Question

вычислите ....................................

Answer 1

Ответ:

2

$\int\limits^{ \pi} _ { \frac{2\pi}{3} } \frac{dx}{ \sin {}^{2} ( \frac{x}{2} - \frac{\pi}{6} ) } = 2 \int\limits^{ \pi } _ { \frac{2\pi}{3} } \frac{d( \frac{x}{2} - \frac{\pi}{6} ) }{ \sin {}^{2} ( \frac{x}{2} - \frac{\pi}{6} ) } = \\ = - 2ctg( \frac{x}{2} - \frac{\pi}{6}) | ^{ \pi} _ { \frac{2\pi}{3} } = \\ = - 2ctg \frac{\pi}{3} + 2ctg \frac{\pi}{6} = - \frac{2 \sqrt{3} }{3} + 2 \sqrt{3} = \\ = \frac{ - 2 \sqrt{3} + 6 \sqrt{3} }{3} = \frac{4 \sqrt{3} }{3}$

4.

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