Предмет: Математика, автор: vlado123

Решите пожалуйста, нужно вычислить определенные интегралы

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Ответы

Автор ответа: QDominus

$\int \limits_{1}^{4} \frac{{e}^{3 \sqrt{x} } }{ \sqrt{x} } dx = \int \limits_{1}^{4} \frac{t}{ \sqrt{x} } \times \frac{2 \sqrt{x} }{3t} dt = \left. \frac{2}{3}t \right |_{x = 1}^{x = 4} = \\ {e}^{3 \sqrt{x} } = t, \: dx = \frac{2 \sqrt{x} }{3 {e}^{3 \sqrt{x} } } dt \\ = \left. \frac{2}{3} {e}^{3 \sqrt{x} } \right |_{1}^{4} = \frac{2}{3} ( {e}^{3 \sqrt{4} } - {e}^{3 \sqrt{1} } ) = \frac{2}{3} ( {e}^{6} - {e}^{3} )$

$\int \limits_{0}^{ \sqrt{3} } \frac{dx}{ \sqrt{(4 - {x}^{2}) {}^{3} } } = \int \limits_{0}^{ \sqrt{3} } \frac{2 \cos(t)dt }{8 \cos {}^{3} (t) } = \frac{1}{4}\int \limits_{0}^{ \sqrt{3} } \frac{1}{ \cos {}^{2} (t) } = \\ x = 2 \sin(t) , \: dx = 2 \cos(t) dt \rightarrow (4 - (2 \sin(t) ) {}^{2} ) {}^{ \frac{3}{2} } = \\ = 8 \cos {}^{3} (t) , \: u = \arcsin \frac{x}{2} \\ = \frac{1}{4} \tan(t) = \frac{1}{4} \tan( \arcsin \frac{x}{2} ) = \left.\frac{x}{4 \sqrt{4 - {x}^{2} } } \right|_{0}^{ \sqrt{3} } = \\ = \frac{ \sqrt{3} }{4 \sqrt{4 - 3} } = \frac{ \sqrt{3} }{4}$

$\int \limits_{0}^{1} {x}^{2} \arctg xdx = \left.\frac{ {x}^{3} }{3} \arctg x \right|_{0}^{1} - \int \limits_{0}^{1} \frac{ {x}^{3} }{3(1 + {x}^{2})} dx = \\ = \frac{1}{3} \arctg(1) - \frac{1}{3} \int \limits_{0}^{1} \frac{ {x}^{3} }{1 + {x}^{2} } dx = \frac{\pi}{12} - \frac{1}{3}\int \limits_{0}^{1} \frac{ {x}^{2} }{2t}dt = \\ {x}^{2} + 1 = t, \: dx = \frac{1}{2x} dt \\ = \frac{\pi}{12} - \frac{1}{6} \int \limits_{0}^{1} \frac{t - 1}{t} dt = \frac{\pi}{12} - \frac{1}{6} (\left.t\right|_{x = 0}^{x = 1 } - \left. ln(t) \right|_{x = 0}^{x = 1 }) = \\ = \frac{\pi}{12} - \frac{1}{6} (\left. {x}^{2} + 1\right|_{0}^{ 1 } - \left. ln( {x}^{2} + 1) \right|_{0}^{ 1 }) = \frac{\pi}{12} - \frac{1}{6} (1 - ln(2) ) = \\ = \frac{\pi}{12} - \frac{1 - ln(2) }{6}$