Integrate [cosec(30°+x)-cosec(60°+x)] dx in terms of tan x

[Aurelius120]

Homework Statement

Evaluate the integral:
$$\int\frac{\left(1-\frac{1}{\sqrt 3}\right)(\cos x-\sin x)}{1+\frac{2}{\sqrt 3}(\sin 2x)}dx$$

Relevant Equations

$$\sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$$
$$\int \csc(x) dx=\ln(\csc(x)-\cot(x))$$

I proceeded as follows
$$\int\frac{2(\sqrt3-1)(cosx-sinx)}{2(\sqrt3+2sin2x)}dx$$
$$\int\frac{(cos(\pi/6)-sin(\pi/6))(cosx-sinx)}{(sin(\pi/3)+sin2x)}dx$$
$$\frac{1}{2}\int\frac{cos(\pi/6-x)-sin(\pi/6+x)}{sin(\pi/6+x)cos(\pi/6-x)}dx$$
$$\frac{1}{2}\int cosec(\pi/6+x)-sec(\pi/6-x)dx$$
Which leads to the integral in the question:

$$\frac{1}{2}\int\left[\csc\left(\frac{\pi}{6}+x\right)-\csc\left(\frac{\pi}{3}+x\right)\right]dx$$
Now the correct answer is:$$\frac{1}{2}\log\left|\frac{tan(\pi/12+x/2)}{tan(\pi/6+x/2)}\right|$$

How to reach the correct answer from the obtained integral?

 

[anuttarasammyak]

From your effort it seems that
$$\int csc (x+\alpha)dx =\log |\tan(\frac{x+\alpha}{2})|+C$$
or by transformation of integral variable
$$\int csc\ x dx =\log |\tan(\frac{x}{2})|+C$$
Have you investigated it by differentiating the both sides?

 

[Aurelius120]

From your effort it seems that
$$\int csc (x+\alpha)dx =\log |\tan(\frac{x+\alpha}{2})|+C$$
or by transformation of integral variable
$$\int csc\ x dx =\log |\tan(\frac{x}{2})|+C$$
Have you investigated it by differentiating the both sides?

Yes I tried by I cannot reach why

$$\frac{1}{2}\int\left[\csc\left(\frac{\pi}{6}+x\right)-\csc\left(\frac{\pi}{3}+x\right)\right]dx$$

$$=\frac{1}{2}\ln\left[\frac{\csc(\pi/6+x)-cot(\pi/6+x)}{\csc(\pi/3+x)-\cot(\pi/3+x)}\right]+C$$ is equal to the given answer

 

[anuttarasammyak]

With a=x/2,$$\frac{1}{\sin x}-\frac{\cos x}{\sin x}=\frac{1- \cos 2a}{\sin 2a}=\frac{1-\cos^2 a+\sin^2 a}{2\sin a \cos a}=\tan a$$