Question

Допоможіть знайти визначений інтеграл:

Answer 1

Ответ:

$\int\limits^{ \frac{1}{2} } _ { - \frac{1}{2} } \frac{dx}{ \sqrt{ {(1 - {x}^{2} )}^{3} } } \\ \\ x = \sin(t) \\ dx = \cos(t) dt \\ t = arsinx \\ t_1 = arcsin \frac{1}{2} = \frac{\pi}{6} \\ t_2 = arcsin( - \frac{1}{2} ) = - \frac{\pi}{6} \\ \\ \int\limits^{ \frac{\pi}{6} } _ { - \frac{\pi}{6} } \frac{ \cos(t) dt}{ \sqrt{ (1 - \sin {}^{2} (t)) {}^{3} }} = \int\limits^{ \frac{\pi}{6} } _ { \frac{\pi}{6} } \frac{ \cos(t) dt}{ \cos {}^{3} (t) } = \\ = \int\limits^{ \frac{\pi}{6} } _ { - \frac{\pi}{6} } \frac{dt}{ \cos {}^{2} (t) } = tgt|^{ \frac{\pi}{6} } _ { - \frac{\pi}{6} } = tg \frac{\pi}{6} - tg( - \frac{\pi}{6} ) = \\ = \frac{ \sqrt{3} }{3} + \frac{ \sqrt{3} }{3} = \frac{2 \sqrt{3} }{3} = 1.16$