Упр.362 ГДЗ Колмогоров 10-11 класс (Алгебра)

а) $$\int\limits_{-\pi}^{2\pi}\sin\frac{x}{3}\,dx=-3\cos\frac{x}{3}\Big|_{-\pi}^{2\pi}=-3\left(\cos\frac{2\pi}{3}-\cos\left(-\frac{\pi}{3}\right)\right)$$

$$=-3\left(-\frac12-\frac12\right)=3.$$

б) $$\int\limits_{-2}^{2}\frac{dx}{\sqrt{2x+5}}=\int\limits_{-2}^{2}(2x+5)^{-\frac12}\,dx=\sqrt{2x+5}\Big|_{-2}^{2}$$

$$=\sqrt{9}-\sqrt{1}=3-1=2.$$

в) $$\int\limits_{0}^{3\pi}\frac{dx}{\cos^2\frac{x}{9}}=9\tan\frac{x}{9}\Big|_{0}^{3\pi}=9\left(\tan\frac{\pi}{3}-\tan 0\right)=9\sqrt{3}.$$

г) $$\int\limits_{-2}^{6}\frac{dx}{\sqrt{x+3}}=2\sqrt{x+3}\Big|_{-2}^{6}=2(\sqrt{9}-\sqrt{1})=4.$$

Ответ: а) $$3$$; б) $$2$$; в) $$9\sqrt{3}$$; г) $$4$$.