finding definite integral involving inverse of cot function finding $\displaystyle \int^{\pi}_{-\frac{\pi}{3}}\bigg[\cot^{-1}\bigg(\frac{1}{2\cos x-1}\bigg)+\cot^{-1}\bigg(\cos x - \frac{1}{2}\bigg)\bigg]dx$
Attempt:
\begin{align}
& \int^{\frac{\pi}{3}}_{-\frac{\pi}{3}}\bigg[\cot^{-1}\bigg(\frac{1}{2\cos x-1}\bigg)+\cot^{-1}\bigg(\cos x - \frac{1}{2}\bigg)\bigg] \, dx \\[10pt]
+ {} & \int^\pi_{\frac{\pi}{3}}\bigg[\cot^{-1}\bigg(\frac{1}{2\cos x-1}\bigg)+\cot^{-1}\bigg(\cos x - \frac{1}{2}\bigg)\bigg] \, dx
\end{align}
as we break because $\displaystyle \cos x- \frac 1 2 =0$ at $\displaystyle x= \frac \pi 3$
wan,t be able to go further, could some help me
 A: I see some tricks in here. Ideas, as yet unfinished, I'm out of time now but will try to come back later.
Consider $cot^{-1}(b) = \theta \leftrightarrow cot(\theta) = b$
and then $tan(\theta ) = 1/b$ so $\theta = tan^{-1}(1/b)$
Also $(cos(x) - 1/2 ) = (2 cos(x) -1)/2$
$$f(x) = cot^{-1}(\frac{1}{2cos(x) -1}) + cot^{-1}(\frac{2 cos(x) -1}{2})$$
If it weren't for that 2 in the second expression denominator this would be really interesting with a possible simplification (check the original problem again?)
$$f(x) = tan^{-1}(2cos(x) -1) + tan^{-1}(\frac{2}{2 cos(x) -1})$$
This is reminiscent of the tangent addition formula:
If$\ tan(\theta_1) = a $ and $tan(\theta_2) = b$ then 
$tan(\theta_1 + \theta_2)= \frac{a+b}{ab}$  
Let $ \theta_1 = tan^{-1}(2cos(x) -1)$ and $\theta_2 = tan^{-1}(\frac{2}{2 cos(x) -1})$
So a = $(2cos(x) -1)$, $b = \frac{2}{2 cos(x) -1}$, and $ab = 2$
$$tan(\theta_1 + \theta_2) = 1/2((2cos(x) -1) + \frac{2}{2 cos(x) -1}) $$
$$f(x) = \theta_1 + \theta_2 = tan^{-1}( 1/2((2cos(x) -1) + \frac{2}{2 cos(x) -1})) $$
This should be able to go somewhere from here -- will try more later.
