Question

Помогите! Пожалуйста. ​

Answer 1

Ответ:

$\frac{5\sqrt{3}-18}{12}$

Объяснение:

$=\int\limits^\frac{\pi}{4} _\frac{\pi}{3} {\frac{2}{sin^2x} } \, dx +\int\limits^\frac{\pi}{4} _\frac{\pi}{3} {cos2x} } \, dx=-2ctgx|^\frac{\pi}{4} _\frac{\pi}{3} +\frac{1}{2}\int\limits^\frac{\pi}{4} _\frac{\pi}{3} {cos2x} } \, d(2x)=\\ \\ =(-2ctgx+\frac{1}{2}sin2x)|^\frac{\pi}{4} _\frac{\pi}{3}=-2ctg\frac{\pi}{4}+\frac{1}{2}sin\frac{\pi}{2}-(-2ctg\frac{\pi}{3}+\frac{1}{2}sin\frac{2\pi}{3})=$

$=-2+\frac{1}{2}+\frac{2\sqrt{3} }{3} -\frac{\sqrt{3} }{4} =-\frac{3}{2}+\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{4}=\frac{-18+8\sqrt{3}-3\sqrt{3}}{12} = \\ \\ =\frac{5\sqrt{3}-18}{12}$