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Homework Help: Solving a definite integral without integrating

  1. Apr 12, 2010 #1
    Orthonormal basis [tex]\big(\frac{1}{\sqrt{2}}, cos(2x)\big)[/tex]

    [tex]\pi \int_{-\pi}^{\pi} sin^4(x)dx=\frac{3\pi}{4}[/tex]

    Solve the integral with out integrating.

    [tex]sin^4(x)=\big[\frac{1-cos(2x)}{2}\big]^2=\frac{1}{4} - \frac{2cos(2x)}{4} + \frac{cos^2(2x)}{4}[/tex]

    I know I could rewrite [tex]cos^2(2x)[/tex] but then I have a 4x but for the test basis cos 4x is not included.

    I couldn't answer it so can someone post the step by step answer so I can see why I couldn't do it?
     
    Last edited: Apr 12, 2010
  2. jcsd
  3. Apr 12, 2010 #2

    Mark44

    Staff: Mentor

    Edit: Changed the values of the inner products.
    You need to think of this integral as an inner product. Since your basis is orthonormal, <1/sqrt(2), 1/sqrt(2)> = 1 and <cos(2x), cos(2x)> = 1, while <1/sqrt(2), cos(2x)> = 0.

    Rewrite the integrand as sin^2(x) sin^2(x) = ((1/2)(1 - cos(2x))^2, and expand this, then look at the individual products as described in the previous paragraph.
     
    Last edited: Apr 12, 2010
  4. Apr 12, 2010 #3
    What does that line mean?
     
  5. Apr 12, 2010 #4
    I think you've made a typo - shouldn't your 0s and 1s be the other way round?
     
  6. Apr 12, 2010 #5

    Mark44

    Staff: Mentor

    Good catch! Apparently my fingers got disconnected from my brain. I went back and edited what I wrote.
     
  7. Apr 12, 2010 #6
    An orthonormal basis is a set of mutually orthogonal unit vectors that span the vector space.
     
  8. Apr 12, 2010 #7

    Mark44

    Staff: Mentor

    And the context here is apparently an inner product space with {1/sqrt(2), cos(2x)} as an orthonormal basis and the inner product defined as follows:
    [tex]<f, g> = \int_{-\pi}^{\pi} fg~dx[/tex]

    Dustin, it would have been helpful to include this information as part of the complete problem description or relevant equations.
     
  9. Apr 12, 2010 #8
    [tex]\pi*\big[\frac{1}{2}-\frac{cos(2x)}{2}\big]^2[/tex]

    [tex]\pi*\big[\frac{1}{\sqrt{2}} * \frac{1}{\sqrt{2}}-\frac{cos(2x)}{2}\big]^2[/tex]

    [tex]\pi*\big[\frac{1}{\sqrt{2}}-\frac{1}{2}\big]^2=\pi*[\frac{2-\sqrt{2}}{2\sqrt{2}}]^2[/tex]

    [tex]\pi*\big[\frac{3-\sqrt{2}}{4}\big]\neq\frac{3\pi}{4}[/tex]

    Why didn't this work?
     
    Last edited: Apr 12, 2010
  10. Apr 12, 2010 #9
    We didn't need to prove that [tex]\big(\frac{1}{\sqrt{2}},cos(2x)}\big)[/tex] was an orthonormal basis.

    This is a standard orthonormal basis [tex]\big(\frac{1}{\sqrt{2}},cos(x),cos(2x),...,cos(nx),...sin(x),sin(2x),...,sin(nx)\big)[/tex]
     
  11. Apr 12, 2010 #10

    Mark44

    Staff: Mentor

    For starters, sin2(x) [itex]\neq[/itex] (1/2)(1 + cos(2x))

    Second, is there any connection between any of the lines above and the one below it? Is there any connection between what you have here and the problem you're trying to solve?

    Third, are you trying to convince us that cos(2x) = 1? (see third line) That's not true.

    Fourth, is there any connection between the work above and the integral in your first post?

    Fifth, is there any connection between your orthogonal basis and what you have here?
     
    Last edited: Apr 12, 2010
  12. Apr 12, 2010 #11
    Typo. I know sin^2 is the negative.
     
  13. Apr 12, 2010 #12

    Mark44

    Staff: Mentor

    Nor am I asking you to prove it, just follow SOP for the forum by including information that might be relevant in the problem, such as the fact that the integral represents the inner product.
     
  14. Apr 12, 2010 #13
    I am still at the problem that after taking the coefficients and doing the arithmetic the answer wasn't correct.
     
  15. Apr 12, 2010 #14

    Mark44

    Staff: Mentor

    In post 8 you asked why what you did didn't work. I gave you 5 reasons. If you have any specific questions about any of them, fire away.
     
  16. Apr 12, 2010 #15
    I changed the (+) to (-). I took the coefficients, subtracted, and then squared which should be the correct answer.

    I have done this many of times and I obtained the correct answer but I can't this one to work.
     
  17. Apr 12, 2010 #16

    Mark44

    Staff: Mentor

    Did you not see these?


    You seem to be under the (false) impression that you are supposed to simplify the following expression. That is NOT what this problem is about, and even so, you have a mistake in that work (my third point above).
    [tex]\pi*\big[\frac{1}{2} - \frac{cos(2x)}{2}\big]^2[/tex]
     
  18. Apr 12, 2010 #17
    How should it be worked it if it shouldn't be simplified?
     
  19. Apr 12, 2010 #18

    Mark44

    Staff: Mentor

    What is the problem you are trying to solve? If you don't remember, go back to your first post.
     
  20. Apr 12, 2010 #19
    The integral of sin^4
     
  21. Apr 12, 2010 #20

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You've already correctly said that [itex]\sin^4(x)=\frac{1}{4}-\frac{\cos(2x)}{2}+\frac{\cos^2(2x)}{4}[/itex], so now you have

    [tex]\begin{aligned}\int_{-\pi}^{\pi}\sin^4(x)dx &=\int_{-\pi}^{\pi}\left(\frac{1}{4}-\frac{\cos(2x)}{2}+\frac{\cos^2(2x)}{4}\right)dx \\ &= \int_{-\pi}^{\pi}\frac{1}{4}dx-\int_{-\pi}^{\pi}\frac{\cos(2x)}{2}dx+\int_{-\pi}^{\pi}\frac{\cos^2(2x)}{4}dx\end{aligned}[/tex]

    Now, compare these integrals to the inner products that you know. For example, the first one is [itex]1/2[/itex] times the inner product of [itex]\frac{1}{\sqrt{2}}[/itex] with itself; which gives you [itex]\frac{\pi}{2}[/itex]...how about the second one?...And the third?
     
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