Question Number 76442 by aliesam last updated on 27/Dec/19
![](https://www.tinkutara.com/question/10819.png)
Answered by Tanmay chaudhury last updated on 28/Dec/19
![x=tana dx=sec^2 ada sec^(−1) (((1+tan^2 a)/(1−tan^2 a)))→sec^(−1) (sec2a)=2a ∫((2a)/(tana ×a))×sec^2 ada 2∫((d(tana))/(tana))→2ln(tana) so 2∣lnx∣_e ^e^2 =so answer is 2(lne^2 −lne)=2(2−1)=2](https://www.tinkutara.com/question/Q76549.png)
$${x}={tana}\:\:\:\:{dx}={sec}^{\mathrm{2}} {ada} \\ $$$${sec}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{tan}^{\mathrm{2}} {a}}{\mathrm{1}−{tan}^{\mathrm{2}} {a}}\right)\rightarrow{sec}^{−\mathrm{1}} \left({sec}\mathrm{2}{a}\right)=\mathrm{2}{a} \\ $$$$\int\frac{\mathrm{2}{a}}{{tana}\:×{a}}×{sec}^{\mathrm{2}} {ada} \\ $$$$\mathrm{2}\int\frac{{d}\left({tana}\right)}{{tana}}\rightarrow\mathrm{2}\boldsymbol{{ln}}\left(\boldsymbol{{tana}}\right) \\ $$$$\boldsymbol{{so}}\:\:\mathrm{2}\mid\boldsymbol{{lnx}}\mid_{{e}} ^{{e}^{\mathrm{2}} } ={so}\:{answer}\:{is}\:\mathrm{2}\left({lne}^{\mathrm{2}} −{lne}\right)=\mathrm{2}\left(\mathrm{2}−\mathrm{1}\right)=\mathrm{2} \\ $$
Commented by aliesam last updated on 28/Dec/19
![god bless you sir](https://www.tinkutara.com/question/Q76550.png)
$${god}\:{bless}\:{you}\:{sir} \\ $$
Commented by Tanmay chaudhury last updated on 28/Dec/19
![thank you sir](https://www.tinkutara.com/question/Q76553.png)
$${thank}\:{you}\:{sir} \\ $$