Нужно 22ое и 23е задание. Пропустил тему, легкие сам решил, а вот эти вот- засада....

Решите задачу:

$S= \int\limits^{-\frac{\pi}{4}}_{\frac{\pi}{4}}} {cosx} \, dx =sinx|_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\, =sin\frac{\pi}{4}-sin(-\frac{\pi}{4})=2sin\frac{\pi}{4}=2\cdot \frac{\sqrt2}{2}=\sqrt2\\\\2)\; S= \int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}} {sinx} \, dx =-cosx|_{\frac{\pi}{4}}^{\frac{\pi}{3}}=-cos\frac{\pi}{3}+cos\frac{\pi}{4}=-\frac{1}{2}+\frac{\sqrt2}{2}=\frac{\sqrt2-1}{2}\\\\3)\; S= \int\limits^{\frac{\pi}{12}}_{\frac{\pi}{18}} {sin6x} \, dx =-\frac{1}{6}cos6x|_{\frac{\pi}{18}}^{\frac{\pi}{12}}=$

$=-\frac{1}{6}(cos\frac{6\pi }{12}-cos\frac{6\pi }{18})=-\frac{1}{6}(cos\frac{\pi}{2}-cos\frac{\pi}{3})=-\frac{1}{6}(0-\frac{1}{2})=\frac{1}{12}\\\\4)\; S= \int\limits^{\frac{\pi}{12}}_{\frac{\pi}{24}} {cos4x} \, dx =\frac{1}{4}sin4x|_{\frac{\pi}{24}}^{\frac{\pi}{12}}=\frac{1}{4}(sin\frac{\pi}{3}-sin\frac{\pi}{6})=\\\\=\frac{1}{4}(\frac{\sqrt3}{2}-\frac{1}{2})=\frac{\sqrt3-1}{8}$