Упр.21.13 ГДЗ Мордковича 11 класс профильный уровень (Алгебра)

Задача

21-13.
а) $$\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \frac{\sin^2\left(x-\frac{3\pi}{2}\right)\cos(2\pi-x)}{\tan^2\left(x-\frac{\pi}{2}\right)\cos^2\left(x-\frac{3\pi}{2}\right)}\,dx$$
б) $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\tan^2\left(x-\frac{\pi}{2}\right)\cot^2\left(\frac{3\pi}{2}+x\right)}{\cos^2(\pi-x)+\sin^2\left(\frac{\pi}{2}-x\right)+\cos(\pi+x)\cos(2\pi-x)}\,dx$$
в) $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin(\pi-x)}{\tan(\pi-x)}\cdot\frac{\cot\left(\frac{\pi}{2}-x\right)}{\tan\left(\frac{\pi}{2}+x\right)}\cdot\frac{\cos(2\pi-x)}{\sin(-x)}\,dx$$
г) $$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \frac{\sin\left(\frac{3\pi}{2}-x\right)\tan(\pi-x)}{\cos^2\left(\frac{\pi}{2}-x\right)\cos(\pi+x)\tan\left(\frac{3\pi}{2}+x\right)}\,dx$$

Подробный ответ

Используем тригонометрические формулы:

$$\sin\left(x-\frac{3\pi}{2}\right)=\cos x,\quad \cos(2\pi-x)=\cos x,\quad \tg\left(x-\frac{\pi}{2}\right)=-\ctg x,\quad \cos\left(x-\frac{3\pi}{2}\right)=\sin x.$$

а)

$$\int_{\pi/6}^{\pi/4}\frac{\sin^2\left(x-\frac{3\pi}{2}\right)\cos(2\pi-x)}{\tg^2\left(x-\frac{\pi}{2}\right)\cos^2\left(x-\frac{3\pi}{2}\right)}\,dx = \int_{\pi/6}^{\pi/4}\frac{\cos^2 x\cdot \cos x}{\ctg^2 x\cdot \sin^2 x}\,dx$$

$$= \int_{\pi/6}^{\pi/4}\cos x\,dx = \sin x\Big|_{\pi/6}^{\pi/4} = \sin\frac{\pi}{4}-\sin\frac{\pi}{6} = \frac{\sqrt2}{2}-\frac12 = \frac{\sqrt2-1}{2}.$$

б)

$$\int_{\pi/4}^{\pi/3} \frac{\tg^2\left(x-\frac{\pi}{2}\right)\ctg^2\left(\frac{3\pi}{2}+x\right)} {\cos^2(\pi-x)+\sin^2\left(\frac{\pi}{2}-x\right)+\cos(\pi+x)\cos(2\pi-x)}\,dx$$

$$= \int_{\pi/4}^{\pi/3} \frac{\ctg^2 x\cdot \tg^2 x}{\cos^2 x+\cos^2 x-\cos x\cdot \cos x}\,dx = \int_{\pi/4}^{\pi/3}\frac{dx}{\cos^2 x}$$

$$= \tg x\Big|_{\pi/4}^{\pi/3} = \tg\frac{\pi}{3}-\tg\frac{\pi}{4} = \sqrt3-1.$$

в)

$$\int_{\pi/4}^{\pi/3} \frac{\sin(\pi-x)}{\tg(\pi-x)} \cdot \frac{\ctg\left(\frac{\pi}{2}-x\right)}{\tg\left(\frac{\pi}{2}+x\right)} \cdot \frac{\cos(2\pi-x)}{\sin(-x)}\,dx$$

$$= \int_{\pi/4}^{\pi/3} \frac{\sin x}{-\tg x} \cdot \frac{-\tg x}{-\ctg x} \cdot \frac{\cos x}{-\sin x}\,dx = \int_{\pi/4}^{\pi/3}\frac{\cos x}{\ctg x}\,dx$$

$$= \int_{\pi/4}^{\pi/3}\sin x\,dx = -\cos x\Big|_{\pi/4}^{\pi/3} = \cos\frac{\pi}{4}-\cos\frac{\pi}{3} = \frac{\sqrt2}{2}-\frac12 = \frac{\sqrt2-1}{2}.$$

г)

$$\int_{\pi/4}^{\pi/3} \frac{\sin\left(\frac{3\pi}{2}-x\right)\tg(\pi-x)} {\cos^2\left(\frac{\pi}{2}-x\right)\cos(\pi+x)\tg\left(\frac{3\pi}{2}+x\right)}\,dx$$

$$= \int_{\pi/4}^{\pi/3} \frac{-\cos x\cdot(-\tg x)}{\sin^2 x\cdot(-\cos x)\cdot(-\ctg x)}\,dx = \int_{\pi/4}^{\pi/3}\frac{\tg x}{\sin^2 x\cdot \ctg x}\,dx$$

$$= \int_{\pi/4}^{\pi/3}\frac{dx}{\cos^2 x} = \tg x\Big|_{\pi/4}^{\pi/3} = \sqrt3-1.$$