Question

помогите решить интеграл

Answer 1

Ответ:

сделаем замену:

$4 - {x}^{2} = t \\ - 2xdx = dt \\ xdx = - \frac{1}{2} dt \\ {x}^{2} = 4 - t$

$\int\limits \frac{ {x}^{3}dx }{ \sqrt{4 - {x}^{2} } } =\int\limits \frac{ {x}^{2} \times xdx}{ \sqrt{4 - {x}^{2} } } = \\ = - \frac{1}{2} \int\limits \frac{(4 - t)dt}{ \sqrt{t} } = - \frac{1}{2} \int\limits(4 {t}^{ - \frac{1}{2} } - {t}^{ \frac{1}{2} }) dt = \\ = - \frac{1}{2} (4 \times \frac{ {t}^{ \frac{1}{2} } }{ \frac{1}{2} } - \frac{ {t}^{ \frac{3}{2} } }{ \frac{3}{2} } ) + c = \\ = - \frac{1}{2} \times 4 \times 2 \sqrt{t} + \frac{1}{2} \times \frac{2}{3} t \sqrt{t} + C = \\ = - 4 \sqrt{t} + \frac{1}{3} \sqrt{ {t}^{3} } + C = \\ = - 4 \sqrt{4 - {x}^{2} } + \frac{1}{3} \sqrt{ {(4 - {x}^{2} )}^{3} } + C= \\ = \sqrt{4 - {x}^{2} } ( - 4 + \frac{1}{3} (4 - {x}^{2} ) + C = \\ = \sqrt{4 - {x}^{2} } ( - 4 + \frac{4}{3} - \frac{ {x}^{2} }{3} ) + c = \\ = \sqrt{4 - {x}^{2} } ( - \frac{8}{3} - \frac{ {x}^{2} }{3} ) + C = \\ = - \frac{1}{3} \sqrt{4 - {x}^{2} } \times ( {x}^{2} + 8) + C$