Given that, \( \int_{0}^{1 / 2} \frac{d x}{\left(1+x^{2}\right) \sqrt{1-\chi^{2}}} \)
Let \( x=\sin \theta \) then \( d x=\cos \theta d \theta \) and limits become \( 0 \rightarrow 0, \frac{1}{2} \rightarrow \frac{\Pi}{6} . \) So,
\[
\begin{array}{l}
\int_{0}^{\pi / 6} \frac{\cos \theta d \theta}{\left(1+\sin ^{2} \theta\right) \sqrt{1-\sin ^{2} \theta}} \\
=\int_{0}^{\pi / 6} \frac{\cos \theta d \theta}{\left(1+\sin ^{2} \theta\right) \cos \theta} \\
=\int_{0}^{\pi / 6} \frac{\sec ^{2} \theta}{1+2 \tan ^{2} \theta} d \theta
\end{array}
\]
Let \( u=\sqrt{2} \tan \theta \) then \( d u=\sqrt{2} \sec ^{2} \theta d \theta \) and limits becomes
\( 0 \rightarrow 0, \frac{\Pi}{6} \rightarrow \sqrt{\frac{2}{3}} \). So, \( \frac{1}{\sqrt{2}} \int_{0}^{\sqrt{2 / 3}} \frac{1}{1+u^{2}} d u \)
We know that \( \int \frac{1}{1+x^{2}} d x=\tan ^{-1} \chi . \) So,
\[
\begin{array}{l}
\left.\frac{1}{\sqrt{2}} \tan ^{-1} u\right|_{0} ^{\sqrt{2 / 3}}=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)-0 \\
=\frac{1}{\sqrt{2}} \tan ^{-1}\left(\sqrt{\frac{2}{3}}\right)
\end{array}
\]