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Int 1/(sin^3x+cos^3x)

  1. Jul 30, 2011 #1
    1. The problem statement, all variables and given/known data
    I am asked to integrate 1/[(sin[itex]^{3}[/itex]x)+(cos[itex]^{3}[/itex]x)] dx. Indefinite integration.


    2. Relevant equations



    3. The attempt at a solution
    I tried pretty much everything I could think of. Wrote the integral in terms of sin3x and cos3x, and tried to simplify. Even tried a few more identities, and representation in complex form. Found this prob in an undergrad book, so I didn't try DUIS and such. After simplifying I get stuck and I have no idea how to proceed further.

    Any help would be appreciated.
     
  2. jcsd
  3. Jul 30, 2011 #2
    Try starting with partial fractions
     
  4. Jul 30, 2011 #3
    ehem i deleted my own message, woop.
    for me it seems to work after changing the denominator to u, and then partial fractions, then some hefty integration which can be looked up i believe.
     
  5. Jul 30, 2011 #4
    substitute : tan (x) = t and then try to solve it
     
  6. Jul 30, 2011 #5
    OK I substituted sinx=u, and then got the whole integral in terms of u. Splitting it into partial fractions, I get two terms: 1/[itex]\sqrt{1-u^{2}}[/itex] and another term 1/u[itex]^{3}[/itex]+(1-u[itex]^{2}[/itex])[itex]^{3/2}[/itex]. Am stumped about the second term. Actually I don't even know what I did is right!
     
  7. Jul 30, 2011 #6
    let u = (sin^3 (x) + cos^3(x))
     
  8. Jul 30, 2011 #7
    Well if I put the denominator as u, then du=[3sin^2(x)cos(x)-3cos^2(x)sin(x)]dx, right? Now what? Do I take the whole term out and perform integration w.r.t u? That gives some unexpected answers...
    Sorry I haven't brushed up my integration in a while, so please bear with me.
    Btw just for fun, I entered this integral in WolframAlpha(tried it for the first time), and the result is not pretty, containing complex terms and inverse tangent hyperbolic functions!
     
  9. Jul 31, 2011 #8
    scrap that. let tan(x) = u, substitute and solve using the residue theorem.
    edited the theorem name
     
    Last edited: Jul 31, 2011
  10. Jul 31, 2011 #9
    Hmmm....did try that. Putting tan(x)=u, I get integral (1-u[itex]^{2}[/itex])[itex]^{3/2}[/itex]/(1+u[itex]^{2}[/itex])(1+u[itex]^{3}[/itex]).
    I have no idea how to use residue theorem here, considering that the region has not been specified. Should I simply take up a unit circle to be the region? Also, if that were the case then I would end up with complex terms I guess.
     
  11. Jul 31, 2011 #10
    if you end up with complex terms then you are doing something wrong
     
  12. Jul 31, 2011 #11
    That doesn't answer my question of how to select the region when none is given. If it were arbitrary, wouldn't the answer be dependent on the region?
     
  13. Jul 31, 2011 #12
    apart from an abitrary multipicative factor, the region over which you integrate will make no difference to the final result
     
  14. Jul 31, 2011 #13
    I see where you are getting at. But unfortunately, I have no idea how to use residue theorem here, especially when you say that the answer won't have complex terms.
    Thanks for helping me out though. Much appreciated! :smile:
     
  15. Jul 31, 2011 #14

    ehild

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    Try this way (although it will be still very ugly):

    sin3(x)+cos3(x)=(sin(x)+cos(x))(sin2(x)+sin(x)cos(x)+cos2(x))
    And use that sin(x)+cos(x) = √2 sin(x+π/4).

    ehild
     
    Last edited: Jul 31, 2011
  16. Aug 1, 2011 #15
    You could try the substitution u = tan(x/2) where sinx = 2u/(u2 + 1) and cosx = (1 - u2)/(u2 + 1). Looks like it will turn into some nasty partial fractions though, but it shouldn't have any complex terms in it.
     
  17. Aug 2, 2011 #16
    Hmm...tried that out and did get some pretty ugly fractions. Got stuck at a point, but that's nothing I cant figure out when I get free.
    Thanks for all your help guys!
     
  18. Aug 2, 2011 #17

    ehild

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    The denominator becomes √2 sin(x+π/4)(1+0.5 sin(2x))

    Introduce the new variable u=x+π/4.

    sin(2x)=sin(2u-pi/2)=-cos(2u).
    1+0.5 sin(2x)=0.5(2-cos(2u))=0.5(1+2sin2(u))

    The new form of the integral is

    [tex]I=\int{\frac{\sqrt{2}}{sin(u)(1+2 sin^2(u))}du}[/tex]

    Substitute [tex]sin(u)=\frac{2 tan(u/2)}{1+tan^2(u/2)}[/tex],
    the integrand becomes

    [tex]\frac{(1+tan^2(u/2))^3}{\sqrt{2}tan(u/2)(1+10tan^2(u/2)+tan^4(u/2))}[/tex]

    Let be t a new variable:

    t=tan(u/2), du = 2/(1+t2)dt
    .

    [tex]I=\int{\sqrt{2}\frac{(1+t^2)^2}{t(t^4+10t^2+1)}dt}=
    \sqrt{2}\int{(\frac{1}{t}-\frac{8t}{t^4+10t^2+1})dt}[/tex]

    It is straightforward to proceed further.

    Could somebody check it, please?

    ehild
     
  19. Aug 3, 2011 #18
    this equality is incorrect unless the sign of sin(x)cos(x) is minus.
    at the very end then how would you go about integrating an integrand which has a polynomial of higher degree in the denominator than the numerator?
     
  20. Aug 3, 2011 #19

    ehild

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    You are right, I made that mistake while doing the derivation second time (first time it was minus).
    This will change the integral to

    [tex]I=\int{\frac{\sqrt{2}}{sin(u)(1+2 cos^2(u))}du}[/tex]

    unless I am mistaken again.

    ehild
     
  21. Aug 3, 2011 #20

    tiny-tim

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    using ehild's :smile: substitution u = x + π/4, sinu = (sinx + cosx)/√2, sin2x = -cos2u, du = dx …

    ∫ dx/(sin3x + cos3x)

    = ∫ dx/((sinx + cosx)3 - 3sinxcosx(sinx + cosx))

    = ∫ du/sinu(2√2sin2u + (3/√2)cos2u)

    = ∫ du/sinu(3/√2 + (2√2 - 3√2)sin2u)

    = ∫ du/√2sinu(3/2 - sin2u)

    = ∫ du/√2sinu(√(3/2) - sinu)(√(3/2) + sinu)

    then use partial fractions to give three separate integrands

    = ∫ du/√2sinu + (1/2√2)∫ du/(√(3/2) - sinu) - (1/2√2)∫ du/(√(3/2) + sinu)

    (eugh :yuck: … that needs checking :redface:)

    then use tan(u/2) = t
     
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