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ÇáÒÇÆÑ ÇáßÑíã:ÊÓÊØíÚ ãÔÇåÏÉ ßá ÇáãæÇÖíÚ ÈÏæä ÔÑØ ÇáÚÖæíÉ áßä ãÔÇÑßÊß ÊÍÊÇÌ ÚÖæíÉ ÇáãäÊÏì
ÕÝÍÉ 5 ãä 8 ÇáÃæáìÇáÃæáì 12345678 ÇáÃÎíÑÉÇáÃÎíÑÉ
ÇáäÊÇÆÌ 49 Åáì 60 ãä 86

ÇáãæÖæÚ: ÓáÓáÉ ÇáÊßÇãá

  1. Top | #49

    ÊÇÑíÎ ÇáÊÓÌíá
    May 2014
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    19175
    ÇááÞÈ
    ÇÓÊÇÐ ãÊãíÒ ( ÎÈíÑÚÇã)
    ãÚÏá ÇáãÔÇÑßÇÊ
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá

    $\large {\color{Red} \int (\frac{\tan2x-tanx}{\tan2x+tanx})dx=\int (\frac{\frac{2tanx}{1-tan^{2}x}-tanx}{\frac{2tanx}{1-tan^{2}x}+tanx})dx=\int (\frac{2tanx+tan^{3}x-tanx}{2tanx-tan^{3}x+tan})dx}$
    ${\color{Red} =\int (\frac{sec^{2}x}{3-tan^{2}x})dx=\int (\frac{dy}{3-y^{2}})=\frac{1}{\sqrt{3}}\tanh^{-1}(1/\sqrt{3})y+c}$

  2. The Following User Says Thank You to Ìáíá ÇáÒíÇÏí For This Useful Post:

     (06-19-2016)

  3. Top | #50

    ÊÇÑíÎ ÇáÊÓÌíá
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    ÇÓÊÇÐ ãÊãíÒ ( ÎÈíÑÚÇã)
    ãÚÏá ÇáãÔÇÑßÇÊ
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá

    $\large {\color{Blue} \int sin^{3}xcos(x/2))dx=8\int sin^{3}(x/2)cos^{4}(x/2)dx=8\int sin(x/2)cos^{4}(x/2)[1-cos^{2}(x/2)]dx}$
    $ {\color{Blue} =8[\frac{-2}{a}cos^{a}(x/2)+\frac{2}{7}cos^{7}(x/2)]+c}$
    ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ Ìáíá ÇáÒíÇÏí ; 06-18-2016 ÇáÓÇÚÉ 06:44 PM

  4. The Following User Says Thank You to Ìáíá ÇáÒíÇÏí For This Useful Post:

     (06-19-2016)

  5. Top | #51

    ÊÇÑíÎ ÇáÊÓÌíá
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá

    $\LARGE {\color{DarkBlue} \int \frac{\sec x}{\sqrt{\sin(2x+2)+sin2}}dx=\int \frac{\sec x} {\sqrt{sin2xcos2+sin2cos2x+sin2}}dx}$
    ${\color{DarkBlue} =\int \frac{\sec x}{\sqrt{2sinxcosxcos2+2sin2cos^{2}x}}dx=\int {\frac{\sec^{2}x}{\sqrt{2}\sqrt{cos2tanx+sin2}}}dx }$
    ${\color{DarkBlue} =\frac{\sqrt{2}}{cos2}\sqrt{cos2tanx+sin2}+c}$
    ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ Ìáíá ÇáÒíÇÏí ; 06-18-2016 ÇáÓÇÚÉ 09:24 PM

  6. The Following User Says Thank You to Ìáíá ÇáÒíÇÏí For This Useful Post:

     (06-19-2016)

  7. Top | #52

    ÊÇÑíÎ ÇáÊÓÌíá
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    ÇÓÊÇÐ ãÊãíÒ ( ÎÈíÑÚÇã)
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá

    $\LARGE {\color{DarkBlue} \int \frac{1}{\sqrt[3]{sin^{11}xcosx}}dx=\int \frac{csc^{4}x}{\sqrt[3]{cotx}}dx} $
    ${\color{DarkBlue} =\int \frac{csc^{2}x(1+cot^{2}x)}{\sqrt[3]{cotx}}dx=-\int \frac{1+u^{2}}{\sqrt[3]{u}}du}$
    ${\color{DarkBlue} =-\frac{3}{2}u^{\frac{2}{3}}-\frac{3}{8}u^{\frac{8}{3}}+c}$
    ${\color{DarkBlue} u=cotx}$

  8. The Following 2 Users Say Thank You to Ìáíá ÇáÒíÇÏí For This Useful Post:

     (06-19-2016),  (06-19-2016)

  9. Top | #53

    ÊÇÑíÎ ÇáÊÓÌíá
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá

    $Q46)\int e^{x}(\frac{x+2}{x+4})^{2}dx$

    $Q47)\int e^{arctanx}(\frac{1+x+x^{2}}{1+x^{2}})dx$
    "ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
    ãä ãæÇÖíÚ


  10. Top | #54

    ÊÇÑíÎ ÇáÊÓÌíá
    Oct 2011
    ÑÞã ÇáÚÖæíÉ
    541
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá

    ÇÞÊÈÇÓ ÇáãÔÇÑßÉ ÇáÃÕáíÉ ßÊÈÊ ÈæÇÓØÉ ÝáÇÍ ÇáäÇÕÑí ãÔÇåÏÉ ÇáãÔÇÑßÉ
    $Q46)\int e^{x}(\frac{x+2}{x+4})^{2}dx$

    $Q47)\int e^{arctanx}(\frac{1+x+x^{2}}{1+x^{2}})dx$

    \[Q46)\int e^{x}\left ( \frac{x+2}{x+4} \right )^{2}dx=\int e^{x}\left ( 1-\frac{2}{x+4} \right )^{2}dx\\=\int e^{x}\left ( 1-\frac{4}{x+4}+\frac{4}{\left ( x+4 \right )^{2}} \right )dx\\=\int \left ( e^{x}-4\times \frac{(x+4)e^{x}-e^{x}}{\left ( x+4 \right )^{2}} \right )dx\\=e^{x}-\frac{4e^{x}}{x+4}+C=\frac{xe^{x}}{x+4}+C\]

  11. The Following 4 Users Say Thank You to ÍÇÌã ÇáÑÈíÚí For This Useful Post:

     (06-21-2016),  (06-20-2016),  (06-20-2016),  (06-20-2016)

  12. Top | #55

    ÊÇÑíÎ ÇáÊÓÌíá
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá

    \[Q47)\int e^{arctan}\left ( \frac{1+x+x^{2}}{1+x^{2}} \right )dx=\\\int \left ( e^{arctanx}+\frac{xe^{arctanx}}{1+x^{2}} \right )dx=\\\int\left ( e^{arctanx}\left ({x}' \right ) +x{\left (e^{arctanx} \right )}'\right )dx=\\xe^{arctanx}+C\]

  13. The Following 3 Users Say Thank You to ÍÇÌã ÇáÑÈíÚí For This Useful Post:

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  14. Top | #56

    ÊÇÑíÎ ÇáÊÓÌíá
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá

    $Q48)\int \frac{1+4cosx+2sinx}{1+2cosx+sinx}dx$

    $Q49)\int \frac{4sinx+2cosx}{2cosx+sinx}dx$

    $Q50)\int \frac{coshx+sinhx sinx}{1+cosx}dx$

    $Q51)\int \frac{(x^3+x)arctanx}{1+x^2}dx$

    ÇáÊÚÏíá ÇáÃÎíÑ Êã ÈæÇÓØÉ ÝáÇÍ ÇáäÇÕÑí ; 08-25-2016 ÇáÓÇÚÉ 12:40 AM
    "ÇáÑíÇÖíÇÊ åí Êáß ÇáãÊÚÉ ÇáÊí íÈÍË ÚäåÇ ÇáÃÐßíÇÁ æíÍÇæáæä ÇÓÊßÔÇÝ ÃÓÑÇÑåÇ æÍá ãÌåæáÇÊåÇ"
    ãä ãæÇÖíÚ


  15. The Following User Says Thank You to ÝáÇÍ ÇáäÇÕÑí For This Useful Post:

     (08-22-2016)

  16. Top | #57

    ÊÇÑíÎ ÇáÊÓÌíá
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá


  17. Top | #58

    ÊÇÑíÎ ÇáÊÓÌíá
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    ÑÏ: ÓáÓáÉ ÇáÊßÇãá


  18. The Following 2 Users Say Thank You to ÍÇÌã ÇáÑÈíÚí For This Useful Post:

     (08-22-2016),  (08-25-2016)

  19. Top | #59

  20. Top | #60

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