Упр.23.16 ГДЗ Мордкович Семенов 11 класс (Алгебра)

а) ?((-1; v3); dx / (1 + x^2)); в) ?((0; 1); (2dx / (1 + x^2) — ?/2) dx);
б) ?((-0,5; 1); dx / v(1 — x^2)); г) ?((-v3/2; v2/2); 3dx / v(1 — x^2)). Найдите значение выражения:
а) cos(20°)cos(40°)cos(80°);
б) sin(10°)cos(20°)cos(40°).

а) Используем формулу первообразной:

$$\int \frac{dx}{1+x^2}=\arctg x + C.$$

Тогда

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

$$ \arctg \sqrt{3}=\frac{\pi}{3}, \qquad \arctg(-1)=-\frac{\pi}{4}. $$

$$ \int\limits_{-1}^{\sqrt{3}} \frac{dx}{1+x^2} = \frac{\pi}{3}+\frac{\pi}{4} = \frac{7\pi}{12}. $$

б) Используем формулу:

$$\int \frac{dx}{\sqrt{1-x^2}}=\arcsin x + C.$$

$$ \int\limits_{-0{,}5}^{1} \frac{dx}{\sqrt{1-x^2}} = \arcsin x \Big|_{-0{,}5}^{1} = \arcsin 1-\arcsin(-0{,}5). $$

$$ \arcsin 1=\frac{\pi}{2}, \qquad \arcsin(-0{,}5)=-\frac{\pi}{6}. $$

$$ \int\limits_{-0{,}5}^{1} \frac{dx}{\sqrt{1-x^2}} = \frac{\pi}{2}+\frac{\pi}{6} = \frac{2\pi}{3}. $$

в)

$$ \int\limits_{0}^{1}\left(\frac{2}{1+x^2}-\frac{\pi}{2}\right)dx = \left(2\arctg x-\frac{\pi}{2}x\right)\Big|_{0}^{1}. $$

$$ \left(2\arctg 1-\frac{\pi}{2}\right)-\left(2\arctg 0-0\right) = \left(2\cdot\frac{\pi}{4}-\frac{\pi}{2}\right)-0 = 0. $$

г)

$$ \int\limits_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{2}}{2}} \frac{3\,dx}{\sqrt{1-x^2}} = 3\arcsin x \Big|_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{2}}{2}}. $$

$$ 3\arcsin\frac{\sqrt{2}}{2}-3\arcsin\left(-\frac{\sqrt{3}}{2}\right) = 3\cdot\frac{\pi}{4}-3\cdot\left(-\frac{\pi}{3}\right) = \frac{7\pi}{4}. $$

а)

$$ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{8\sin 20^\circ \cos 20^\circ \cos 40^\circ \cos 80^\circ}{8\sin 20^\circ}. $$

$$ 8\sin 20^\circ \cos 20^\circ = 4\sin 40^\circ, $$

$$ 4\sin 40^\circ \cos 40^\circ = 2\sin 80^\circ, $$

$$ 2\sin 80^\circ \cos 80^\circ = \sin 160^\circ. $$

Значит,

$$ \cos 20^\circ \cos 40^\circ \cos 80^\circ = \frac{\sin 160^\circ}{8\sin 20^\circ} = \frac{\sin 20^\circ}{8\sin 20^\circ} = \frac18. $$

б)

$$ \sin 10^\circ \cos 20^\circ \cos 40^\circ = \frac{8\sin 10^\circ \cos 10^\circ \cos 20^\circ \cos 40^\circ}{8\cos 10^\circ}. $$

$$ 8\sin 10^\circ \cos 10^\circ = 4\sin 20^\circ, $$

$$ 4\sin 20^\circ \cos 20^\circ = 2\sin 40^\circ, $$

$$ 2\sin 40^\circ \cos 40^\circ = \sin 80^\circ. $$

Тогда

$$ \sin 10^\circ \cos 20^\circ \cos 40^\circ = \frac{\sin 80^\circ}{8\cos 10^\circ} = \frac{\sin 80^\circ}{8\sin 80^\circ} = \frac18. $$

Ответ

$$ \text{а) } \frac{7\pi}{12}; \quad \text{б) } \frac{2\pi}{3}; \quad \text{в) } 0; \quad \text{г) } \frac{7\pi}{4}; \quad \text{а) } \frac18; \quad \text{б) } \frac18. $$