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The most direct solution for a definite integral

  1. Jan 1, 2014 #1
    Hi, I'm wondering if I have the most direct solution for this integral or if there is a more efficient way of solving this. I haven't seen a double substitution deployed on one of these problems yet, so I thought perhaps this was not necessary.

    1. The problem statement, all variables and given/known data

    Using the substitution t = tan x, or otherwise, evaluate:
    [itex]\int^{\frac{\pi}{3}}_{0} \frac{1}{9 - 8sin^2x} dx[/itex]


    2. Relevant equations



    3. The attempt at a solution

    [itex]\int^{\frac{\pi}{3}}_{0} \frac{1}{9 - 8sin^2x} dx[/itex]

    [itex]= \int^{\frac{\pi}{3}}_{0} \frac{1}{1 + 8cos^2x} dx[/itex]

    [itex]= \int^{\frac{\pi}{3}}_{0} \frac{1}{4cos2x + 5} dx[/itex]

    [itex]let\ t = tanx[/itex]

    [itex]cos2x = \frac{1 - t^2}{1 + t^2}[/itex] (trig identity)

    [itex]so \int^{\frac{\pi}{3}}_{0} \frac{1}{4cos2x + 5} dx[/itex] [itex]= \int^{\frac{\pi}{3}}_{0} \frac{1}{4(\frac{1 - t^2}{1 + t^2}) + 5} dx[/itex]

    [itex]= \int^{\frac{\pi}{3}}_{0} \frac{1 + t^2}{t^2 +9} dx[/itex]

    [itex]= \int^{\frac{\pi}{3}}_{0} \frac{sec^2x}{tan^2x +9} dx[/itex]

    [itex] t = tanx → dt = sec^2x\ dx[/itex]

    [itex]so \int^{\frac{\pi}{3}}_{0} \frac{sec^2x}{tan^2x + 9} dx[/itex] [itex]= \int^{\sqrt 3}_{0} \frac{1}{t^2 +9} dt[/itex]

    [itex]let\ t = 3tanθ[/itex] [itex]so\ dt = 3sec^2θ\ dθ[/itex]

    [itex]so \int^{\sqrt 3}_{0} \frac{1}{t^2 +9} dt[/itex] [itex]= \int^{atan\frac{1}{\sqrt 3}}_{0} \frac{3sec^2θ}{9tan^2θ +9} dθ[/itex]

    [itex]= \int^{π/6}_{0} \frac{3sec^2θ}{9sec^2θ} dθ[/itex]

    [itex]= \int^{π/6}_{0} \frac{3}{9} dθ[/itex]

    [itex]= \frac{π}{18}[/itex]
     
  2. jcsd
  3. Jan 1, 2014 #2
    I haven't checked your entire work ,but what was quite evident is as follows-

    You could have avoided the second substitution by properly performing the first substitution and changing the limits .When you changed the variable from 'x' to 't' ,you should also change the limits .When the variable is 'x' ,the limits correspond to 'x' .When the variable is 't' ,the limits should correspond to 't' .

    Since t=tanx ,dt = sec2xdx or dx = dt/1+t2

    When x=0 →t=0 and when x=π/3 → t = √3

    The item in red should have been written as [itex]= \int^{\sqrt{3}}_{0} \frac{1}{t^2 +9} dt[/itex]

    You will find this integral quite easy to handle without making further substitutions.

    Hope this helps
     
    Last edited: Jan 1, 2014
  4. Jan 1, 2014 #3

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You result is correct, but you overcomplicated the integration.

    very good... Substitute [itex]\cos^2x=\frac{1}{1+\tan^2x}[/itex]

    and then use the substitution

    [itex]let\ t = tanx[/itex] x=arctan(t), [itex]dx=\frac{1}{1+t^2}dt[/itex]. You directly arrive at

    You can write the integral in form [itex]= \frac{1}{9}\int^{\sqrt 3}_{0} \frac{1}{\left(\frac{t}{3}\right)^2 +1} dt[/itex] which is a basic integral - do you recognize it?
    ∫(1/(1+x2)dx=arctan(x)

    Edit: Oh, Tania beat me:smile:

    ehild
     
    Last edited: Jan 1, 2014
  5. Jan 1, 2014 #4
    Thanks for your helpful comments, I can see I need to familiarize myself more thoroughly with some of these basic integrals.
     
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