How do I evaluate this improper integral $\int_{-1}^{1}\frac{dx}{(2-x)\sqrt{1-x^{2}}}$ Given integral is $$\int_{-1}^{1}\frac{dx}{(2-x)\sqrt{1-x^{2}}}.$$ I tried to split it up at $0$, but I donot know what to do ahead.
Thanks.
 A: HINT:
Let $\arcsin x=y\implies x=\sin y$ and $\cos y=+\sqrt{1-x^2}$ (why?)
$$I=\int_{-\pi/2}^{\pi/2}\dfrac{dy}{2-\sin y}$$
Method $\#1:$  Use $\sin2u=\dfrac{2\tan u}{1+\tan^2u}$ and set $\tan\dfrac y2=t$
Method $\#2:$  Use $\displaystyle\int_a^bf(y)\ dy=\int_a^bf(a+b-y)\ dy,$
and then $2I=4\int_{-\pi/2}^{\pi/2}\dfrac{dy}{4-\sin^2y}=4\int_{-\pi/2}^{\pi/2}\dfrac{\csc^2y\ dy}{4(1+\cot^2y)-1}$
Set $\cot y=v$
A: If you don't know complex analysis, we can use this way. Put $x=\sin\left(u\right)
 $ to get $$\int_{-\pi/2}^{\pi/2}\frac{1}{2-\sin\left(u\right)}du
 $$ and now take $v=\tan\left(\frac{u}{2}\right)
 $ $$=\int_{-1}^{1}\frac{1}{v^{2}-v+1}dv=\int_{-1}^{1}\frac{1}{\left(v-\frac{1}{2}\right)^{2}+\frac{3}{4}}dv\overset{s=v-1/2}{=}\int_{-3/2}^{1/2}\frac{1}{s^{2}+\frac{3}{4}}ds
 $$ $$=\frac{4}{3}\int_{-3/2}^{1/2}\frac{1}{\frac{4s^{2}}{3}+1}ds\overset{w=2s/\sqrt{3}}{=}\frac{2}{\sqrt{3}}\int_{-3/\sqrt{3}}^{1/\sqrt{3}}\frac{1}{w^{2}+1}ds=
 $$ $$=\frac{\pi}{\sqrt{3}}\approx1.8138.
 $$
A: Let $\displaystyle I =\int_{-1}^{1}\frac{1}{(2-x)\sqrt{1-x^2}}dx....(1)$
Noe Let $x = -t\;,$ Then $dx = -dt$ and changing Limit, we get
$\displaystyle I = \int_{1}^{-1}\frac{1}{(2+t)\sqrt{1-t^2}}\times -dt = \int_{-1}^{1}\frac{1}{(2+x)\sqrt{1-x^2}}......(2)$
Now Add $(1)$ and $(2)\;,$ we get $\displaystyle 2I = \int_{-1}^{1}\left[\frac{1}{2-x}+\frac{1}{2+x}\right]\cdot \frac{1}{\sqrt{1-x^2}}dx = 4\int_{-1}^{1}\frac{1}{(4-x^2)\sqrt{1-x^2}}dx$
So $\displaystyle 2I = 8\int_{0}^{1}\frac{1}{(4-x^2)\sqrt{1-x^2}}dx$
Now Put $x=\sin \phi$ and $dx = \cos \phi\phi$ and Changing Limit, we get
$\displaystyle 2I = 8\int_{0}^{\frac{\pi}{2}}\frac{1}{4-\sin^2 \phi}d\phi. = \int_{0}^{\frac{\pi}{2}}\frac{\sec^2 \phi}{4(1+\tan^2 \phi)-1}d\phi = 8\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 \phi}{3+4\tan^2 \phi}d\phi$
Now Put $\tan \phi = z\;,$ Then $\sec^2 \phi d\phi = dx$ and changing limit, we get
$\displaystyle 2I = 8\int_{0}^{\infty}\frac{1}{3+4z^2}dz = 2\int_{0}^{\infty}\frac{1}{z^2+\left(\frac{\sqrt{3}}{2}\right)^2}dz = \frac{4}{\sqrt{3}}\cdot \frac{\pi}{2}\Rightarrow I = \frac{\pi}{\sqrt{3}}$
