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Integration by substitution of sqrt cos theta.sin cube theta

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data

    To show that [tex]\int_{0}^ \frac{\pi}{2}\sqrt{cos\theta}sin^3(\theta) d\theta[/tex] = 8/21

    3. The attempt at a solution
    The above expression was simplified as
    [tex]\int_{0}^ \frac{\pi}{2}\sqrt{cos\theta}sin^2(\theta) sin(\theta) d\theta[/tex]
    [tex]\int_{0}^ \frac{\pi}{2}\sqrt{cos\theta}(1-cos^2(\theta)) sin\theta d\theta[/tex]
    I have tried using integration by substitution method.
    Let [tex]cos\theta = t^2[/tex]
    or,[tex] sin\theta d\theta = 2tdt[/tex]

    also changing the limits, when [tex]\theta = 0[/tex] , t becomes 1
    and when [tex]\theta = \frac{\pi}{2} [/tex], t becomes 0

    therefore the expression will look like this.
    [tex]\int_{1}^ 0 t.(t^4-1)2t.dt[/tex]

    Am I going in to the right direction or should I use any other method like integration by parts.
  2. jcsd
  3. Dec 30, 2011 #2
    Let t = cos θ. See if you can figure it out from there.
  4. Dec 30, 2011 #3


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    Distribute [itex]\sqrt{\cos(\theta)}[/itex] through (1 - cos2(θ)) .

    and remember that [itex]\displaystyle \sqrt{x}=x^\frac{1}{2}\,.[/itex]

    ... and use Ivan92's suggested substitution.
  5. Dec 31, 2011 #4
    Why don't you just integrate what you have and find out?
  6. Dec 31, 2011 #5
    Thanks to everyone who responded, especially Harrisonized, you were right.
    Guess I was heading in to the right path.
    [tex]\int_{1}^ 0 t(t^4-1)2t dt[/tex]

    --->[tex] 2.\frac{t^7}{7}-2.\frac{t^3}{3}[/tex] when integrated.

    --->[tex] 0-\frac{6-14}{21}[/tex] when variable substituted with the limits.
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