Question Number 38202 by prof Abdo imad last updated on 22/Jun/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}} \sqrt{\mathrm{1}−{e}^{−\mathrm{2}{x}} }{dx} \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
![changement e^(−x) =t give x=−ln(t) and I =− ∫_1 ^e^(−1) t(√(1−t^2 )) (dt/t) = ∫_(1/e) ^1 (√(1−t^2 ))dt after we use the chang.t=sinθ I = ∫_(arcsin(e^(−1) )) ^(π/2) cosθ.cossθ dθ = ∫_(arcsin(e^(−1) )) ^(π/2) ((1+cos(2θ))/2)dθ =(1/2)( (π/2) −arcsin(e^(−1) )) +(1/4)[sin(2θ)]_(arcsin(e^(−1) )) ^(π/2) =(π/4) −(1/2)arcsin(e^(−1) )−(1/4)sin(2arcsin(e^(−1) )).](https://www.tinkutara.com/question/Q38298.png)
$${changement}\:{e}^{−{x}} ={t}\:{give}\:{x}=−{ln}\left({t}\right)\:{and} \\ $$$${I}\:=−\:\int_{\mathrm{1}} ^{{e}^{−\mathrm{1}} } {t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\frac{{dt}}{{t}} \\ $$$$=\:\int_{\frac{\mathrm{1}}{{e}}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:\:{after}\:{we}\:{use}\:{the}\:{chang}.{t}={sin}\theta \\ $$$${I}\:=\:\int_{{arcsin}\left({e}^{−\mathrm{1}} \right)} ^{\frac{\pi}{\mathrm{2}}} {cos}\theta.{coss}\theta\:{d}\theta \\ $$$$=\:\int_{{arcsin}\left({e}^{−\mathrm{1}} \right)} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\frac{\pi}{\mathrm{2}}\:−{arcsin}\left({e}^{−\mathrm{1}} \right)\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}\theta\right)\right]_{{arcsin}\left({e}^{−\mathrm{1}} \right)} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}{arcsin}\left({e}^{−\mathrm{1}} \right)−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{arcsin}\left({e}^{−\mathrm{1}} \right)\right). \\ $$