ГДЗ по Алгебре 11 Класс Номер 11.20 Углубленный Уровень Мерзляк, Номировский — Подробные Ответы

\(
\text{Вычислите определённые интегралы:}
\)
1) \(\int_{\frac{5\pi}{4}}^{\frac{15\pi}{4}} \cot^2\left(\frac{x}{5}\right) \, dx\);
2) \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(2x) \, dx\);
3) \(\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin(7x) \cos(3x) \, dx\);
4) \(\int_{-2}^{-1} \frac{x^2 — e^x}{x^2 e^x} \, dx.\)

1)
\(
\int_{\frac{15\pi}{4}}^{\frac{5\pi}{4}} \cot^2 x \, dx = \int_{\frac{15\pi}{4}}^{\frac{5\pi}{4}} \left(\frac{1}{\sin^2 x} — 1\right) \, dx = -5 \cot x — x \Big|_{\frac{15\pi}{4}}^{\frac{5\pi}{4}} =
\)
\(
\left(-5 \cot \frac{3\pi}{4} — \frac{15\pi}{4}\right) — \left(-5 \cot \frac{\pi}{4} — \frac{5\pi}{4}\right) = 10 — \frac{20 — 5\pi}{2} = \frac{20 — 5\pi}{2};
\)
Ответ: \(\frac{20 — 5\pi}{2}\).

2)
\(
\int_{-\pi}^{\pi} \cos^2 2x \, dx = \int_{-\pi}^{\pi} \frac{1 + \cos 4x}{2} \, dx = \frac{1}{2} \int_{-\pi}^{\pi} \left(x + \frac{1}{4} \sin 4x\right) \, dx =
\)
\(
\frac{1}{2} \left(\frac{\pi}{2} + \frac{1}{4} \sin 2\pi\right) — \frac{1}{2} \left(-\frac{\pi}{2} + \frac{1}{4} \sin (-2\pi)\right) = \frac{\pi}{4} + \frac{\pi}{2};
\)
Ответ: \(\frac{\pi}{2}\).

3)
\(
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin 7x \cos 3x \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{2} (\sin 4x + \sin 10x) \, dx =
\)
\(
\frac{1}{2} \left(-\frac{1}{8} \cos 4x — \frac{1}{10} \cos 10x\right) \Big|_{\frac{\pi}{4}}^{\frac{\pi}{2}} = \frac{1}{8} \cos \pi — \frac{1}{20} \cos 5\pi — \left(\frac{1}{8} \cos \frac{\pi}{2} — \frac{1}{20} \cos \frac{5\pi}{2}\right) =
\)
\(
\frac{1}{8} + \frac{1}{20} — \frac{5}{20} = \frac{1}{5};
\)
Ответ: \(\frac{1}{5}\).

4)
\(
\int_{-2}^{-1} x^2 — e^x \, dx = \int_{-2}^{-1} \left(\frac{1}{e^x} — \frac{1}{x^2}\right) \, dx = -e^{-x} + \frac{1}{x^{-1}} \Big|_{-2}^{-1} =
\)
\(
\left(-e^{-1} + 1\right) — \left(-e^{-2} + \frac{1}{2}\right) = e^2 — e — \frac{1}{2};
\)
Ответ: \(e^2 — e — \frac{1}{2}\).

1)
\(
\int_{\frac{5\pi}{4}}^{\frac{15\pi}{4}} \cot^2\left(\frac{x}{5}\right) \, dx
\)

Сделаем замену переменной:
\(
t = \frac{x}{5} \implies dx = 5 dt
\)

Пределы интегрирования:
\(
x = \frac{5\pi}{4} \Rightarrow t = \frac{5\pi}{4} \cdot \frac{1}{5} = \frac{\pi}{4}
\)
\(
x = \frac{15\pi}{4} \Rightarrow t = \frac{15\pi}{4} \cdot \frac{1}{5} = \frac{3\pi}{4}
\)

Тогда интеграл:
\(
\int_{\frac{5\pi}{4}}^{\frac{15\pi}{4}} \cot^2\left(\frac{x}{5}\right) dx = 5 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cot^2 t \, dt
\)

Используем формулу:
\(
\cot^2 t = \csc^2 t — 1
\)
и
\(
\int \cot^2 t \, dt = -\cot t — t + C
\)

Вычисляем:
\(
5 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cot^2 t \, dt = 5 \left[-\cot t — t\right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} = 5 \left(-\cot \frac{3\pi}{4} — \frac{3\pi}{4} + \cot \frac{\pi}{4} + \frac{\pi}{4}\right)
\)

Вычислим значения:
\(
\cot \frac{\pi}{4} = 1, \quad \cot \frac{3\pi}{4} = \cot \left(\pi — \frac{\pi}{4}\right) = -\cot \frac{\pi}{4} = -1
\)

Подставляем:
\(
5 \left(-(-1) — \frac{3\pi}{4} + 1 + \frac{\pi}{4}\right) = 5 \left(1 — \frac{3\pi}{4} + 1 + \frac{\pi}{4}\right) = 5 \left(2 — \frac{2\pi}{4}\right) = 5 \left(2 — \frac{\pi}{2}\right) = 10 — \frac{5\pi}{2}
\)

2)
\(
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(2x) \, dx
\)

Используем формулу понижения степени:
\(
\cos^2 \theta = \frac{1 + \cos 2\theta}{2}
\)

Тогда:
\(
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(2x) \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 + \cos 4x}{2} \, dx = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx + \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 4x \, dx
\)

Вычисляем:
\(
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = \pi
\)
\(
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 4x \, dx = \left.\frac{\sin 4x}{4}\right|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \frac{\sin 2\pi — \sin (-2\pi)}{4} = 0
\)

Итого:
\(
\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^2(2x) \, dx = \frac{1}{2} \pi + \frac{1}{2} \cdot 0 = \frac{\pi}{2}
\)

3)
\(
\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin(7x) \cos(3x) \, dx
\)

Используем формулу:
\(
\sin A \cos B = \frac{1}{2} \left(\sin (A + B) + \sin (A — B)\right)
\)

Тогда:
\(
\sin(7x) \cos(3x) = \frac{1}{2} \left(\sin 10x + \sin 4x\right)
\)

Интеграл:
\(
\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin(7x) \cos(3x) \, dx = \frac{1}{2} \int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin 10x \, dx + \frac{1}{2} \int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin 4x \, dx
\)

Вычисляем каждый:
\(
\int \sin kx \, dx = -\frac{\cos kx}{k} + C
\)

Подставляем пределы:
\(
\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin 10x \, dx = \left[-\frac{\cos 10x}{10}\right]_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} = -\frac{\cos \left(-\frac{5\pi}{2}\right)}{10} + \frac{\cos (-5\pi)}{10}
\)

\(
\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin 4x \, dx = \left[-\frac{\cos 4x}{4}\right]_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} = -\frac{\cos (-\pi)}{4} + \frac{\cos (-2\pi)}{4}
\)

Вычисляем косинусы:
\(
\cos \left(-\frac{5\pi}{2}\right) = \cos \frac{5\pi}{2} = 0
\)
\(
\cos (-5\pi) = \cos 5\pi = \cos \pi = -1
\)
\(
\cos (-\pi) = \cos \pi = -1
\)
\(
\cos (-2\pi) = \cos 2\pi = 1
\)

Подставляем:
\(
\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin 10x \, dx = -\frac{0}{10} + \frac{-1}{10} = -\frac{1}{10}
\)
\(
\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin 4x \, dx = -\frac{-1}{4} + \frac{1}{4} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
\)

Итог:
\(
\int_{-\frac{\pi}{2}}^{-\frac{\pi}{4}} \sin(7x) \cos(3x) \, dx = \frac{1}{2} \left(-\frac{1}{10} + \frac{1}{2}\right) = \frac{1}{2} \cdot \frac{4}{10} = \frac{2}{10} = \frac{1}{5}
\)

4)
\(
\int_{-2}^{-1} \frac{x^2 — e^x}{x^2 e^x} \, dx
\)

Разобьем интеграл:
\(
\frac{x^2 — e^x}{x^2 e^x} = \frac{x^2}{x^2 e^x} — \frac{e^x}{x^2 e^x} = \frac{1}{e^x} — \frac{1}{x^2}
\)

Тогда:
\(
\int_{-2}^{-1} \frac{x^2 — e^x}{x^2 e^x} \, dx = \int_{-2}^{-1} e^{-x} \, dx — \int_{-2}^{-1} \frac{1}{x^2} \, dx
\)

Вычисляем каждый интеграл:
\(
\int e^{-x} \, dx = -e^{-x} + C
\)
\(
\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -x^{-1} + C = -\frac{1}{x} + C
\)

Подставляем пределы:
\(
\int_{-2}^{-1} e^{-x} \, dx = \left[-e^{-x}\right]_{-2}^{-1} = -e^{1} + e^{2} = e^{2} — e
\)
\(
\int_{-2}^{-1} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_{-2}^{-1} = -\frac{1}{-1} + \frac{1}{-2} = 1 — \frac{1}{2} = \frac{1}{2}
\)

Итог:
\(
\int_{-2}^{-1} \frac{x^2 — e^x}{x^2 e^x} \, dx = (e^{2} — e) — \frac{1}{2} = e^{2} — e — \frac{1}{2}
\)

Ответы:

1)
\(
10 — \frac{5\pi}{2}
\)

2)
\(
\frac{\pi}{2}
\)

3)
\(
\frac{1}{5}
\)

4)
\(
e^{2} — e — \frac{1}{2}
\)