Question

$\int\limits^\pi_0 {sinxcos^2x} \, dx$
Знайти інтеграл

Answer 1

Ответ:

$\int^\pi_0 sinxcos^2xdx = \\ = |t = cosx,dt = - sinxdx,cos\pi = - 1,cos0 = 1| = \\ = - \int^ {- 1}_1 {t}^{2} dt = \int^ {1}_{ - 1 }{t}^{2} dt = \\ = \frac{ {t}^{3} }{3} | ^ {1}_ {- 1} = \frac{1}{3} - ( - \frac{1}{3} ) = \frac{2}{3}$