Question

помогите решить интегралы пожалуйста

Answer 1

Ответ:

a

$\int\limits^{ \pi} _ { \frac{\pi}{3} }(4 - 3 \cos(x)) dx = (4x - 3 \sin(x)) |^{ \pi} _ { \frac{\pi}{3} } = \\ = 4\pi - 3 \sin(\pi) - \frac{4\pi}{3} + 3 \sin( \frac{\pi}{3} ) = \\ = \frac{8\pi}{3} - 0 + \frac{3 \sqrt{3} }{2} = \frac{8\pi}{3} + \frac{3 \sqrt{3} }{2}$

б

$\int\limits^{ \frac{\pi}{3} } _ { \frac{\pi}{6} }(2x - \frac{3}{ \cos {}^{2} (x) } )dx = ( {x}^{2} - 3tgx)| ^{ \frac{\pi}{3} } _ { \frac{\pi}{6} } = \\ = \frac{ {\pi}^{2} }{9} - 3tg \frac{\pi}{3} - \frac{ {\pi}^{2} }{36} + 3tg \frac{\pi}{6} = \\ = \frac{4 {\pi}^{2} - \pi {}^{2} }{36} - 3 \times \sqrt{3} + 3 \times \frac{ \sqrt{3} }{3} = \\ = \frac{ {\pi}^{2} }{12} - 3 \sqrt{3} + \sqrt{3} = \frac{ {\pi}^{2} }{12} - 2 \sqrt{3}$

в

$\int\limits^{ 3 } _ {1} \frac{4 {x}^{2} - 3x + 2}{ {x}^{2} } dx = \int\limits^{ 3 } _ {1}(4 - \frac{3}{x} + 2 {x}^{ - 2} )dx = \\ = (4x - 3 ln( |x| ) - \frac{2}{x} ) | ^{ 3 } _ {1} = \\ = 12 - 3 ln(3) - \frac{2}{3} - 4 + 3 ln(1) + 2 = \\ = 10 - \frac{2}{3} - 3 ln(3) = 9 \frac{1}{3} - 3 ln(3)$