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Do I integrate it using Trigonometry substitution?

by caseyjay
Tags: integrate, substitution, trigonometry
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caseyjay
#1
Oct28-09, 05:15 AM
P: 19
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.
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tiny-tim
#2
Oct28-09, 02:54 PM
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Hi caseyjay!
Quote Quote by caseyjay View Post
May I know what am I doing wrong here?
Nothing …

(cosecθ + cotθ)2 = (secθ + 1)2/tan2θ = (secθ + 1)/(secθ - 1)
caseyjay
#3
Oct28-09, 11:19 PM
P: 19
Quote Quote by tiny-tim View Post
Hi caseyjay!

Nothing …

(cosecθ + cotθ)2 = (secθ + 1)2/tan2θ = (secθ + 1)/(secθ - 1)
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.

l'Hôpital
#4
Oct29-09, 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.
tiny-tim
#5
Oct29-09, 03:09 AM
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tiny-tim's Avatar
P: 26,160
yeah …
as l'Hôpital says, 2lnx = lnx2


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