What is the integration of \[ \int_0^{\pi /2} {\frac{{\tan x}}{{1 + {m^2}{{\tan }^2}x}}} dx\] ~~~~~[NCERT Exemp. Q. 30,Page 165 ]

Let \[ I = \int_0^{\pi /2} {\frac{{\tan xdx}}{{1 + {m^2}{{\tan }^2}x}}} dx\]

\[ = \int_0^{\pi /2} {\frac{{\frac{{\sin x}}{{\cos x}}}}{{1 + {m^2} \cdot \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}}} dx\]

\[ = \int_0^{\pi /2} {\frac{{\frac{{\sin x}}{{\cos x}}}}{{\frac{{{{\cos }^2}x + {m^2}{{\sin }^2}x}}{{{{\cos }^2}x}}}}} dx\]


\[ = \int_0^{\pi /2} {\frac{{\sin x\cos xdx}}{{1 – {{\sin }^2}x + {m^2}{{\sin }^2}x}}} dx\]


\[ = \int_0^{\pi /2} {\frac{{\sin x\cos x}}{{1 – {{\sin }^2}x\left( {1 – {m^2}} \right)}}} dx\]

Let’s put \[{\sin ^2}x = t\]

\[ \Rightarrow \] \[2\sin x\cos xdx = dt\]

therefore,\[I = \frac{1}{2}\int_0^1 {\frac{{dt}}{{1 – t\left( {1 – {m^2}} \right)}}} \]


\[ = \frac{1}{2}\left[ { – \log \left| {1 – t\left( {1 – {m^2}} \right)} \right| \cdot \frac{1}{{1 – {m^2}}}} \right]_0^1\]


\[ = \frac{1}{2}\left[ { – \log \left| {1 – 1 + {m^2}} \right| \cdot \frac{1}{{1 + {m^2}}} + \log |1| \cdot \frac{1}{{1 – {m^2}}}} \right]\]


\[ = \frac{1}{2}\left[ { – \log \left| {{m^2}} \right| \cdot \frac{1}{{1 – {m^2}}}} \right] = \frac{2}{2} \cdot \frac{{\log m}}{{\left( {{m^2} – 1} \right)}}\]

 

\[ = \log \frac{m}{{{m^2} – 1}}\]

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