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Integration with trig

  1. Oct 12, 2006 #1
    (1)[tex]\int\sin^3{x}\cos^3{x}dx[/tex]
    (2)[tex]=\int\sin^3{x}(1-\sin^2{x})\cos{x}dx[/tex]
    (3a) let u=sin(x), du=cos(x)dx
    (3b)[tex]=\int(u^3-u^5)du[/tex]
    (4)[tex]=\frac{u^4}{4} -\frac{u^6}{6} + C[/tex]
    (5)[tex]=\frac{1}{4}\sin^4{x} -\frac{1}{6}\sin^6{x} + C[/tex]

    If at step (2) I pull off a sin(x) and go from there, I end up with [tex]\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} + C[/tex]
    I tried to prove that this is equvalent to step (5) by using sin^2(A)+cos^2(A)=1, but this turned step (5) into [tex]\frac{1}{12} +\frac{3}{4}\cos^4{x} -\frac{1}{6}\cos^6{x} + C[/tex]. And so I was bewildered and befuddled.

    Then I decided to plug the original problem into a website's Integrator. And the integrator told me that the answer really is [tex]\frac{1}{192}(-9\cos(2x) +\cos(6x))[/tex].

    Now I am just :confused:

    Help, please, and be blessed!

    (Last little bit of frustration: I typed up this post before I logged in, and even the browser's back button couldn't help me retrieve it. Latex is cool but twice?)
     
  2. jcsd
  3. Oct 12, 2006 #2
    So your doing [tex] \int \sin x(\sin^{2} x - \sin^{4} x) dx [/tex]?

    That is equaled to [tex] \int(u^{3}-u^{5}) du [/tex]
     
  4. Oct 12, 2006 #3
    yes, because what I did was
    [tex]\int\sin^3{x}(1-\sin^2{x})(\cos{x})dx[/tex]
    [tex]=\int(\sin^3{x}-\sin^5{x})(\cosx)dx[/tex]
    [tex]=\int(u^3-u^5)du[/tex]
    if u=sinx, du=cosxdx
     
    Last edited: Oct 12, 2006
  5. Oct 12, 2006 #4
    ok. so what are you having problems with then? [tex] \int \sin x(\sin^{2} x - \sin^{4} x) dx [/tex] is equivalent do what you did. you say you pulled off a [tex] \sin x [/tex]. Did you mean factor it like I did in the integral?
     
  6. Oct 12, 2006 #5
    No, I mean that instead of my original step (2), I did this:
    [tex]\int(1-\cos^2{x})(\cos^3{x})(\sin{x})dx[/tex]
    [tex]=\int(\cos^3{x}-\cos^5{x})(-1)(-\sin{x})[/tex]
    let u=cos(x), du=-sin(x), so the integral becomes
    [tex]\int(u^3-u^5)(-1)du[/tex]
    [tex]=\int(u^5-u^3)du[/tex]
    [tex]=\frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} + C[/tex]
     
  7. Oct 12, 2006 #6
    [tex] \int(1-\cos^{2} x)(\cos^{3} x)\sin x dx [/tex]

    [tex] \int(\cos^{3} x - \cos^{5}x) \sin x dx [/tex]

    [tex] u = \cos x, du = -\sin x [/tex].

    [tex] -\int u^{3} - u^{5} du = -(\frac{\cos^{4}x}{4}-\frac{\cos^{6}x}{6}) [/tex]
    [tex] \int u^{3} - u^{5}du = (\frac{\cos^{4}x}{4}-\frac{\cos^{6}x}{6}) [/tex]
     
  8. Oct 12, 2006 #7
    That's exactly it--
    If I set u=sin(x) I seem to get a different answer that if I set u=cos(x).
    But why?
     
  9. Oct 12, 2006 #8
    They are equivalent because of the constant of integration. The constants are [tex] C_{1}, C_{2} [/tex]
     
  10. Oct 16, 2006 #9
    But if I wanted to evaluate the integral between a and b, wouldn't the constants cancel?
     
  11. Oct 16, 2006 #10
    The integral should be [tex] \frac{\cos^{6}x}{6}-\frac{\cos^{4}x}{4} [/tex]
     
  12. Oct 16, 2006 #11

    HallsofIvy

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    Yes, they would. Did you try it?
    For example, you could have
    [tex]\int_0^{\frac{pi}{2}}sin^4x cos^3xdx= \frac{1}{4}sin^4x- \frac{1}{6}sin^6x+ C\left|_0^{\frac{\pi}{2}}[/tex]
    or
    [tex]\int_0^{\frac{pi}{2}}sin^4x cos^3xdx= \frac{1}{6}cos^6x-\frac{1}{4}cos^4x+ C\left|_0^{\frac{\pi}{2}}[/tex]
    Those are exactly the same. For a slightly harder example try between 0 and [itex]\pi/4[/itex].
     
  13. Oct 16, 2006 #12
    the integral between the limits a and b would be the same, even though the actual functions are not equal to each other. thats the key idea. For example:

    [tex] \int_{0}^{1} \frac{1}{4}\sin^4{x} -\frac{1}{6}\sin^6{x} = \int_{0}^{1} \frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} [/tex] even though [tex] \frac{1}{4}\sin^4{x} -\frac{1}{6}\sin^6{x} [/tex] does not equal [tex] \frac{1}{6}\cos^6{x} -\frac{1}{4}\cos^4{x} [/tex]
     
    Last edited: Oct 16, 2006
  14. Oct 16, 2006 #13
    That is so cool!!
    But why, please?

    And thank you very much for your prolonged patience.
     
  15. Oct 16, 2006 #14
    you could look at the graph, and see that the areas are the same (i.e. area below x-axis and area above x-axis). maybe employ some trig identities. Also look at the graphs of sin(x) and cos(x) together.
     
    Last edited: Oct 16, 2006
  16. Oct 16, 2006 #15

    HallsofIvy

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    Pick specific values for a and b, do the calculation yourself and see why!
     
  17. Oct 16, 2006 #16
    Ohhhhhh.... :understanding dawns:
    I love math.:!!)
    Hail PhysicsForums, bastion of nerdiness!

    And yes, HallsofIvy, I did evaluate the integral between 0 and pi/2 (took the easy way out because I just wanted to see what was going on), and I saw the coolness take place before my very eyes, but I still understood it not. Sigh.

    Pictures are great. Graphs are pictures. ==> Graphs are great.

    Thank you, people!
     
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