Question

Знайти визначений інтеграл

Answer 1

Ответ:

замена:

$\sin(x) = t \\ \cos(x) dx = dt \\ t1 = \sin( \frac{\pi}{2} ) = 1 \\ t2 = \sin( \frac{\pi}{6} ) = \frac{1}{2}$

$\int\limits^{ 1 } _ { \frac{1}{2} } {t}^{3} dt = \frac{ {t}^{4} }{4} | ^{1} _ { \frac{1}{2} } = \frac{1}{4} - \frac{1}{4} \times \frac{1}{16} = \\ = \frac{1}{4} \times (1 - \frac{1}{16} ) = \frac{1}{4} \times \frac{15}{16} = \frac{15}{64}$