JEE Advance - Mathematics (1984 - No. 36)
Evaluate the following $$\int\limits_0^{{1 \over 2}} {{{x{{\sin }^{ - 1}}x} \over {\sqrt {1 - {x^2}} }}dx} $$
$$\frac{6 + \pi \sqrt{3}}{12}$$
$$\frac{\pi \sqrt{3} - 6}{12}$$
$$\frac{6 - \pi \sqrt{3}}{12}$$
$$\frac{12 - \pi \sqrt{3}}{6}$$
$$\frac{6 - \pi}{12}$$