:
1 6 6

:

  1. Top | #1

    Mar 2014
    18721
    0.00
    11
    Thanked: 11

    \[\int \frac{2}{1\dotplus \sin x-cos x }\]

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

     (06-19-2016)

  3. Top | #2

    Sep 2013
    17034
    0.29
    /
    1,349
    Thanked: 1595

    :

    $\int \frac{2}{1+sinx-cosx}dx$

    $=\int \frac{2sin^{2}\frac{x}{2}+2cos^{2}\frac{x}{2}}{sin ^{2}\frac{x}{2}+cos^{2}\frac{x}{2}+2sin\frac{x}{2} cos\frac{x}{2}-cos^{2}\frac{x}{2}+sin^{2}\frac{x}{2}}dx$

    $=\int \frac{2sin\frac{x}{2}cos\frac{x}{2}+2cos^{2}\frac{ x}{2}-2sin\frac{x}{2}cos\frac{x}{2}+2sin^{2}\frac{x}{2}} {2sin^{2}\frac{x}{2}+2sin\frac{x}{2}cos\frac{x}{2} }dx$

    $=\int \frac{2sin\frac{x}{2}cos\frac{x}{2}+2cos^{2}\frac{ x}{2}}{2sin^{2}\frac{x}{2}+2sin\frac{x}{2}cos\frac {x}{2}}dx-\int \frac{2sin\frac{x}{2}cos\frac{x}{2}+2sin^{2}\frac{ x}{2}}{2sin^{2}\frac{x}{2}+2sin\frac{x}{2}cos\frac {x}{2}}dx$

    $=\int \frac{cos\frac{x}{2}}{sin\frac{x}{2}}dx-\int \frac{cos\frac{x}{2}-sin\frac{x}{2}}{sin\frac{x}{2}+cos\frac{x}{2}}dx$

    $=2ln\left | sin\frac{x}{2} \right |-2ln\left | sin\frac{x}{2}+cos\frac{x}{2} \right |+c$

  4. The Following 2 Users Say Thank You to For This Useful Post:

     (06-21-2016),  (06-20-2016)

  5. Top | #3

    Oct 2011
    541
    0.57
    3,046
    Thanked: 2931

    :

    \[\int \frac{2}{1\dotplus \sin x-cos x }\]
    \[\int \frac{2}{1+sinx-cosx}{\color{Red} \times \frac{1-\left ( sinx-cosx \right )}{1-\left ( sinx-cosx \right )}}dx=2\int\frac{1-sinx+cosx}{2sinxcosx} dx\\=\int \left [ 2csc2x {\color{Red} \times \frac{csc2x+cot2x}{csc2x+cot2x}}+secx{\color{Red} \times \frac{secx+tanx}{secx+tanx}}+cscx {\color{Red} \times \frac{cscx+cotx}{cscx+cotx}}\right ]dx \\=-ln|csc2x+cot2x|+ln|secx+tanx|-ln|cscx+cotx|+C\]
    [URL="http://alnasiry.net/forums"][IMG]http://alnasiry.net/forums/uploaded/2_iraqiflag.gif[/IMG][/URL]


  6. The Following 2 Users Say Thank You to For This Useful Post:

     (06-21-2016),  (06-20-2016)

  7. Top | #4

    Sep 2013
    17034
    0.29
    /
    1,349
    Thanked: 1595

    :



    $\int \frac{2}{1+sinx-cosx}dx$

    $=\int \frac{2}{1+\frac{2tan\frac{x}{2}}{1+tan^{2}\frac{x }{2}}-\frac{1-tan^{2}\frac{x}{2}}{1+tan^{2}\frac{x}{2}}}dx$

    $=\int \frac{1+tan^{2}\frac{x}{2}}{tan\frac{x}{2}+tan^{2} \frac{x}{2}}dx$

    $=\int \frac{sec^{2}\frac{x}{2}}{tan\frac{x}{2}(1+tan \frac{x}{2})}dx$

    $=\int \frac{sec^{2}\frac{x}{2}}{tan\frac{x}{2}}dx-\int \frac{sec^{2}\frac{x}{2}}{(1+tan\frac{x}{2})}dx$

    $=2ln\left | tan\frac{x}{2} \right |-2ln\left | 1+tan\frac{x}{2} \right |+c$

    $=2ln\left | \frac{tan\frac{x}{2}}{1+tan\frac{x}{2}} \right |+c$
    ; 06-20-2016 08:37 AM

  8. The Following 3 Users Say Thank You to For This Useful Post:

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

  9. Top | #5

  10. The Following 2 Users Say Thank You to For This Useful Post:

     (06-24-2016),  (06-25-2016)

  11. Top | #6

    Nov 2014
    19983
    0.03
    119
    Thanked: 131

    :


  12. The Following 3 Users Say Thank You to For This Useful Post:

     (06-25-2016),  (06-25-2016),  (06-25-2016)

: 1 (0 1 )


Powered by vBulletin® Version 4.2.3
Copyright © 2026 vBulletin Solutions, Inc. All rights reserved.
Translate By Almuhajir
Developed By Marco Mamdouh
Style
Zavord