finding definite integral involving inverse of cot function

Question

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

Answer 1

Assuming that the solution requires numerical integration.

Considering $$f(x)=\cot^{-1}\bigg(\frac{1}{2\cos x-1}\bigg)+\cot^{-1}\bigg(\cos x - \frac{1}{2}\bigg)$$ let us compute $$I_1(k)=\int_{-\frac \pi 3}^{\frac \pi 3-10^{-k}}f(x)\,dx\qquad I_2(k)=\int_{\frac \pi 3+10^{-k}}^\pi f(x)\,dx\qquad I(k)=I_1(k)+I_2(k)$$ As a function of $k$, the following results would be obtained $$\left( \begin{array}{cccc} k & I_1(k) & I_2(k) & I(k) \\ 1 & 3.633155522 & -3.703785769 & -0.07063024720 \\ 2 & 3.778692433 & -3.849483982 & -0.07079154961 \\ 3 & 3.792872380 & -3.863664096 & -0.07079171605 \\ 4 & 3.794286526 & -3.865078242 & -0.07079171622 \\ 5 & 3.794427902 & -3.865219618 & -0.07079171622 \\ 6 & 3.794442039 & -3.865233755 & -0.07079171622 \\ 7 & 3.794443453 & -3.865235169 & -0.07079171622 \end{array} \right)$$

Answer 2

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.