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| Интегральное (8) http://mathhelpplanet.com/viewtopic.php?f=19&t=18255 |
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| Автор: | jagdish [ 23 сен 2012, 21:55 ] |
| Заголовок сообщения: | Интегральное (8) |
[math]\displaystyle \int\frac{\sec^2(x)}{\left(\sec (x)+\tan (x)\right)^{\frac{9}{2}}}dx[/math] |
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| Автор: | Avgust [ 24 сен 2012, 22:10 ] |
| Заголовок сообщения: | Re: Интегральное (8) |
One problem: cos(x) = f[tg(x)] - ? |
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| Автор: | Prokop [ 25 сен 2012, 18:43 ] |
| Заголовок сообщения: | Re: Интегральное (8) |
There is a way but it is very long. [math]I = \int {\frac{{\cos ^{{5 \mathord{\left/ {\vphantom {5 2}} \right. \kern-\nulldelimiterspace} 2}} x}}{{\left( {1 + \sin x} \right)^{{9 \mathord{\left/ {\vphantom {9 2}} \right. \kern-\nulldelimiterspace} 2}} }}dx} = \int {\frac{{\cos ^{{3 \mathord{\left/ {\vphantom {3 2}} \right. \kern-\nulldelimiterspace} 2}} x \cdot \cos x}}{{\left( {1 + \sin x} \right)^{{9 \mathord{\left/ {\vphantom {9 2}} \right. \kern-\nulldelimiterspace} 2}} }}dx} = \left\{ {\sin x = t} \right\} = \int {\frac{{\left( {1 - t} \right)^{{3 \mathord{\left/ {\vphantom {3 4}} \right. \kern-\nulldelimiterspace} 4}} }}{{\left( {1 + t} \right)^{{{15} \mathord{\left/ {\vphantom {{15} 4}} \right. \kern-\nulldelimiterspace} 4}} }}dt} = \left\{ {1 + t = s} \right\} = \int {\left( {2 - s} \right)^{{3 \mathord{\left/ {\vphantom {3 4}} \right. \kern-\nulldelimiterspace} 4}} s^{{{ - 15} \mathord{\left/ {\vphantom {{ - 15} 4}} \right. \kern-\nulldelimiterspace} 4}} ds}[/math] Thus came to an integration of the differential binomial Answer [math]- \frac{2}{{77}}\frac{{\cos ^{{7 \mathord{\left/ {\vphantom {7 2}} \right. \kern-\nulldelimiterspace} 2}} x\left( {9 + 2\sin x} \right)}}{{\left( {1 + \sin x} \right)^{{9 \mathord{\left/ {\vphantom {9 2}} \right. \kern-\nulldelimiterspace} 2}} }} + C[/math] |
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| Автор: | jagdish [ 27 сен 2012, 16:19 ] |
| Заголовок сообщения: | Re: Интегральное (8) |
Thanks prokop I have solve it like that way [math]\displaystyle \int \frac{\sec^2 x}{\left(\sec x+\tan x\right)^{\frac{9}{2}}}dx[/math] Let [math]\sec x+\tan x = t[/math], Then [math]\sec x.\left(\sec x +\tan x \right)dx = dt[/math] [math]\displaystyle \sec x.dx = \frac{dt}{t}[/math] So [math]\displaystyle \int \frac{\sec x}{t^{\frac{11}{2}}}dt[/math] Now Using [math]\left(\sec x +\tan x \right).\left(\sec x +\tan x \right) = 1[/math] So [math]\left(\sec x -\tan x \right) = \frac{1}{t}[/math] So [math]\displaystyle \sec xdx = \frac{t^2+1}{2t}[/math] So [math]\displaystyle \int \frac{\sec x}{t^{\frac{11}{2}}}dt = \frac{1}{2}\int \frac{t^2+1}{t^\frac{13}{2}}dt[/math] So [math]\displaystyle = \frac{1}{2}\int t^{-\frac{9}{2}}dt+\frac{1}{2}\int t^{-\frac{13}{2}}dt[/math] [math]\displaystyle = -\frac{1}{7}t^{-\frac{7}{2}}-\frac{1}{11}t^{-\frac{11}{2}}+C[/math] So [math]\displaystyle \int \frac{\sec^2 x}{\left(\sec x+\tan x\right)^{\frac{9}{2}}}dx = -\frac{1}{7}.\left(\sec x +\tan x \right)^{-\frac{7}{2}}-\frac{1}{11}.\left(\sec x +\tan x \right)^{-\frac{11}{2}}+C[/math] To Prokop Sir would you like to explain me the steps Prokop писал(а): There is a way but it is very long. [math]I = \int {\frac{{\cos ^{{5 \mathord{\left/ {\vphantom {5 2}} \right. \kern-\nulldelimiterspace} 2}} x}}{{\left( {1 + \sin x} \right)^{{9 \mathord{\left/ {\vphantom {9 2}} \right. \kern-\nulldelimiterspace} 2}} }}dx} = \int {\frac{{\cos ^{{3 \mathord{\left/ {\vphantom {3 2}} \right. \kern-\nulldelimiterspace} 2}} x \cdot \cos x}}{{\left( {1 + \sin x} \right)^{{9 \mathord{\left/ {\vphantom {9 2}} \right. \kern-\nulldelimiterspace} 2}} }}dx} = \left\{ {\sin x = t} \right\} = \int {\frac{{\left( {1 - t} \right)^{{3 \mathord{\left/ {\vphantom {3 4}} \right. \kern-\nulldelimiterspace} 4}} }}{{\left( {1 + t} \right)^{{{15} \mathord{\left/ {\vphantom {{15} 4}} \right. \kern-\nulldelimiterspace} 4}} }}dt} = \left\{ {1 + t = s} \right\} = \int {\left( {2 - s} \right)^{{3 \mathord{\left/ {\vphantom {3 4}} \right. \kern-\nulldelimiterspace} 4}} s^{{{ - 15} \mathord{\left/ {\vphantom {{ - 15} 4}} \right. \kern-\nulldelimiterspace} 4}} ds}[/math] Thus came to an integration of the differential binomial Answer [math]- \frac{2}{{77}}\frac{{\cos ^{{7 \mathord{\left/ {\vphantom {7 2}} \right. \kern-\nulldelimiterspace} 2}} x\left( {9 + 2\sin x} \right)}}{{\left( {1 + \sin x} \right)^{{9 \mathord{\left/ {\vphantom {9 2}} \right. \kern-\nulldelimiterspace} 2}} }} + C[/math] Thanks |
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| Автор: | Prokop [ 28 сен 2012, 09:05 ] |
| Заголовок сообщения: | Re: Интегральное (8) |
jagdish Very good [math]\begin{gathered}\int {\left({2 - s} \right)^{3/4} s^{- 15/4} ds}= \int {\left( {\frac{{2 - s}}{s}} \right)} ^{3/4} s^{ - 3} ds = \left\{ {y = \frac{2}{s} - 1} \right\} = - \frac{1}{4}\int {y^{3/4} \left({y + 1}\right)dy}= \hfill \\= - \left({\frac{1}{{11}}\left( {\frac{{2 - s}}{s}}\right)^{11/4}+ \frac{1}{7}\left({\frac{{2 - s}}{s}} \right)^{7/4} } \right) + C = - \left( {\frac{1}{{11}}\left( {\frac{{1 - \sin x}}{{1 + \sin x}}} \right)^{11/4} + \frac{1}{7}\left( {\frac{{1 - \sin x}}{{1 + \sin x}}} \right)^{7/4} } \right) + C = \hfill \\ = - \left( {\frac{1}{{11}}\frac{{\cos ^{11/2} x}}{{\left( {1 + \sin x} \right)^{11/2} }} + \frac{1}{7}\frac{{\cos ^{7/2} x}}{{\left( {1 + \sin x} \right)^{7/2} }}}+C \right) = \ldots \hfill \\ \end{gathered}[/math] |
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