Should I Hunt for the Antiderivative?

$\begin{aligned} I & = \int_0^\frac 12 \frac {dx}{(1-x)^2\sqrt{1-x^2}} & \small \color{#3D99F6} \text{Let }x = \sin \theta \implies dx = \cos \theta \\ & = \int_0^\frac \pi 6 \frac {d\theta}{(1-\sin \theta)^2} & \small \color{#3D99F6} \text{Let } t = \tan \frac \theta 2 \implies dt = \frac {\sec^2 \theta}2 d\theta \\ & = \int_0^{2-\sqrt 3} \frac {2\ dt}{\left(1-\frac {2t}{1+t^2}\right)^2 (1+t^2)} \\ & = 2 \int_0^{2-\sqrt 3} \frac {t^2 + 1}{(t-1)^4} dt & \small \color{#3D99F6} \text{By partial fraction decomposition} \\ & = 2 \int_0^{2-\sqrt 3} \frac 1{(t-1)^2} + \frac 2{(t-1)^3} + \frac 2{(t-1)^4} \ dt & \small \color{#3D99F6} \text{Let } u = t - 1 \implies du = dt \\ & = 2 \int_{-1}^{1-\sqrt 3} \frac 1{u^2} + \frac 2{u^3} + \frac 2{u^4} \ du \\ & = 2 \left[\frac 1u + \frac 1{u^2} + \frac 2{3u^3} \right]^{-1}_{1-\sqrt 3} \\ & = 2 \left[-1 + 1 - \frac 23 + \frac {\sqrt 3+1}2 - \frac {\sqrt 3+2}2 + \frac {3\sqrt 3+5}6 \right] \\ & = \sqrt 3 - \frac 23 \end{aligned}$

Then $\min (A+B+C+D) = 1+3+2+3 = \boxed 9$ .

Consider:

$I = \int_{0}^{1/2} \frac{dx}{(1-x)^2\sqrt{1-x^2}}$

Taking $x = \sin(t)$ transforms the integral to:

$I = \int_{0}^{\pi/6}\frac{dt}{(1-\sin{t})^2}$

Multiplying the numerator and denominator by $(1+\sin{t})^2$ gives:

$I = \int_{0}^{\pi/6} \frac{(1+\sin{t})^2}{(1+\sin{t})^2(1-\sin{t})^2}dt$

$I = \int_{0}^{\pi/6} \frac{1+2\sin{t}+\sin^2{t}}{\cos^4{t}}dt$

$I = \int_{0}^{\pi/6} \left(\sec^4{t}+2\tan{t}\sec^3{t}+\tan^2{t}\sec^2{t}\right)dt$

$I = \int_{0}^{\pi/6}\sec^4{t}dt + \int_{0}^{\pi/6}2\tan{t}\sec^3{t}dt + \int_{0}^{\pi/6}\tan^2{t}\sec^2{t}dt$

$I = \int_{0}^{\pi/6}(\left(1+\tan^2{t}\right)\sec^2{t})dt + \int_{0}^{\pi/6}(2\left(\tan{t}\sec{t}\right)\sec^2{t})dt + \int_{0}^{\pi/6}(\tan^2{t}\sec^2{t})dt = I_1 + I_2 + I_3$

$I_1$ and $I_3$ can be evaluated by taking $\tan{t} = z$ while $I_2$ can be evaluated by taking $\sec{t} = z$ . All the integrals transform to:

$I = \int_{0}^{1/\sqrt{3}}\left(1+z^2\right)dz + 2\int_{1}^{2/\sqrt{3}}z^2dz + + \int_{0}^{1/\sqrt{3}}z^2dz$

Evaluating the above and simplifying gives:

$\boxed{I = 1\sqrt{3} - \frac{2}{3}}$

Trivial steps involving working out the change of the variables of integration and evaluation of simple definite integrals have been skipped throughout this solution.