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Integration problem!

  1. Mar 7, 2005 #1
    could you please give me any idea to solve the problem below;

    integral( dx / (3cosx-4sinx) )


    and given a hint to make a subtitution u=tan(x/2), i've tried to write cosx and sinx in the form of cos (x/2) and sin(x/2), but it's seems like i'm not going anywhere, any suggestions?
     
  2. jcsd
  3. Mar 7, 2005 #2
    [tex]\int (\frac{dx}{3cosx - 4sinx})[/tex]

    Does this help???:

    [tex]\int tanxdx = \int \frac{sinx}{cosx}dx[/tex]

    It sounds like a partial fractions to me.

    The Bob (2004 ©)
     
  4. Mar 7, 2005 #3
    Yes, you have to use the half angle formulas (or is it double angle formula?) for sin x and cos x.

    [tex]\cos^2{x} = \frac {1 + \cos{2x}}{2} \ \ \mbox{and} \ \ \sin^2{x} = \frac {1-\cos{2x}}{2}[/tex]
     
  5. Mar 7, 2005 #4

    dextercioby

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    It looks really ugly.

    [tex] I=:\int \frac{dx}{3\cos x-4\sin x} [/tex](1)

    Make the substitution:

    [tex] x=2\arctan u (<=> u=\tan\frac{x}{2}) [/tex] (2)

    ,under which simple trigonometry and differentiation will show that

    [tex] dx=\frac{2 du}{1+u^{2}} [/tex] (3)

    [tex] \sin x= \frac{2u}{1+u^{2}} [/tex] (4)

    [tex] \cos x=\frac{1-u^{2}}{1+u^{2}} [/tex] (5)

    Can u continue from here...?

    Daniel.
     
  6. Mar 7, 2005 #5
    thanks, i got it =)
     
  7. Mar 10, 2005 #6
    Dextercioby, I don't get what u wrote. in the 4th warning. the (2) would u tell me. thanx
     
  8. Mar 10, 2005 #7

    arildno

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    One-D: It's 4'th POST, not WARNING!
    Daniel made a very common and useful change of variables.
    That's all there is to it.
     
  9. Mar 10, 2005 #8

    dextercioby

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    Incidentally i have 4 warnings...:rofl: :uhh:

    Daniel.
     
  10. Mar 10, 2005 #9

    arildno

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    I already knew you were a good and inoffensive boy..:wink:
     
  11. Mar 10, 2005 #10

    dextercioby

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    Thanks for the trust.Marlon feels the same way,though i don't remember any warning gotten from the clashes we've had...:wink:

    Daniel.
     
  12. Mar 12, 2005 #11
    thanx. know i understand. it's only a simple subs. thx anyway.
     
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