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What is the integral of 1/((1+cosx)^2)?

  1. Feb 2, 2013 #1
    Use integration to find the area of the region bounded by the given polar curves

    r = [itex]\frac{3}{(1+cos \theta )}[/itex]

    and

    [itex]\theta[/itex] = [itex]\frac{\pi}{2}[/itex]



    A = [itex]\frac{1}{2}[/itex] [itex]\int[/itex]f([itex]\theta[/itex])[itex]^{2}[/itex]d[itex]\theta[/itex]



    My attempt:

    (from -[itex]\frac{-\pi}{2}[/itex] to [itex]\frac{\pi}{2}[/itex] )

    A = [itex]\frac{1}{2}[/itex][itex]\int[/itex] ([itex]\frac{3}{(1+cos \theta )}[/itex])[itex]^{2}[/itex] d[itex]\theta[/itex]

    A = [itex]\frac{9}{2}[/itex][itex]\int[/itex] ([itex]\frac{1}{(1+cos \theta )}[/itex])[itex]^{2}[/itex] d[itex]\theta[/itex]

    A = [itex]\frac{9}{2}[/itex][itex]\int[/itex] [itex]\frac{1}{(1+cos \theta )^{2}}[/itex])[itex][/itex] d[itex]\theta[/itex]

    →[itex](1+cos \theta )^{2} [/itex] = [itex] cos^{2}\theta + 2cos\theta + 1 [/itex]

    = 1 - sin[itex]^{2}[/itex][itex]\theta[/itex] + 2cos[itex]\theta[/itex] + 1

    = 2 + 2cos [itex]\theta[/itex] - 1/2 + 1/2 cos 2 [itex]\theta[/itex]
    ...?
     
  2. jcsd
  3. Feb 2, 2013 #2

    LCKurtz

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    Perhaps you can use ##1+\cos\theta = 2\cos^2(\frac\theta 2)##.
     
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