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Trigonometric Integrals

  1. Jan 25, 2004 #1
    I've been able to do all of them up until these two:

    1) [tex] \int x\cos^{2}x dx [/tex]

    and

    2) [tex] \int [/tex] from [tex]-\pi[/tex] to [tex]\pi[/tex] of [tex] \sin^{137}x dx [/tex]

    I've been using half angle formulas where (cos x)^2 = (1/2)(1 + cos 2x) and (sin x)^2 = (1/2)(1 - cos 2x)

    I just can't figure those two out. Any help would be highly appreciated
     
  2. jcsd
  3. Jan 25, 2004 #2

    Hurkyl

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    Here's some hints:

    1) Do you know how to integrate [itex]\int x e^x \, dx[/itex]?

    2) A large number (like 137) is often a hint that there's something more interesting at stake. Try some smaller powers and see if you can recognize a pattern, and then prove it.


    P.S. if you were wondering how to make the integral, it's this:

    [tex]
    \int_{-\pi}^{\pi} \sin^{137} x \, dx
    [/tex]
     
  4. Jan 25, 2004 #3


    If this was last semester, yes, but now I can't remember. I know [tex]\int e^x \, dx[/tex] equals [tex]e^x + c[/tex]. Substitution plays a role, but I just can't see how.

    I got nothing. Not even a hunch.
     
  5. Jan 25, 2004 #4

    Hurkyl

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    1) Do you remember integration by parts?


    2) So what smaller powers have you tried and what were the values of the integrals?
     
  6. Jan 25, 2004 #5
    Yes, let me try to solve your equation now.

    [itex]\int x e^x \, dx[/itex]

    u = [tex]x[/tex] du = [tex]dx[/tex]
    dv = [tex]e^x[/tex] v = [tex]e^x[/tex]

    Therefore [itex]\int x e^x \, dx[/itex] = [tex]x e^x - \int e^x dx[/tex] which in turn equals [tex]x e^x - 2x e^x + 2x + c[/tex]

    Correct?

    [tex]\int_{-\pi}^{\pi} \sin^{137} x \, dx[/tex] = [tex]\int_{-\pi}^{\pi} \sin^{135} x \sin^{2} x \, dx[/tex] and [tex]\sin^{2} x = 1/2(1 - \cos 2x)[/tex]

    I'm slowly getting it. Thanks for the help as far.
     
  7. Jan 25, 2004 #6
    Back to #1 [tex] \int x\cos^{2}x dx [/tex]

    u = [tex]x[/tex] du = [tex]dx[/tex]
    dv = [tex]\cos^{2} x[/tex] v = [tex]\sin^{2} x[/tex]

    = [tex]x \sin^{2} x - \int \sin^{2} x dx[/tex]
    = [tex]x \sin^{2} x - (1/2)\int (1 - \cos 2x) dx[/tex]
    = [tex]x \sin^{2} x - (x/2) - ( \sin 2x / 4) + c[/tex]

    how's that look?
     
  8. Jan 25, 2004 #7

    Hurkyl

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    Well, [itex]x e^x - \int e^x \, dx = x e^x - e^x + C[/itex], but I think you knew that and just made a typo. And you should really always include the [itex]dx[/itex] part, so you say that [itex]dv = e^x \, dx[/itex].


    Now, when you attacked your actual problem, you have [itex]dv = \cos^2 x \, dx[/itex]. You got [itex]v = sin^2 x[/itex], but that doesn't work because [itex]dv = 2 \sin x \cos x \, dx[/itex]! Remember this says that the derivative of [itex]v[/itex] has to be [itex]\cos^2 x[/itex], so we need to integrate to find [itex]v[/itex]! In other words:

    [tex]v = \int \cos^2 x \, dx[/tex]



    For problem 2:

    There is a really "obvious" method for this problem that takes advantage of the shape of your integrand and the domain of integration... one you'll recognize with a little practice. I'm trying not to spoil the exercise by just giving you the answer, which is why I'm suggesting to try some small cases; I want you to find:

    [tex]\int_{-\pi}^{\pi} \sin^1 x \, dx[/tex]
    [tex]\int_{-\pi}^{\pi} \sin^2 x \, dx[/tex]
    [tex]\int_{-\pi}^{\pi} \sin^3 x \, dx[/tex]
    [tex]\int_{-\pi}^{\pi} \sin^4 x \, dx[/tex]
    [tex]\int_{-\pi}^{\pi} \sin^5 x \, dx[/tex]

    That should probably be enough to see the relevant pattern.


    (Just to check, you have learned how to integrate odd powers of the sine function right?)
     
  9. Jan 25, 2004 #8


    [tex]v = \int \cos^2 x \, dx[/tex] = [tex]x/2 + \sin 2x/4[/tex] ??

    Basically what you're saying is that no matter the number sine is raised to, the integrand between [tex]-\pi[/tex] and [tex]\pi[/tex] will always be the same?? Therefore I should just use [tex]\int_{-\pi}^{\pi} \sin^2 x \, dx[/tex] and solve that? Yes? Maybe? Not even close? Slowly, but surely
     
  10. Jan 25, 2004 #9
    They all equal zero
     
  11. Jan 25, 2004 #10
    ∫xcos^2x dx = x[0.5(x + sin2x/2)] - ∫[0.5(x + sin2x/2)]dx

    = ... - (x^2/4 - cos[2x]/8)
     
  12. Jan 25, 2004 #11

    Hurkyl

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    Yyou got v right.


    Actually, only the odd powers of sin turn out to integrate to zero (and 137 is an odd power); can you figure out why?
     
  13. Jan 26, 2004 #12
    Simply it is an odd function and they are symmetrical w.r.t Origin or in (I &III) coordinate hence one portion will be above x axis and other below x axis with equal magnitude hence the result would be zero
     
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