Question

помогите решить интеграл

Answer 1

Ответ:

замена:

$x = 2arctgt \\ t = tg \frac{x}{2} \\ \\ \cos(x) = \frac{1 - {t}^{2} }{1 + {t}^{2} } \\ \sin(x) = \frac{2t}{1 + {t}^{2} } \\ dx = \frac{2dt}{1 + {t}^{2} } \\ \\ t1 = tg( \frac{\pi}{2 \times 2} ) = 1 \\ t2 = tg(0) = 0$

$\int\limits^{ 1 } _ {0} \frac{2dt}{1 + {t}^{2} } \times \frac{1}{1 - \frac{1 - {t}^{2} }{1 + { t}^{2} } + \frac{2t}{1 + {t}^{2} } } = \\ = \int\limits^{ 1 } _ {0} \frac{2dt}{t {}^{2} + 1 } \times \frac{1 + {t}^{2} }{1 + {t}^{2} - 1 + {t}^{2} + 2t } = \\ = \int\limits^{ 1 } _ {0} \frac{2dt}{2 {t}^{2} + 2t} = \int\limits^{ 1 } _ {0} \frac{dt}{t(t + 1)} \\ \\ \frac{1}{t(t + 1)} = \frac{A}{t} + \frac{B}{t + 1} \\ 1 = At + A + Bt \\ \\ 0 = A + B \\ 1 = A \\ \\ A= 1 \\ B = - A = - 1 \\ \\ \int\limits^{ 1 } _ {0} \frac{dt}{t} - \int\limits^{ 1 } _ {0} \frac{dt}{t + 1} = \\ = ( ln(t) - ln(t + 1) )| ^{1} _ {0} = ln( \frac{t}{t + 1} ) | ^{1} _ {0} = \\ = ln( \frac{1}{2} ) - 0 = - ln(2)$