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Homework Help: Confusing Trig/rational Integral

  1. May 9, 2005 #1
    [tex] \int (sin(t)-cos(t)) \sqrt{cos^2(t)-sin^2(t)} dt [/tex]

    Is there a trig idendity I can use? I've distributed that root to both terms to get:
    [tex] \int sin(t) \sqrt{cos^2(t)-sin^2(t)} dt -[/tex] [tex] \int cos(t) \sqrt{cos^2(t)-sin^2(t)} [/tex]

    If I take one of the terms and integrate by parts, I'm trying to put [tex] u=\sqrt{cos^2(t)-sin^2(t)} [/tex] and [tex]dv= sin(t) [/tex] or [tex]dv= cos(t) [/tex] but that ugly root's derivative appears inside the [tex] \int vdu [/tex] part.

    Is there a trig identity I'm missing, or some other tactic I could use? or could this just be really really ugly math?
  2. jcsd
  3. May 9, 2005 #2
    Its either

    [tex] cos^2(t) - sin^2(t) = cos(2t) [/tex]**


    [tex] sin^2(t) - cos^2(t) = cos(2t) [/tex]
  4. May 9, 2005 #3
    [tex] cos^2(t) - sin^2(t) = cos(2t) [/tex] is the correct identity (and thanks! It may just help!)

    Your other identity: [tex] sin^2(t) - cos^2(t) = cos(2t) [/tex] is erroneous, but not by much. [tex] sin^2(t) - cos^2(t) [/tex] yields [tex] -cos(2t) [/tex]

    This could actually help as well, however. Thank you for posting both (even if one was inaccurate).
  5. May 9, 2005 #4


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    You needn't that identity.

    [tex] \int \left(\sin t-\cos t\right)\sqrt{\cos^{2}t-\sin^{2}t} \ dt = \int \sin t\sqrt{2\cos^{2}t-1} \ dt -\int \cos t\sqrt{1-2\sin^{2}t} \ dt [/tex]

    Can u take it from here (HINT:eek:bvious substitutions necessary) ?

  6. May 9, 2005 #5
    That would have been much easier dex, I think....
  7. May 9, 2005 #6
    Well it was one of the two, I justw anst sure which one. dex's way is easier though its basically the same thing.
  8. May 10, 2005 #7
    The Method is substitution, but then what's u? I'm guessing the basic trig function outside the root is du.

    But when I derive the function within that root, I get something different. The derivative of either one ends up to be [tex]-4cos(t)sin(t) [/tex].

    Am I choosing the wrong U? I'm integrating with respect to t, and simply inserting the extra parts onto by basic trig function isn't exactly kosher.
  9. May 10, 2005 #8


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    Okay,let's take the first,i'll leave with the second,which is basically similar (maybe other substitution).

    [tex] I=\int \sin t\sqrt{2\cos^{2}t-1} \ dt [/tex] (1)

    becomes after the substitution [itex] \cos t=u;\sin t \ dt= -du [/itex] (2)

    [tex] I=-\int \sqrt{2u^{2}-1} \ du [/tex](3)

    Make the substitution

    [tex]\sqrt{2} \ u=\cosh v;du=\frac{1}{\sqrt{2}} \sinh v \ dv [/tex](4)


    [tex] I=-\frac{\sqrt{2}}{2}\int \sinh^{2}v \ dv=-\frac{\sqrt{2}}{4}\int (\cosh 2v-1) \ dv=-\frac{\sqrt{2}}{8}\sinh 2v+\frac{\sqrt{2}}{4}v +C [/tex] (5)

    And now invert the 2 substitutions

    [tex] I=-\frac{\sqrt{2}}{8}\sinh \left(2 \ \mbox{arg}\cosh\left(\sqrt{2}\cos t\right)\right)+\frac{\sqrt{2}}{4}\mbox{arg}\cosh\left(\sqrt{2}\cos t\right) + C [/tex] (6)

    I'm sure the other one will be simpler.

    Last edited: May 10, 2005
  10. May 10, 2005 #9
    Didn't know that if you took [tex] cos(t) = u [/tex] that you could allow [tex] u^2 [/tex] to simply be [tex] cos^2(t) [/tex].

    Of course, I didn't think about that either.
  11. May 10, 2005 #10


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    The other way around is incorrect,of course,but i simply squared.

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