Substitution! (8)

Put $x=\sin \theta$ such that $\mathrm{d}x=\cos \theta\mathrm{d}\theta$ and use $1-\sin^2\theta=\cos^2 \theta$ such that integration is:-

$\displaystyle\int_0^{\pi/6}\dfrac{\sin^2 \theta\cos \theta~\mathrm{d}\theta}{\cos^3\theta}$ $=\displaystyle\int_0^{\pi/6}\tan^2 \theta\mathrm{d}\theta= \displaystyle\int_0^{\pi/6}(\sec^2\theta-1)\mathrm{d}\theta$ $\large =\left|\tan \theta -\theta\right|_0^{\pi/6}$ $\large =\dfrac{2\sqrt{3}-\pi}{6}$

$\huge \therefore 2\times 3\times 6=\boxed{36}$

$\text {Put } x=\sin{t} \\ \implies dx=dt\cos{t}\\ \implies \large\displaystyle\int_0^{\frac12}\frac{\sin^2t}{(1-\sin^2{t})^{\frac32}}dx \\ \implies\large\displaystyle\int_0^{\frac12}\frac{\sin^2t}{(cos^2t)^{\frac32}}dt\cos{t} \\ \implies\large\displaystyle\int_0^{\frac{\pi}6}\frac{\sin^2t}{cos^2t}dt \\\implies\large\displaystyle\int_0^{\frac{\pi}6}\tan^2tdt \\ \text{Using the Result :} \int\tan^2{x}dx = \tan{x}-x+c , \text {We get } \\ \implies \tan{\dfrac{\pi}6}-\dfrac{\pi}6-0+0 \\ \implies \dfrac{1}{\sqrt3}-\dfrac{\pi}6 = \dfrac{2\sqrt3-\pi}6\\ a=2 , b=3 ,c=6 \\ Ans \color{#3D99F6}{\implies\boxed{36}}$