin^2(\frac{x}{2})}=-2\int\limits^{\displaystyle\frac{\pi}{2}}_{\displaystyle\frac{\pi}{3}}\frac{d(\frac{x}{2})}{sin^2(\frac{x}{2})}=2ctg\frac{x}{2}|^{\displaystyle\frac{\pi}{2}}_{\displaystyle\frac{\pi}{3}}=2ctg\frac{\pi}{4}-2ctg\frac{\pi}{6}=\\=2-2\sqrt3\approx-1,46$

$\displaystyle\int\limits^{\displaystyle\frac{\pi}{3}}_{\displaystyle\frac{\pi}{2}}\frac{dx}{sin^2(\frac{x}{2})}=-2\int\limits^{\displaystyle\frac{\pi}{4}}_{\displaystyle\frac{\pi}{6}}\frac{dt}{sin^2(t)}=2ctgt=2ctg\frac{\pi}{4}-2ctg\frac{\pi}{6}=2-2\sqrt3\approx\\\\\approx-1,46\\\\\\\\\frac{x}{2}=t=\ \textgreater \ dt=\frac{1}{2}dx=\ \textgreater \ dx=2dt\\\\t_1=\frac{\pi}{4}\ t_2=\frac{\pi}{6}$