Evaluate $\int_0^{2\pi}\frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\ (A^2+B^2) <<1$ I need to evaluate the definite integral $\int_0^{2\pi}\frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \ \text{ for }\  A<<1, B<<1, (A^2+B^2) <<1,$

For unresricted (but real) A&B, Wolfram Alpha provides the following indefinite general solution:-
$$\int \frac{\cos(\theta)}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \left( \frac{1}{A^2+B^2}\right) \left[ \frac{2B}{K} \tanh^{-1} \left( \frac{A-(B-1)\tan\left(\frac{\theta}{2}\right)}{K}\right) + F(\theta)\right]$$
where $F(\theta) = A \ln(1 + A\sin\theta+B\cos\theta)+B\theta$,
and $K = \sqrt{A^2 + B^2 -1}$, therefore K is complex for the range of A,B  I am interested in.

In a previous question seeking a solution for the similar, but slightly simpler, definite integral (with numerator $1$ rather than $\cos\theta$) user Dr. MV found a solution given by:
$$\int_0^{2\pi}\frac{1}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{2\pi}{\sqrt{1-A^2-B^2}}\text{   } (\text { for} \sqrt{A^2+B^2}<1) $$.

My question: Can similar solutions be found for these two definite integrals $$\int_0^{2\pi}\frac{\cos\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \tag 1$$
and
$$\int_0^{2\pi}\frac{\sin\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta \tag 2$$?

EDIT
I have taken the solution proposed by user Chappers.
By simultaneous equations in A,B,I,J it turns out that
$$I=\int_0^{2\pi}\frac{\cos\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{B}{A^2+B^2} 2\pi (1-\frac{1}{\sqrt{1-A^2-B^2}})$$ and 
$$J=\int_0^{2\pi}\frac{\sin\theta}{1 + A\sin(\theta) + B\cos(\theta)}\, \mathrm{d}\theta = \frac{A}{A^2+B^2}  2\pi (1-\frac{1}{\sqrt{1-A^2-B^2}})$$. 
These were confirmed in a numerical model.

There's a trick to this: let the cosine one be $I$, the sine one $J$. We add to get something easy to integrate:
$$ BI+AJ = \int_0^{2\pi} \frac{A\sin{\theta}+B\cos{\theta}}{1+A\sin{\theta}+B\cos{\theta}} \, d\theta = \int_0^{2\pi} \left( 1 - \frac{1}{1+A\sin{\theta}+B\cos{\theta}} \right) d\theta, $$
which you can do using the previous answer to get 
$$ BI+AJ = 2\pi - \frac{2\pi}{\sqrt{1-A^2-B^2}}. $$
The harder bit is to come up with a second equation. Remembering our derivatives, we try
$$ AI-BJ = \int_0^{2\pi} \frac{A\cos{\theta}-B\sin{\theta}}{1+A\sin{\theta}+B\cos{\theta}} \, d\theta, $$
which has antiderivative $ \log{(1+A\sin{\theta}+b\cos{\theta})} $, continuous if $A^2+B^2<1$. But this has the same value at both of the endpoints, so the integral is zero, and $AI=BJ$. Now you can solve this simultaneously with the first equation to find the values of $I$ and $J$.
