Evaluate $\left[ -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K}) \right]_0^{2\pi}$ where $K = \sqrt{A^2 + B^2 -1}$ and $A^2 + B^2 << 1$. I am having difficulty in evaluating the following integrand: 
$$\left[ -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K})
\right]_0^{2\pi}$$ 
where $K = \sqrt{A^2 + B^2 -1}$ and   $A^2 + B^2 << 1$.
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To  evaluate the indefinite integral $I$ $$I = \int \frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{    for various}\ A,B <<1.$$
Wolfram Alpha provides the following solution:-
$$I= -(2/K) \tanh^{-1} ( \frac{A-(B-1)\tan(\theta/2)}{K}) $$
where $K = \sqrt{A^2 + B^2 -1}$.
In the simple case when $A=B=0$ the definite integral over the range $0...2\pi$ should obviously be given by:-
$$I_{A=B=0}=\int_0^{2\pi}\frac{1}{1 + 0 + 0}\, \mathrm{d}\theta = 2\pi.$$
This specific result has been confirmed by the restricted ( for $A^2+B^2<1$ ) Definite Solution 
$$I=\int_0^{2\pi}\frac{1}{1 + A\cos(\theta) + B\sin(\theta)}\, \mathrm{d}\theta = 2\pi.$$ 
developed by Dr. MV in his answer to my previous question.
However I would like to know where my evaluation of the integrand is going wrong as I will need in future to apply a similar approach to other, more complex integrands for which I dont yet have Definite Solutions.

We might (naively) approach the Wolfram Alpha solution in the following way, noting that when $A=B=0$ so $K=i$, :-
$$I_{A=B=0}= \left[ -(2/i) \tanh^{-1} ( \frac{\tan(\theta/2)}{i}) \right]_0^{\infty}$$
$$= -(2/i) \tanh^{-1} ( \frac{\tan(2\pi/2)}{i}) +(2/i) \tanh^{-1} ( \frac{\tan(0/2)}{i}) $$
$$ = -(2/i) \tanh^{-1} ( \frac{\tan(\pi)}{i}) +(2/i) \tanh^{-1} ( \frac{\tan(0)}{i}) $$
$$ = -(2/i) \tanh^{-1} ( \frac{0}{i}) +(2/i) \tanh^{-1} ( \frac{0}{i}) $$
$$ = 0.$$
but this is clearly incorrect.  I presume the error comes from trying to integrate across the range $0, 2\pi$ where the $\tan$ function has singularities at $\pi/2$ and $3\pi/2$.

So let us try and break the integration into the three continuous ranges $0...\pi/2$ and $\pi/2...3\pi/2$ and $3\pi/2...2\pi$ thus:-
$$I_{A=B=0}=$$
$$ -(2/i) \tanh^{-1} ( \frac{\tan(2\pi/2)}{i}) - -(2/i) \tanh^{-1} ( \frac{\tan(3\pi/4)}{i}) + $$
$$ -(2/i) \tanh^{-1} ( \frac{\tan(3\pi/4)}{i}) - - (2/i) \tanh^{-1} ( \frac{\tan(\pi/4)}{i}) +$$
$$ -(2/i) \tanh^{-1} ( \frac{\tan(\pi/4)}{i}) - - (2/i) \tanh^{-1} ( \frac{\tan(0)}{i}) $$
leading to 
$$I_{A=B=0}=$$
$$ -(2/i) \tanh^{-1} ( \frac{0}{i}) +(2/i) \tanh^{-1} ( \frac{-1}{i}) + $$
$$ -(2/i) \tanh^{-1} ( \frac{-1}{i}) +(2/i) \tanh^{-1} ( \frac{1}{i}) +$$
$$ -(2/i) \tanh^{-1} ( \frac{1}{i}) +(2/i) \tanh^{-1} ( \frac{0}{i}) $$
Now 
$tanh^{-1}\frac{0}{i} = 0$,  $tanh^{-1}\frac{-1}{i} = \frac{i\pi}{4}$,   and $tanh^{-1}\frac{1}{i} = \frac{-i\pi}{4}$
and so
$$I_{A=B=0}=$$
$$ -(2/i) 0 +(2/i) \frac{i\pi}{4} + $$
$$ -(2/i) \frac{i\pi}{4}+(2/i) \frac{-i\pi}{4} +  $$
$$ -(2/i) \frac{-i\pi}{4}+(2/i) 0    $$
then 
$$I_{A=B=0}= +(2/i) \frac{i\pi}{4} -(2/i) \frac{i\pi}{4}+(2/i) \frac{-i\pi}{4} -(2/i) \frac{-i\pi}{4}  $$
$$I_{A=B=0}=  \frac{2\pi}{4} -\frac{2\pi}{4} - \frac{2\pi}{4} + \frac{2\pi}{4}  $$
$$I_{A=B=0}= 0$$.
So we still get the wrong answer.  Clearly the right answer ($2 \pi$) could be obtained if all the minus signs were converted into plus signs, but I cant't find justification for doing that, as yet.
Any suggestions welcomed!
 A: Following comments by user mickep let us look at correcting the partitions used...
So, to avoid operating across the singularity at $\frac{\pi}{2}$, let us break the integration into the two continuous ranges $0...\pi$ and $\pi...2\pi$ thus:-
$$I_{A=B=0}=$$
$$ -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left(2\pi/2\right)}{i}\right) - -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left(\pi/2\right)}{i}\right) + $$
$$ -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left(\pi/2\right)}{i}\right) - - \left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{\tan\left(0\right)}{i}\right) +$$
leading to 
$$I_{A=B=0}=$$
$$ -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{0}{i}\right) +\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{-\infty}{i}\right)  -\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{+\infty}{i}\right) +\left(\frac{2}{i}\right) \tanh^{-1} \left( \frac{0}{i}\right) $$
Now 
$tanh^{-1}\left(\frac{0}{i}\right) = 0$,  $tanh^{-1}\left(\frac{-\infty}{i}\right) = \frac{i\pi}{2} $,   and $tanh^{-1}\left(\frac{\infty}{i}\right) = \frac{-i\pi}{2} $
and so
$$I_{A=B=0} = 0 +\left(\frac{2}{i}\right) \frac{i\pi}{2} -\left(\frac{2}{i}\right) \frac{-i\pi}{2} + 0 $$
$$I_{A=B=0} = \pi --\pi $$
$$I_{A=B=0} = 2\pi . $$
which is the expected answer.
