Evaluate $\int_{-1}^{1} \cot^{-1} \left(\frac{1}{\sqrt{1-x^2}}\right) \cot^{-1}\left(\raise{3pt}\frac{x}{\sqrt{1-(x^2)^{|x|}}}\right)$ 
$$\int_{-1}^{1} \left(\cot^{-1} \dfrac{1}{\sqrt{1-x^2}}\right) \left(\cot^{-1}\dfrac{x}{\sqrt{1-(x^2)^{|x|}}}\right)= \dfrac{\pi^2(\sqrt a-\sqrt b )}{\sqrt c}$$
  , where a,b, c are natural numbers and are there in their lowest form, then find the value of a+b+c.

Using $\int_a^b f(x)dx= \int_a^bf(a+b-x{) dx}$, I got: 
$$2I =2\pi\int_{0}^1 \cot^{-1}\left(\dfrac{1}{\sqrt{1-x^2}}\right) dx$$
Then, letting $x = \sin \theta$
$I = \pi\displaystyle\int_{0}^{\pi/2}\arctan (\cos\theta) \cos \theta d\theta$
After this I tried integration by parts but it gets really complicated with that? How do I continue? 
EDIT:
Please note that arccot(x) + arccot(-x)= $\pi$ $\ne 0$
Principal range of $\cot^{-1}x$ considered in the question is $(0,\pi)$

Using the property $$\int_a^b f(x)=\int_a^b f(a+b-x)$$
The integral changes to $$I=\pi\int_0^1 \arctan \sqrt{1-x^2} dx$$
You might know the property that $$\int_a^b f(x)+\int_{f(a)}^{f(b)} f^{-1}(x)  dx=-af(a)+bf(b)$$
Let $$J=\int_0^1 \arctan \sqrt{1-x^2} dx+\int_{\frac {\pi}{4}}^0 \sqrt {1-\tan ^2x } dx$$  Using above property $J=0$ Hence $$\pi\int_0^1 \arctan \sqrt{1-x^2} dx=\pi\int_0^{\frac {\pi}{4}} \sqrt {1-\tan ^2x } dx=I$$
Using the substitution $\tan x=\sin \theta$ in the right integral we get $$I=\pi\int_0^{\frac {\pi}{2}} \frac {\cos ^2\theta}{1+\sin^2\theta} d\theta=\pi\int_0^{\frac {\pi}{2}}\left( -1+\frac {2\sec^2\theta}{1+2\tan^2\theta}\right)  d\theta$$
Hope you can continue further 
