Question Number 76442 by aliesam last updated on 27/Dec/19
Answered by Tanmay chaudhury last updated on 28/Dec/19
$${x}={tana}\:\:\:\:{dx}={sec}^{\mathrm{2}} {ada} \\ $$$${sec}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{tan}^{\mathrm{2}} {a}}{\mathrm{1}−{tan}^{\mathrm{2}} {a}}\right)\rightarrow{sec}^{−\mathrm{1}} \left({sec}\mathrm{2}{a}\right)=\mathrm{2}{a} \\ $$$$\int\frac{\mathrm{2}{a}}{{tana}\:×{a}}×{sec}^{\mathrm{2}} {ada} \\ $$$$\mathrm{2}\int\frac{{d}\left({tana}\right)}{{tana}}\rightarrow\mathrm{2}\boldsymbol{{ln}}\left(\boldsymbol{{tana}}\right) \\ $$$$\boldsymbol{{so}}\:\:\mathrm{2}\mid\boldsymbol{{lnx}}\mid_{{e}} ^{{e}^{\mathrm{2}} } ={so}\:{answer}\:{is}\:\mathrm{2}\left({lne}^{\mathrm{2}} −{lne}\right)=\mathrm{2}\left(\mathrm{2}−\mathrm{1}\right)=\mathrm{2} \\ $$
Commented by aliesam last updated on 28/Dec/19
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by Tanmay chaudhury last updated on 28/Dec/19
$${thank}\:{you}\:{sir} \\ $$