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Integral with sin() and cos()

  1. Apr 11, 2005 #1
    Hi,

    I've been having troubles solving this integral:

    [tex]
    \int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx
    [/tex]

    Here's how I tried it:

    [tex]
    t = \cos x
    [/tex]

    [tex]
    dt = - \sin x dx
    [/tex]

    [tex]
    dx = \frac{dt}{-\sin x}
    [/tex]

    [tex]
    x = \arccos t
    [/tex]

    [tex]
    dx = -\frac{1}{\sqrt{1-t^2}} dt
    [/tex]

    [tex]
    -\frac{1}{\sin x} = -\frac{1}{\sqrt{1-t^2}}
    [/tex]

    [tex]
    \sin x = \sqrt{1-t^2}
    [/tex]

    [tex]
    \int \frac{\sin x.\cos x}{\sin^4 x + x.\cos^4 x} dx = - \int \frac{\cos x}{\sin^4 x + x.\cos^4 x} . \left( -\sin x \right) dx = - \int \frac{t}{\left(1-t^2\right)^2 + t^4.\arccos t} dt
    [/tex]

    Well, I don't know at all what to do with this now...

    Does anybody have any idea?

    PS: Why doesn't it insert newline, when I write "\\" in the LaTeX code?
     
  2. jcsd
  3. Apr 11, 2005 #2
    in the [itex] cos^4 x [/itex] in the denominator how did you manage to sibstitute arcSin t into it
    you know cos x dx = dt but [itex] cos^4 x [/itex] is NOT arccos t

    be careful of your substitutions!
     
  4. Apr 11, 2005 #3

    dextercioby

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    Is this your integral?

    [tex] \int \frac{\sin x\cos x}{\sin^{4}x+x\cos^{4}x} \ dx [/tex]...?

    If so,then i got good news and bad news

    Bad:Maple & Mathematica can't crack it.
    Good:You can forget about it.


    Daniel.
     
  5. Apr 11, 2005 #4
    I have no [itex]\arcsin x[/itex] nor [itex]\cos x dx = dt[/itex] there and I'm quite sure about the substitutions..
     
  6. Apr 11, 2005 #5
    I know, I tried Maple and Mathematica too..So I have some doubts about whether the x in the denominator should be there..Anyway, if I ever met such an integral (with variables outside trigonometric functions), is there any way to solve it?
     
  7. Apr 11, 2005 #6

    dextercioby

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    Very good,but that still doesn't help.

    Daniel.
     
  8. Apr 11, 2005 #7

    dextercioby

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    If it isn't there,the integral is easy.But in general this type of integrals involving "x" & transcendental functions are very seldom expressible in terms of "common" special functions.

    Daniel.
     
  9. Apr 11, 2005 #8
    What did you reply with this to? :smile:
     
  10. Apr 11, 2005 #9
    if there were no x in the denominator then it could be solved, it doesnt look as stubborn as the other integral

    then [tex] \int \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x} \ dx = - \frac{1}{2} ArcTan (Cos(2x)) + C[/tex]
     
  11. Apr 11, 2005 #10
    Yes, I tried it without the 'x' and it is easy. (BTW: Maple and Mathematica give different results. I got the Mathematica's one :smile: )
     
  12. Apr 11, 2005 #11
    Yes, that's the one I got too. Maple gives the same excepting the sign :)
     
  13. Apr 11, 2005 #12

    dextercioby

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    Nope,Maple gives the same answer

    [tex] -\frac{1}{2} \arctan\left(\cos 2x\right) +C [/tex]

    Daniel.
     
  14. Apr 11, 2005 #13
    int( (sin(x)*cos(x)) / ( (sin(x))^4 + (cos(x))^4 ),x); [ENTER]
    [tex]
    1/2\,\arctan \left( -1+2\, \left( \cos \left( x \right) \right) ^{2}
    \right)
    [/tex]
     
  15. Apr 11, 2005 #14

    dextercioby

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    My Maple is integrated in an ancient version of SWP.I guess it's less than 5.0 (i think,not too sure,though).So that's why it may be different from your answer.

    I think your Maple is screwed up :tongue2:

    Daniel.
     
  16. Apr 11, 2005 #15

    dextercioby

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    Nope,the sign in the front of arctan is crucial.They're not the same function.

    Daniel.
     
  17. Apr 11, 2005 #16
    I have Maple 9.01 ;) Strange.
     
  18. Apr 11, 2005 #17
    haha, I just noticed the negative sign there and deleted my post :/

    Didn't see it there before
     
  19. Apr 11, 2005 #18
    hmmmmmm

    I have Maple 9.52 and it gives me the correct answer (with the minus sign)
     
    Last edited: Apr 11, 2005
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