Question

Вычислите интеграл
$\int\limits{\frac{dx}{9sin^2x+16 } \,$

Answer 1

Ответ:

$\int\limits \frac{dx}{ 9\sin^{2} (x) + 16 }$

тригонометрическая замена:

$\sin( x) = \frac{2t}{1 + {t}^{2} } \\ dx = \frac{2dt}{1 + {t}^{2} } \\ t = tg( \frac{x}{2} )$

$\int\limits \frac{2dt}{1 + {t}^{2} } \times \frac{1}{9 \times \frac{2t}{1 + {t}^{2} } + 16} = \\ = \int\limits \frac{2t}{1 + {t}^{2} } \times \frac{1 + {t}^{2} }{18t + 16(1 + {t}^{2}) } = \\ = \int\limits \frac{2dt}{18t + 16 + 16 {t}^{2} } = \int\limits \frac{dt}{8 {t}^{2} + 9t + 8} \\ \\ 8 {t}^{2} + 9 t + 8 = {(2 \sqrt{2}t) }^{2} + 2 \times 2 \sqrt{2}t \times \frac{9}{4 \sqrt{2} } + \frac{81}{32} + \frac{175}{32} = \\ = {(2 \sqrt{2}t + \frac{9}{2 \sqrt{2} } ) }^{2} + {( \frac{5 \sqrt{5} }{4 \sqrt{2} } )}^{2} \\ \\ \int\limits \frac{dt}{ {(2 \sqrt{2}t + \frac{9}{2 \sqrt{2} } ) }^{2} + {( \frac{5 \sqrt{5} }{4 \sqrt{2} } )}^{2}} = \\ = \frac{1}{2 \sqrt{2} } \int\limits \frac{d(2 \sqrt{2}t + \frac{9}{4 \sqrt{2} }) }{ {(2 \sqrt{2}t + \frac{9}{2 \sqrt{2} } ) }^{2} + {( \frac{5 \sqrt{5} }{4 \sqrt{2} } )}^{2}} = \\ = \frac{1}{2 \sqrt{2} } \times \frac{1}{ \frac{5 \sqrt{5} }{4 \sqrt{2} } } arctg( \frac{2 \sqrt{2} t + \frac{9}{4 \sqrt{2} }}{ \frac{5 \sqrt{5} }{4 \sqrt{2} } } ) + c = \\ = \frac{2}{5 \sqrt{5} } arctg( \frac{16t + 9}{5 \sqrt{5} } ) + c = \\ \\ = \frac{2}{5 \sqrt{5} } arctg( \frac{16tg( \frac{x}{2} ) + 9}{5 \sqrt{5} } ) + c$