- #1
caseyjay
- 20
- 0
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.
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.
