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| definite Integral http://mathhelpplanet.com/viewtopic.php?f=19&t=15351 |
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| Автор: | jagdish [ 14 мар 2012, 13:37 ] |
| Заголовок сообщения: | definite Integral |
[math]\int_{0}^{3}\sin \left(\frac{\pi}{2}+\arccos \left(\frac{x-1}{\sqrt{x+1}}\right)\right)dx[/math] |
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| Автор: | Yurik [ 14 мар 2012, 13:55 ] |
| Заголовок сообщения: | Re: definite Integral |
[math]\begin{gathered} \int_0^3 {\sin } \left( {\frac{\pi }{2} + \arccos \left( {\frac{{x - 1}}{{\sqrt {x + 1} }}} \right)} \right)dx = \int_0^3 {\cos } \left( {\arccos \left( {\frac{{x - 1}}{{\sqrt {x + 1} }}} \right)} \right)dx = \int_0^3 {\frac{{x - 1}}{{\sqrt {x + 1} }}} dx = \hfill \\ = \left| \begin{gathered} t = \sqrt {x + 1} \,\, = > \,\,\,x = {t^2} - 1 \hfill \\ dx = 2tdt,\,\,t\left( 0 \right) = 1\,\,t\left( 3 \right) = 2 \hfill \\ \end{gathered} \right| = 2\int_1^2 {\frac{{t\left( {{t^2} - 2} \right)}}{t}dt} = 2\left. {\left( {\frac{{{t^3}}}{3} - 2t} \right)} \right|_1^2 = 2\left( {\frac{8}{3} - 4 - \frac{1}{3} + 2} \right) = \frac{2}{3} \hfill \\ \end{gathered}[/math] |
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