Предмет: Математика, автор: Smeile2016

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Автор ответа: Miroslava227

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Ответ:

$\int\limits^{ \frac{\pi}{2} } _ {0} \frac{dx}{2 + \cos(x) } \\$

Тригонометрическая замена:

$\cos(x) = \frac{1 - {t}^{2} }{1 + {t}^{2} } \\ dx = \frac{2dt}{1 + {t}^{2} } \\ \\ t = tg( \frac{x}{2} ) \\ t_1 = tg (\frac{\pi}{4} ) = 1 \\ t_2 = tg(0) = 0$

$\int\limits^{ 1 } _ {0} \frac{2dt}{1 + t {}^{2} } \times \frac{1}{2 + \frac{1 - {t}^{2} }{1 + {t}^{2} } } = \\ = \int\limits^{ 1 } _ {0} \frac{2dt}{1 + t {}^{2} } \times \frac{1 + {t}^{2} }{2 + 2 {t}^{2} + 1 - {t}^{2} } = \\ = \int\limits^{ 1 } _ {0} \frac{2dt}{ {t}^{2} + 3} = 2\int\limits^{ 1 } _ {0} \frac{dt}{t {}^{2} + {( \sqrt{3} )}^{2} } = \\ = \frac{2}{ \sqrt{3} } arctg( \frac{t}{ \sqrt{3} } ) | ^{ 1 } _ {0} = \\ = \frac{2}{ \sqrt{3} } arctg( \frac{1}{ \sqrt{3} } ) - \frac{2}{ \sqrt{3} } arctg(0) = \\ = \frac{2}{ \sqrt{3} } \times \frac{\pi}{6} = \frac{\pi}{3 \sqrt{3} } = \frac{\pi \sqrt{3} }{9}$

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