Question

Помогите с решением.

Answer 1

Ответ:

$\int _{ - \frac{\pi}{6} }^{0} \cos(3x) dx = ( \frac{1}{3} \sin(3x) )| _{ - \frac{\pi}{6} }^{0} = \frac{1}{3} \times \sin(3 \times 0) - ( \frac{1}{3} \times \sin(3 \times ( - \frac{\pi}{6} ) ) = 0 - ( \frac{1}{3} \times \sin( - \frac{\pi}{2} ) ) = - ( \frac{1}{3} \times ( - 1)) = - ( - \frac{1}{3} ) = \frac{1}{3}$