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Do I integrate it using Trigonometry substitution? |
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| Oct28-09, 05:15 AM | #1 |
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Do I integrate it using Trigonometry substitution?
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\rig ht)}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. |
| Oct28-09, 02:54 PM | #2 |
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Hi caseyjay!
![]() (cosecθ + cotθ)2 = (secθ + 1)2/tan2θ = (secθ + 1)/(secθ - 1)
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| Oct28-09, 11:19 PM | #3 |
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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. |
| Oct29-09, 12:29 AM | #4 |
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Do I integrate it using Trigonometry substitution?
The two outside of the ln can become a power by log rules.
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| Oct29-09, 03:09 AM | #5 |
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yeah …
as l'Hôpital |
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