$\begin{align}  & \int_{\tfrac{a}{2}}^{a}{\frac{1}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx} \\ & x=\frac{1}{u},\text{ }u=\frac{1}{x},\text{ }\frac{du}{dx}=-\frac{1}{{{x}^{2}}},\text{ }dx=-{{x}^{2}}du \\ & \text{Changing the limits }{{u}_{1}}=\tfrac{2}{a},\text{ }{{u}_{2}}=\tfrac{1}{a} \\ & \int_{\tfrac{a}{2}}^{a}{\frac{1}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}=\int_{\tfrac{2}{a}}^{\tfrac{1}{a}}{\frac{1}{{{x}^{2}}\sqrt{1-\frac{1}{{{u}^{2}}}}}(-{{x}^{2}}du)} \\ & \int_{\tfrac{a}{2}}^{a}{\frac{1}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}=-\int_{2}^{\tfrac{1}{a}}{\frac{1}{\sqrt{1-\frac{1}{{{u}^{2}}}}}du}=-\int_{\tfrac{2}{a}}^{\tfrac{1}{a}}{\frac{u}{\sqrt{{{u}^{2}}-1}}} \\ & \int_{\tfrac{a}{2}}^{a}{\frac{1}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}=-\left[ {{\sin }^{-1}}\frac{{\scriptstyle{}^{1}/{}_{u}}}{a} \right]_{\tfrac{2}{a}}^{\tfrac{1}{a}}=-\left[ {{\sin }^{-1}}\frac{1}{au} \right]_{\tfrac{2}{a}}^{\tfrac{1}{a}} \\ & \int_{\tfrac{a}{2}}^{a}{\frac{1}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}=-\left[ {{\sin }^{-1}}\frac{1}{a(\tfrac{1}{a})}-{{\sin }^{-1}}\frac{1}{a(\tfrac{2}{a})} \right] \\ & \int_{\tfrac{a}{2}}^{a}{\frac{1}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}=-\left[ {{\sin }^{-1}}1-{{\sin }^{-1}}\tfrac{1}{2} \right] \\ & \int_{\tfrac{a}{2}}^{a}{\frac{1}{{{x}^{2}}\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}=-\left[ \frac{\pi }{2}-\frac{\pi }{6} \right]=-\frac{\pi }{3} \\\end{align}$