Do I integrate it using Trigonometry substitution?

[caseyjay]

Dear all,

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

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

u=tan\left(\frac{x}{2}\right)

and therefore

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

and

dx=\frac{2}{1+u^{2}}du

Eventually I can rewrite my integral as:

2\int\frac{1}{u^{2}-4u+1}du

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

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

And then by letting

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

After that I substitute u and du into the equation and I obtain

\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

And that will give me:

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

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

\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

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.

 

[tiny-tim]

Hi caseyjay!

May I know what am I doing wrong here?

Nothing …

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

 

[caseyjay]

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]

The two outside of the ln can become a power by log rules.

 

[tiny-tim]

yeah …

as l'Hôpital says, 2lnx = lnx² ​