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

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

Так как $$\cos\frac{2\pi}{3}=-\frac12,\qquad \cos\left(-\frac{\pi}{3}\right)=\cos\frac{\pi}{3}=\frac12,$$ то

$$\int_{-\pi}^{2\pi}\sin\frac{x}{3}\,dx=-3\left(-\frac12-\frac12\right)=3.$$

б) $$\int_{-2}^{2}\frac{dx}{\sqrt{2x+5}}=\int_{-2}^{2}(2x+5)^{-\frac12}\,dx.$$

Положим $$u=2x+5,$$ тогда $$du=2\,dx,\quad dx=\frac{du}{2}.$$ Получаем

$$\int_{-2}^{2}(2x+5)^{-\frac12}\,dx=\left.\sqrt{2x+5}\right|_{-2}^{2}=\sqrt{9}-\sqrt{1}=3-1=2.$$

в) $$\int_{0}^{3\pi}\frac{dx}{\cos^2\frac{x}{9}}=\int_{0}^{3\pi}\sec^2\frac{x}{9}\,dx.$$

Так как $$\int \sec^2\frac{x}{9}\,dx=9\tan\frac{x}{9},$$ то

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

г) $$\int_{-2}^{6}\frac{dx}{\sqrt{x+3}}=\int_{-2}^{6}(x+3)^{-\frac12}\,dx.$$

Тогда

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

Ответ

$$3;\ 2;\ 0;\ 4.$$