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

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

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

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

\[ = \log \frac{m}{{{m^2} – 1}}\]

Buy Best Mathematics E-Books Visit : https://mathstudy.in/

Buy Mathematics Formula Book for Class XI,XII,JEE and other Engineering Competition Exam https://mathstudy.in/product/mathematics-formula-book/

Buy Mathematics Workbook for Class XII ( Fully Solved ) : https://mathstudy.in/product/work-book-class-xii-c-b-s-e-fully-solved/

Buy Mathematics Chapter Tests for Class XII ( Fully Solved) : https://mathstudy.in/product/mathematics-chapter-tests-class-xii-c-b-s-e/

Buy Objective Type Question Bank Class XII (Fully Solved ) : https://mathstudy.in/product/objective-type-question-bank-for-mathematics-class-xii-c-b-s-e/