Question

\[\int\frac{e^{m \tan^{- 1} x}}{\left( 1 + x^2 \right)^{3/2}} \text{ dx}\]

Answer 1

\[\text{We have}, \]

\[I = \int\frac{e^{m \tan^{- 1} x}}{\left( 1 + x^2 \right)^\frac{3}{2}}\text{ dx}\]

\[\text{ Putting tan}^{- 1} x = t \Rightarrow x = \tan t\]

\[ \Rightarrow \frac{1}{1 + x^2} \text{ dx}= dt\]

\[ \Rightarrow dx = \left( 1 + x^2 \right)dt\]

\[ \Rightarrow dx = \left( 1 + \tan^2 t \right)dt\]

\[ \therefore I = \int\frac{e^{mt}}{\left( 1 + \tan^2 t \right)^\frac{3}{2}}\left( 1 + \tan^2 t \right)dt\]

\[ = \int\frac{e^{mt} dt}{\sqrt{1 + \tan^2 t}}\]

\[ = \int {e_{II}}^{mt} \cos_I t \text{ dt}\]

\[ = \cos t\frac{e^{mt}}{m} - \int\left( - \sin t \right)\frac{e^{mt}}{m} \text{ dt}\]

\[ = \cos t\frac{e^{mt}}{m} + \frac{1}{m}\int e^{mt} \text{ sin t dt }\]

\[ = \cos t\frac{e^{mt}}{m} + \frac{1}{m} I_1 . . . . . \left( 1 \right)\]

\[\text{ Where,} \]

\[ I_1 = \int {e_{II}}^{mt} \sin_I t  \text{  dt}\]

\[ = \sin t\frac{e^{mt}}{m} - \int\cos t\frac{e^{mt}}{m}dt\]

\[ I_1 = \sin t\frac{e^{mt}}{m} - \frac{1}{m}I . . . . . \left( 2 \right)\]

\[\text{ from} \left( 1 \right)\text{  and }\left( 2 \right)\]

\[I = \cos t\frac{e^{mt}}{m} + \frac{1}{m} \left[ \sin t\frac{e^{mt}}{m} - \frac{1}{m}I \right]\]

\[ \Rightarrow I = \cos t\frac{e^{mt}}{m} + \frac{\text{ sin t e}^{mt}}{m^2} - \frac{1}{m^2} I\]

\[ \Rightarrow I + \frac{I}{m^2} = \frac{e^{mt} \left( m \cos t + \sin t \right)}{m^2}\]

\[ \Rightarrow I = \frac{e^{mt} \left( m \cos t + \sin t \right)}{1 + m^2} + C\]

\[ \Rightarrow I = \frac{e^{mt}}{\sqrt{1 + m^2}} \left[ \cos t\frac{m}{\sqrt{1 + m^2}} + \sin t\frac{1}{\sqrt{1 + m^2}} \right] + C\]

\[\text{ Let }  \frac{m}{\sqrt{1 + m^2}} = \cos \theta\]

\[\text{ Then, }\sin\theta = \frac{1}{\sqrt{1 + m^2}}\]

\[ \Rightarrow \cot\theta = m\]

\[ \Rightarrow \theta = \cot^{- 1} m\]

\[ \therefore I = \frac{e^{mt}}{\sqrt{1 + m^2}} \left\{ \cos t \cos \theta + \sin t \sin \theta \right\} + C\]

\[ = \frac{e^{mt}}{\sqrt{1 + m^2}} \left\{ \cos \left( t - \theta \right) \right\} + C\]

\[ = \frac{e^{mt}}{\sqrt{1 + m^2}} \left\{ \cos \left( \tan^{- 1} x - \cot^{- 1} m \right) \right\} + C\]