Question Number 68370 by mhmd last updated on 09/Sep/19
Answered by MJS last updated on 09/Sep/19
![∫(dt/( (√(e^t −1))))= [u=(√(e^t −1)) → dt=((2(√(e^t −1)))/e^t )du] =2∫(du/(u^2 +1))=2arctan u =2arctan (√(e^t −1)) [2arctan (√(e^t −1))]_a ^(2ln 2) =(π/6) 2arctan (√3) −2arctan (√(e^a −1)) =(π/6) (π/3)−arctan (√(e^a −1)) =(π/(12)) arctan (√(e^a −1)) =(π/4) ... a=ln 2](https://www.tinkutara.com/question/Q68373.png)
$$\int\frac{{dt}}{\:\sqrt{\mathrm{e}^{{t}} −\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{u}=\sqrt{\mathrm{e}^{{t}} −\mathrm{1}}\:\rightarrow\:{dt}=\frac{\mathrm{2}\sqrt{\mathrm{e}^{{t}} −\mathrm{1}}}{\mathrm{e}^{{t}} }{du}\right] \\ $$$$=\mathrm{2}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2arctan}\:{u}\:=\mathrm{2arctan}\:\sqrt{\mathrm{e}^{{t}} −\mathrm{1}} \\ $$$$\left[\mathrm{2arctan}\:\sqrt{\mathrm{e}^{{t}} −\mathrm{1}}\overset{\mathrm{2ln}\:\mathrm{2}} {\right]}_{{a}} =\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{2arctan}\:\sqrt{\mathrm{3}}\:−\mathrm{2arctan}\:\sqrt{\mathrm{e}^{{a}} −\mathrm{1}}\:=\frac{\pi}{\mathrm{6}} \\ $$$$\frac{\pi}{\mathrm{3}}−\mathrm{arctan}\:\sqrt{\mathrm{e}^{{a}} −\mathrm{1}}\:=\frac{\pi}{\mathrm{12}} \\ $$$$\mathrm{arctan}\:\sqrt{\mathrm{e}^{{a}} −\mathrm{1}}\:=\frac{\pi}{\mathrm{4}} \\ $$$$… \\ $$$${a}=\mathrm{ln}\:\mathrm{2} \\ $$