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**DeMath** Hint:

$\displaystyle \int\limits_{ - 1}^1 {\frac{{\sqrt {1 - {x^2}} }}

{{1 + {x^2}}}dx} = \int\limits_{ - 1}^1 {\frac{{1 - {x^2}}}

{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}dx} = \int\limits_{ - 1}^1 {\frac{{2 - \left( {1 + {x^2}} \right)}}

{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}dx} =$

$\displaystyle = - \int\limits_{ - 1}^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} + 2\int\limits_{ - 1}^1 {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}} = - \pi + 2\int\limits_{ - 1}^1 {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}} .$

Now try to use this substitution $\displaystyle x = \cos \theta, \, dx = - \sin \theta \, d\theta$

$\displaystyle 2\int\limits_{ - 1}^1 {\frac{{dx}}

{{\left( {1 + {x^2}} \right)\sqrt {1 - {x^2}} }}} = \left\{ \begin{gathered}

x = \cos \theta, \hfill \\

dx = - \sin \theta\,d\theta \hfill \\

\end{gathered} \right\} = - 2\int\limits_\pi ^0 {\frac{{d\theta}}

{{1 + {{\cos }^2}\theta}}} = \int\limits_{ - \pi }^\pi {\frac{{d\theta}}

{{1 + {{\cos }^2}\theta}}} .$