Упр.21.11 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)

а) $$\int_{0}^{\frac{\pi}{4}}\left(1+\operatorname{tg}^{2}x\right)\,dx$$; в) $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\left(\operatorname{ctg}^{2}x+1\right)\,dx$$; б) $$\int_{\frac{\pi}{3}}^{\frac{\pi}{4}}\left(3-3\operatorname{ctg}^{2}x\right)\,dx$$; г) $$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\left(1+2\operatorname{tg}^{2}x\right)\,dx$$.

а) $$\int_{0}^{\pi/4}(1+\tg^2 x)\,dx=\int_{0}^{\pi/4}\frac{dx}{\cos^2 x}=\tg x\Big|_{0}^{\pi/4}=\tg\frac{\pi}{4}-\tg 0=1.$$

Ответ: $$1.$$

б) $$\int_{\pi/3}^{\pi/4}(3-3\ctg^2 x)\,dx=\int_{\pi/3}^{\pi/4}\left(6-\frac{3}{\sin^2 x}\right)\,dx=(6x+3\ctg x)\Big|_{\pi/3}^{\pi/4}.$$

$$\begin{aligned} (6x+3\ctg x)\Big|_{\pi/3}^{\pi/4} &=\left(6\cdot\frac{\pi}{4}+3\ctg\frac{\pi}{4}\right)-\left(6\cdot\frac{\pi}{3}+3\ctg\frac{\pi}{3}\right)\\ &=\frac{3\pi}{2}+3-2\pi-\sqrt{3}\\ &=3-\frac{\pi}{2}-\sqrt{3}. \end{aligned}$$

Ответ: $$3-\frac{\pi}{2}-\sqrt{3}.$$

в) $$\int_{\pi/4}^{\pi/3}(\ctg^2 x+1)\,dx=\int_{\pi/4}^{\pi/3}\frac{dx}{\sin^2 x}=(-\ctg x)\Big|_{\pi/4}^{\pi/3}.$$

$$-\ctg\frac{\pi}{3}+\ctg\frac{\pi}{4}=-\frac{\sqrt{3}}{3}+1.$$

Ответ: $$1-\frac{\sqrt{3}}{3}.$$

г) $$\int_{\pi/6}^{\pi/4}(1+2\tg^2 x)\,dx=\int_{\pi/6}^{\pi/4}\left(\frac{2}{\cos^2 x}-1\right)\,dx=(2\tg x-x)\Big|_{\pi/6}^{\pi/4}.$$

$$\begin{aligned} (2\tg x-x)\Big|_{\pi/6}^{\pi/4} &=\left(2\tg\frac{\pi}{4}-\frac{\pi}{4}\right)-\left(2\tg\frac{\pi}{6}-\frac{\pi}{6}\right)\\ &=2-\frac{\pi}{4}-\frac{2\sqrt{3}}{3}+\frac{\pi}{6}\\ &=2-\frac{2\sqrt{3}}{3}-\frac{\pi}{12}. \end{aligned}$$

Ответ: $$2-\frac{2\sqrt{3}}{3}-\frac{\pi}{12}.$$