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.

Try using the substitution $\displaystyle t = \tan \frac{\theta}{2}$, this is a handy substitution to make when there are trigonometric functions that you cannot simplify very easily. In our case we have that,
$$\frac{\mathrm{d}t}{\mathrm{d}\theta} = \frac{1}{2} \sec^2\frac{\theta}{2} = \frac{1}{2} \left(1 + \tan^2 \frac{\theta}{2} \right) = \frac{1}{2}(1 + t^2) \Rightarrow \frac{2\mathrm{d}t}{1+t^2} = \mathrm{d}\theta  $$
The usefulness of this substitution is that we can now use properties of a right angled triangle that has angle $\theta /2$ to say that,
$$\sin \theta = 2\sin \frac{\theta}{2} \cos \frac{\theta}{2} = 2 \frac{1}{\sqrt{1+t^2}} \frac{t}{\sqrt{1+t^2}}= \frac{2t}{1+t^2} $$
Putting this into our integral we have,
$$\int_{-\pi/2}^{\pi/2} \frac{\mathrm{d}\theta}{2 - \sin \theta} = \int_{-1}^1 \frac{2}{1+t^2} \cdot \frac{1}{2  - \frac{2t}{1+t^2}} \ \mathrm{d}t = \int_{-1}^1 \frac{1}{t^2 - t+ 1} \ \mathrm{d}t $$
Now we have no more square roots, we have an irreducible quadratic at the bottom so we can complete the square to obtain,
$$\int_{-1}^1 \frac{1}{t^2 - t + 1} \ \mathrm{d} t = \int_{-1}^1 \frac{1}{(t - \frac{1}{2})^2 + \frac{3}{4}} \ \mathrm{d}t = \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2}{\sqrt{3}}\left(t - \frac{1}{2}\right) \right) \Big|_{-1}^1 = \frac{\pi}{\sqrt{3}} $$
