Question

Помогите решить ..............................

Answer 1

Ответ:

$2)\ \ \displaystyle \int\limits^{\pi/3}_{\pi /4}\, \frac{1}{sin^2x}\, dx=-ctgx\Big|_{\pi /4}^{\pi /3}=-ctg\frac{\pi}{3}+ctg\frac{\pi}{4}=-\frac{\sqrt3}{3}+1=\frac{3-\sqrt3}{3}\\\\\\3)\ \ \int\limits^{\pi /4}_0\, \frac{1}{cos^2x}\, dx=tgx\Big|_{0}^{\pi /4}=tg\frac{\pi}{4}-tg\, 0=1-0=1\\\\\\4)\ \ \displaystyle \int\limits^{16}_{3}\, \frac{1}{\sqrt{x}}\, dx=3\cdot 2\sqrt{x}\Big|_9^{16}=6\cdot (\sqrt{16}-\sqrt{9})=6\cdot (4-3)=6$