- #1

- 20

- 0

## Main Question or Discussion Point

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\right)}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.

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\right)}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.