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Find the integral of annoying

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\int sin^{2}x dx[/itex]

    2. Relevant equations



    3. The attempt at a solution

    I don't see why. Clearly it is designed for me to shoot myself before finishing.

    u = (sinx)^2
    du = 2sinxcosx dx, magically equal to sin2x dx
    v = x
    dv = dx

    (-xcos2x/2) + (sin2x/4)

    =

    (-2xcos2x+sin2x)/(4) + C = wrong
     
  2. jcsd
  3. Sep 26, 2011 #2

    SammyS

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    Since you just reviewed the double angle formula for sin, what's the double angle formula for cos?

    cos(2x) = ? = ? = ?

    Yes, there are three forms. One will work quite nicely here.
     
  4. Sep 26, 2011 #3
    Firstly, let me apologize for my attitude. Secondly, thanks for the tip, I'll work it again and post back.
     
  5. Sep 26, 2011 #4

    SammyS

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    At times, getting frustrated can lead to good things.
     
  6. Sep 26, 2011 #5
    [itex]\int sin^{2}x dx[/itex]

    [itex]u = sin^{2}x[/itex]
    [itex]du = sin2x dx[/itex]
    [itex]v = x[/itex]
    [itex]dv = dx[/itex]

    [itex]xsin^{2}x - \int xsin2x dx[/itex]

    [itex]u = x[/itex]
    [itex]du = dx[/itex]
    [itex]v = -\frac{cos2x}{2}[/itex]
    [itex]dv = sin2x dx[/itex]

    [itex]xsin^{2}x - [-\frac{xcos2x}{2} - \int -\frac{cos2x}{2} dx ][/itex]
    [itex]xsin^{2}x - [-\frac{xcos2x}{2} + \frac{sin2x}{4}][/itex]

    [itex]xsin^{2}x + \frac{xcos2x}{2} - \frac{sin2x}{4}[/itex]

    [itex] \frac{2xsin^{2}x + xcos2x}{2} - \frac{sin2x}{4}[/itex]


    [itex] \frac{2xsin^{2}x + x(cos^{2}x - sin^{2}x)}{2} - \frac{sin2x}{4}[/itex]


    [itex] \frac{2xsin^{2}x + xcos^{2}x - xsin^{2}x}{2} - \frac{sin2x}{4}[/itex]


    [itex] \frac{xsin^{2}x + xcos^{2}x}{2} - \frac{sin2x}{4}[/itex]


    [itex] \frac{x(sin^{2}x + cos^{2}x)}{2} - \frac{sin2x}{4}[/itex]


    [itex] \frac{x}{2} - \frac{sin2x}{4}[/itex]
     
  7. Sep 26, 2011 #6

    Mark44

    Staff: Mentor

    There's a much shorter way of doing this, following Sammy's hint.

    sin2(x) = 1/2 * (1 - cos(2x))

    This will give you an integral that you can do without having to resort to integration by parts.
     
  8. Sep 26, 2011 #7

    SammyS

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    Mark44's expression comes from solving the following for sin2(x) .

    cos(2x) = 1 - 2 sin2(x) .

    It's also true that

    cos(2x) = cos2(x) - sin2(x)

    cos(2x) = 2 cos2(x) - 1
     
  9. Sep 26, 2011 #8
    th_smiley-bangheadonwall.gif

    Welcome to the wonderful world of integrals! There are times when I want to kick small, defenseless animals because of this stuff. I usually listen to some thrash metal while studying math. I don't know why, but it helps!

    Try to remember as many identities as you possibly can. It makes life a lot easier.
     
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