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Integration question

  1. Jan 28, 2005 #1
    My text book did the following without giving an explanination. If anyone can fill me in it would be much appreciated.

    the integration of (1+2*sin(x)^2)^2 from pi+pi/6 to pi+(3*pi)/6 is equal to
    the integration of (1-2*sin(x)^2)^2 from pi/6 to (3*pi)/6.

    thanks
     
  2. jcsd
  3. Jan 28, 2005 #2

    dextercioby

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    Is that what u have to show:
    [tex]\int_{\pi+\frac{\pi}{6}}^{\pi+\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx =
    \int_{\frac{\pi}{6}}^{\frac{3\pi}{6}} (1+2\sin^{2}x)^{2} dx [/tex]

    ???If,so make a substitution...An obvious one.

    Daniel.
     
  4. Jan 28, 2005 #3
    the second one is 1-sinx^2 not 1+sin^2...
     
  5. Jan 28, 2005 #4
    Hey Daniel could you please show me how that'd be solved? im just curious. I have an idea but I dont see it.

    Thanks.
     
  6. Jan 28, 2005 #5

    dextercioby

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    Try [tex] x=u-\pi [/tex]

    Daniel.
     
  7. Jan 28, 2005 #6
    Im only in calc2 right now and even that doesn't help me but I guess I will learn more as I finish the rest of my calc courses.

    with x=u-pi x'=1 (if pi is a constant)

    sorry i still dont see it lol
     
  8. Jan 29, 2005 #7
    me neither...
     
  9. Jan 29, 2005 #8

    dextercioby

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    [tex] dx=du [/tex]


    Of course,but to wrote everything in terms of "u",u need to make the substitution both in the argument of [itex] \sin [/itex] and in the limits of integration.

    Daniel.
     
  10. Jan 29, 2005 #9
    Do you have any trig identity in mind?
    Also, are we on the same page? Are we all talking about how (1+2*sin(x)^2)^2 from pi+pi/6 to pi+(3*pi)/6 is equal to (1-2*sin(x)^2)^2 from pi/6 to (3*pi)/6.
     
  11. Feb 1, 2005 #10
    Looking at the two equations, there is a lot of reason to be suspicious of the integrals being the same.
    [tex]1-2sin^2u=cos2u; 1+2sin^2u=2-cos2u[/tex] (This might make integration a little easier.)

    [tex]\int_{\pi/6}^{\pi/2}cos^2(2u)du =\pi/6-\sqrt3/16=.415[/tex]

    While the other integral is [tex]3/2\pi-15/16\sqrt3 =6.336[/tex]

    P.S. It should have been noted that sin(u-Pi) = -sin(u), but for the square, sin^2(u-Pi) = sin^2(u). Thus, as dextercioby suggests, the substitution z=u-Pi, reduces the integral between 7Pi/6 and 3Pi/2 to :

    [tex]\int_\frac{\pi}{6}^\frac{\pi}{2}(1+2\sin^2z)^2dz[/tex]

    This has a different sign than the other integral.
     
    Last edited: Feb 1, 2005
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