# Do I integrate it using Trigonometry substitution?

## Main Question or Discussion Point

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
Homework Helper
Hi caseyjay! May I know what am I doing wrong here?
Nothing …

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

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

tiny-tim
as l'Hôpital says, 2lnx = lnx2 