What is the integration of \[\int_0^{1/2} {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 – {x^2}} }}} \] ~~~~~[NCERT Exemp. Q. 34,Page 165 ]

What is the integration of \[\int_0^{1/2} {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 – {x^2}} }}} \] ~~~~~[NCERT Exemp. Q. 34,Page 165 ]

Let \[I = \int_0^{1/2} {\frac{{dx}}{{\left( {1 + {x^2}} \right)\sqrt {1 – {x^2}} }}} \]

Let’s put \[x = \sin \theta \]

\[ \Rightarrow \] \[ dx = \cos \theta d\theta \]
As\[ x \to 0,\] then \[ \theta \to 0\]

and \[ x \to \frac{1}{2},\] then \[ \theta \to \frac{\pi }{6}\]

therefore,\[I = \int_0^{\pi /6} {\frac{{\cos \theta }}{{\left( {1 + {{\sin }^2}\theta } \right)\cos \theta }}} d\theta = \int_0^{\pi /6} {\frac{1}{{1 + {{\sin }^2}\theta }}} d\theta \]

\[ = \int_0^{\pi /6} {\frac{1}{{{{\cos }^2}\theta \left( {{{\sec }^2}\theta + {{\tan }^2}\theta } \right)}}} d\theta \]

\[ = \int_0^{\pi /6} {\frac{{{{\sec }^2}\theta }}{{{{\sec }^2}\theta + {{\tan }^2}\theta }}} d\theta \]

\[ = \int_0^{\pi /6} {\frac{{{{\sec }^2}\theta }}{{1 + {{\tan }^2}\theta + {{\tan }^2}\theta }}} d\theta \]

\[ = \int_0^{\pi /6} {\frac{{{{\sec }^2}\theta }}{{1 + 2{{\tan }^2}\theta }}} d\theta \]

Again, Let’s put \[\tan \theta = t\]

\[ \Rightarrow \] \[{\sec ^2}\theta d\theta = dt\]
As \[\theta \to 0,\] then \[t \to 0\]

and \[\theta \to \frac{\pi }{6}\], then \[t \to \frac{1}{{\sqrt 3 }}\]

therefore,\[ I = \int_0^{1/\sqrt 3 } {\frac{{dt}}{{1 + 2{t^2}}}} = \frac{1}{2}\int_0^{1/\sqrt 3 } {\frac{{dt}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + {t^2}}}} \]


\[ = \frac{1}{2} \cdot \frac{1}{{1/\sqrt 2 }}\left[ {{{\tan }^{ – 1}}\frac{t}{{\frac{1}{{\sqrt 2 }}}}} \right]_0^{1/\sqrt 3 } = \frac{1}{{\sqrt 2 }}\left[ {{{\tan }^{ – 1}}(\sqrt 2 t)} \right]_0^{1/\sqrt 3 }\]


\[ = \frac{1}{{\sqrt 2 }}\left[ {{{\tan }^{ – 1}}\sqrt {\frac{2}{3}} – 0} \right] = \frac{1}{{\sqrt 2 }}{\tan ^{ – 1}}\left( {\sqrt {\frac{2}{3}} } \right)\]

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