Integrating $I=\int^{\pi}_0\frac{x}{1-\cos{\beta}\sin{x}}dx$ without Weierstrass Substitution. I'm having trouble integrating the below without using the Weierstrass Substitution.$$I=\int^{\pi}_0\frac{x}{1-\cos{\beta}\sin{x}}dx$$
Links like: Finding $\int \frac{dx}{a+b \cos x}$ without Weierstrass substitution. end up suggesting the same substitution, so my question isn't a duplicate as the answers in those links are unsatisfactory.)

Using definite integral properties, I simplified it to:
$$\frac{2I}{\pi}=\int^{\pi}_0\frac{1}{1-\cos{\beta}\sin{x}}dx$$
How can I solve this without using the W-Sub?  $$t=\tan{\frac{x}{2}}$$
 A: By symmetry of $\displaystyle \sin{x}$ around $\displaystyle \frac{\pi}{2}$, the domain of integration can be halved, with the integral doubled to compensate.
$\displaystyle I = \pi \int_0^{\frac{\pi}{2}} \frac{1}{1-\cos{\beta}\sin{x}}\,\text{d}x = \pi \int_0^{\frac{\pi}{2}} \frac{1+\cos{\beta}\sin{x}}{1-\cos^2{\beta}\sin^2{x}}\,\text{d}x$
The integral splits into two components, which is easily dealt with termwise.
$\displaystyle I_1 = \pi \int_0^\frac{\pi}{2} \frac{1}{1-\cos^2{\beta}\sin^2{x}}\,\text{d}x = \pi \int_0^\frac{\pi}{2} \frac{\sec^2{x}\, \text{d}x}{1+\sin^2{\beta}\tan^2{x}}$
After the substitution $\displaystyle t=\tan{x}$, the result is $\displaystyle \frac{\pi^2}{2\sin{\beta}}$
$\displaystyle I_2 = \pi \int_0^\frac{\pi}{2} \frac{\cos{\beta}\sin{x}}{1-\cos^2{\beta}\sin^2{x}}\,\text{d}x = \pi \int_0^\frac{\pi}{2} \frac{\cos{\beta}\sin{x}}{\sin^2{\beta} + \cos^2{\beta}\cos^2{x}}\,\text{d}x$
After the substitution $\displaystyle c = \cos{x}$, the result is $\displaystyle \pi \frac{\tan^{-1}{(\cot{\beta})}}{\sin{\beta}}$
Full simplification of the integral depends entirely on the location of the value of $\displaystyle \beta$ on the unit circle. I will leave the rest to you.
A: Let $I$ be the integral given by 
$$I=\int_0^\pi \frac{1}{1-\cos(\beta)\sin(x)}\,dx \tag 1$$
where $\beta \ne 2 \ell \pi$ for integer $\ell$.
Enforcing the substitution $x\to x+\pi/2$ and exploiting symmetry reveals
$$\begin{align}
I&=2\int_0^{\pi/2}\frac{1}{1-\cos(\beta)\cos(x)}\,dx\\\\
&=2\int_0^{\pi/2}\frac{1}{1-\cos(\beta)(\cos^2(x/2)-\sin^2(x/2))}\,dx\\\\
&=\int_0^{\pi/2}\frac{1}{\sin^2(\beta/2)\cos^2(x/2)+\cos^2(\beta/2)\sin^2(x/2)}\,dx\\\\
&=\int_0^{\pi/4}\frac{2\csc^2(\beta/2)\sec^2(x)}{1+\cot^2(\beta/2)\tan^2(x)}\,dx\\\\
&=\int_{x=0}^{x=\pi/4} \frac{4\csc(\beta)}{1+\cot^2(\beta/2)\tan^2(x)}d(\cot(\beta/2)\tan(x))\\\\
&=4\csc(\beta)\arctan(\cot(\beta/2))\\\\
&=2(\pi-\beta)\csc(\beta)
\end{align}$$
