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.

HINT: set $$t=\tan(x/2)$$, $$\sin(x)=\frac{2t}{1+t^2}$$, $$\cos(t)=\frac{1-t^2}{1+t^2}$$ and $$dx=\frac{2dt}{1+t^2}$$
