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Integral (cos x)^2?

  1. Jan 2, 2006 #1
    integral (cos x)^2??
    the answer is 1/2x + 1/4 sin2x
    pls help...thanx.....
     
  2. jcsd
  3. Jan 2, 2006 #2

    VietDao29

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    Homework Helper

    To integrate something like sin2x dx, cos2x dx, we use Power-reduction formulas, ie:
    [tex]\cos ^ 2 x = \frac{1 + \cos(2x)}{2} \quad \mbox{and} \quad \sin ^ 2 x = \frac{1 - \cos(2x)}{2}[/tex]
    Can you go from here?
     
  4. Jan 2, 2006 #3

    LeonhardEuler

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    Gold Member

    You can do this using trig identities:
    [tex]\cos{2\theta}=\cos^2{\theta}-\sin^2{\theta} = 2\cos^2{\theta}-1[/tex]
    Now solve for [itex]\cos^2{\theta}[/itex]:
    [tex]\cos^2{\theta}=\frac{1+\cos{2\theta}}{2}[/tex]
    So
    [tex]\int\cos^2{x}dx=\frac{1}{2}\int 1+\cos{2x}dx[/tex]
     
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