If the value of the integral $\int_{0}^{1 / 2} \frac{x^{2}}{\left(1-x^{2}\right)^{3 / 2}} d x$ is
$\frac{\mathrm{k}}{6}$, then $\mathrm{k}$ is equal to :
Correct Option: 1
$\int_{0}^{1 / 2} \frac{\left(\left(x^{2}-1\right)+1\right)}{\left(1-x^{2}\right)^{3 / 2}} d x$
$\int_{0}^{1 / 2} \frac{d x}{\left(1-x^{2}\right)^{3 / 2}}-\int_{0}^{1 / 2} \frac{d x}{\sqrt{1-x^{2}}}$
$\int_{0}^{1 / 2} \frac{x^{-3}}{\left(x^{-2}-1\right)^{3 / 2}} d x-\left(\sin ^{-1} x\right)_{0}^{1 / 2}$
Let $\mathrm{x}^{-2}-1=\mathrm{t}^{2} \Rightarrow \mathrm{x}^{-3} \mathrm{~d} \mathrm{x}=-\mathrm{tdt}$
$\int_{\infty}^{\sqrt{3}-t d t} \frac{\pi}{t^{3}}-\frac{\pi}{6}=\int_{\sqrt{3}}^{\infty} \frac{d t}{t^{2}}-\frac{\pi}{6}=\frac{1}{\sqrt{3}}-\frac{\pi}{6}=\frac{k}{6}$
$\mathrm{k}=2 \sqrt{3}-\pi$
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