Evaluate the integral $\int_{-1}^1 \frac{1}{(2-x)\sqrt{1-x^2}}\,dx$ I started with the substitution $x=\sin(\theta)$ with $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$
And I got the integral $$\int_{-\pi/2}^{\pi/2} \frac{1}{2-\sin(\theta)}\,d\theta$$
which I am again not able to evaluate.
Even integration by parts is not working.

Here is a slightly different approach.
Let 
$$I = \int_{-1}^1 \frac{dx}{(2 - x) \sqrt{1 - x^2}} \qquad (1)$$
Enforcing a substitution of $x \mapsto - x$ gives
$$I = \int_{-1}^1 \frac{dx}{(2 + x) \sqrt{1 - x^2}} \qquad (2)$$
Adding (1) to (2) one obtains
$$I = 2 \int_{-1}^1 \frac{dx}{(4 - x^2) \sqrt{1 - x^2}} = 4 \int_0^1 \frac{dx}{(4 - x^2) \sqrt{1 - x^2}},$$
since the integrand is even between symmetric limits. 
Now set $x = \sin \theta$. As $dx = \cos \theta \, d\theta$, doing so yields
$$I = 4 \int_0^{\pi/2} \frac{d\theta}{4 - \sin^2 \theta}.$$
Taking advantage of the identity $\sin^2 \theta + \cos^2 \theta = 1$, the dominator in the above integral can be rewritten as
$$I = 4 \int_0^{\pi/2} \frac{d\theta}{3 \sin^2 \theta + 4 \cos^2 \theta} = 4 \int_0^{\pi/2} \frac{\sec^2 \theta}{3 \tan^2 \theta + 4} \, d\theta.$$
Setting $u = \tan \theta, du = \sec^2 \theta \, d\theta$ one has
\begin{align}
I &= 4 \int_0^\infty \frac{du}{3u^2 + 4}\\
&= \frac{4}{3} \int_0^\infty \frac{du}{u^2 + \left (\frac{2}{\sqrt{3}} \right )^2}\\
&= \frac{4}{3} \left [\frac{\sqrt{3}}{2} \tan^{-1} \left (\frac{u \sqrt{3}}{2} \right ) \right ]_0^\infty\\
&= \frac{\pi}{\sqrt{3}}.
\end{align} 
