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Do I integrate it using Trigonometry substitution?

  1. Oct 28, 2009 #1
    Dear all,

    I would like to evaluate [tex]\int\frac{1}{1-2sin\left(x\right)}dx[/tex]

    Firstly, I make use of the Weierstrass substitution method by letting:

    [tex]u=tan\left(\frac{x}{2}\right)[/tex]

    and therefore

    [tex]sin\left(x\right)=\frac{2u}{1+u^{2}}[/tex]

    and

    [tex]dx=\frac{2}{1+u^{2}}du[/tex]

    Eventually I can rewrite my integral as:

    [tex]2\int\frac{1}{u^{2}-4u+1}du[/tex]

    Since the denominator of the integrand cannot be factorised, I try using trigonometry substitution by first rewriting the integral as

    [tex]2\int\frac{1}{\left(u-2\right)^{2}-\sqrt{3}^{2}}du[/tex]

    And then by letting

    [tex]u=2+\sqrt{3}sec\left(\theta\right)[/tex]
    [tex]du=\sqrt{3}sec\left(\theta\right)tan\left(\theta\right)d\theta[/tex]

    After that I substitute [tex]u[/tex] and [tex]du[/tex] into the equation and I obtain

    [tex]\frac{2}{\sqrt{3}}\int\frac{sec\left(\theta\right)}{tan\left(\theta\right)}d\theta=\frac{2}{\sqrt{3}}\int csc\left(\theta\right)d\theta[/tex]

    And that will give me:

    [tex]\frac{2}{\sqrt{3}}ln\left|csc(\theta)+cot(\theta)\right|+C[/tex]

    But now if I replace [tex]\theta[/tex] with [tex]sec^{-1}\frac{u-2}{\sqrt{3}}[/tex] I am unable to obtain the answer which is given as:

    [tex]\frac{1}{\sqrt{3}}ln\left|\frac{tan\left(\frac{x}{2}\right)-2-\sqrt{3}}{tan\left(\frac{x}{2}\right)-2+\sqrt{3}}\right|+C[/tex]

    From the answer given, it seems to me that I should use partial fraction instead of trigonometry substitution. However I am pretty sure if I use trigonometry substitution, I should get the answer.

    May I know what am I doing wrong here?

    Thank you very much in advance.
     
  2. jcsd
  3. Oct 28, 2009 #2

    tiny-tim

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    Hi caseyjay! :smile:
    Nothing …

    (cosecθ + cotθ)2 = (secθ + 1)2/tan2θ = (secθ + 1)/(secθ - 1) :wink:
     
  4. Oct 28, 2009 #3
    Hi Tiny-Tim,

    May I know why do you square cosecθ + cotθ? Did I miss out any "squares" in my working? I do not know from where and why did you square cosecθ + cotθ.
     
  5. Oct 29, 2009 #4
    The two outside of the ln can become a power by log rules.
     
  6. Oct 29, 2009 #5

    tiny-tim

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    yeah …

    as l'Hôpital :smile: says, 2lnx = lnx2 :wink:
     
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