Question

приклади №1, №2, №3 потрібно обов"язково рішити вказуванням: t = ..., dt = ...
Наприклад: t = х^2+1, dt =2х*dх і т.д.

Answer 1

Ответ:

1.

$\int\limits \frac{dx}{ \sqrt[3]{2 - 5x} } \\ \\ 2 - 5x = t \\ - 5dx = dt \\dx = - \frac{dt}{5} \\ \\ - \frac{1}{5} \int\limits \frac{dt}{ \sqrt[3]{t} } = - \frac{1}{5} \int\limits {t}^{ - \frac{1}{3} } dt = - \frac{1}{5} \times \frac{ {t}^{ \frac{2}{ 3 } } }{ \frac{2}{3} } + C= \\ = - \frac{3}{10} \sqrt[3]{ {t}^{2} } + C= - \frac{3}{10} \sqrt[3]{ {(2 - 5x)}^{2} } + C$

2.

$\int\limits \frac{dx}{5 +4 \cos(x) } \\ \\ \text{тригонометрическая замена:} \\ \\ t = tg( \frac{x}{2} ) \\ \cos(x) = \frac{1 - {t}^{2} }{1 + {t}^{2} } \\ dx = \frac{2dt}{1 + t {}^{2} } \\ \\ \int\limits \frac{2dt}{1 + {t}^{2} } \times \frac{1}{5 + \frac{4 \times (1 - t {}^{2} )}{1 + t {}^{2} } } = \\ = \int\limits \frac{2dt}{1 + t {}^{2} } \times \frac{1 + {t}^{2} }{5(1 + {t}^{2} ) + 4 - 4 {t}^{2} } = \\ = \int\limits \frac{2dt}{5 + 5 {t}^{2} + 4 - 4 {t}^{2} } = 2\int\limits \frac{dt}{ {t}^{2} + 9 } = \\ = 2\int\limits \frac{dt}{ {t}^{2} + {3}^{2} } = \frac{2}{3} arctg( \frac{t}{3} ) + C = \\ = \frac{2}{3} arctg( \frac{ tg( \frac{x}{2} )}{3}) + C$

3.

$\int\limits \frac{dx}{1 - 2x - {x}^{2} } \\ \\ \text{Выделяем квадрат:} \\ 1 - 2x - {x}^{2} = - ( {x}^{2} + 2x - 1) = \\ = - ( {x}^{2} + 2x + 1 - 2) = - ( {(x + 1)}^{2} - 2) = \\ = 2 - {(x + 1)}^{2} = {( \sqrt{2}) }^{2} - {(x + 1)}^{2} \\ \\ \int\limits \frac{dx}{ {( \sqrt{2} )}^{2} - {(x + 1)}^{2} } = \int\limits \frac{d(x + 1)}{ {( \sqrt{2}) }^{2} - {(x + 1)}^{2} } = \\ = \frac{1}{2 \sqrt{2} } ln | \frac{ \sqrt{2} - (x + 1) }{ \sqrt{2} + (x + 1)} | + C = \frac{1}{2 \sqrt{2} } ln | \frac{ \sqrt{2} - 1 - x }{ \sqrt{2} + 1 + x } | + C$