Do I integrate it using Trigonometry substitution?by caseyjay Tags: integrate, substitution, trigonometry 

#1
Oct2809, 05:15 AM

P: 19

Dear all,
I would like to evaluate [tex]\int\frac{1}{12sin\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(u2\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\leftcsc(\theta)+cot(\theta)\right+C[/tex] But now if I replace [tex]\theta[/tex] with [tex]sec^{1}\frac{u2}{\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
Oct2809, 02:54 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

Hi caseyjay!
(cosecθ + cotθ)^{2} = (secθ + 1)^{2}/tan^{2}θ = (secθ + 1)/(secθ  1) 



#3
Oct2809, 11:19 PM

P: 19

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. 



#4
Oct2909, 12:29 AM

P: 256

Do I integrate it using Trigonometry substitution?
The two outside of the ln can become a power by log rules.



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