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Question Number 38202 by prof Abdo imad last updated on 22/Jun/18

$$findthevalueof{\int }_0^1{e}^{-x}\sqrt{1-e^{-2x}}dx$$

Commented by math khazana by abdo last updated on 23/Jun/18

$$changement{e}^{-x}=tgivex=-ln\left(t\right)and$$$$I=-{\int }_1^{e^{-1}}t\sqrt{1-t^2}\frac{dt}{t}$$$$={\int }_{\frac{1}{e}}^1\sqrt{1-t^2}dtafterweusethechang.t=sin\theta$$$$I={\int }_{arcsin\left(e^{-1}\right)}^{\frac{\pi }{2}}cos\theta .coss\theta d\theta$$$$={\int }_{arcsin\left(e^{-1}\right)}^{\frac{\pi }{2}}\frac{1+cos\left(2\theta \right)}{2}d\theta$$$$=\frac{1}{2}(\frac{\pi }{2}-arcsin\left(e^{-1}\right))+\frac{1}{4}{\left[sin\left(2\theta \right)\right]}_{arcsin\left(e^{-1}\right)}^{\frac{\pi }{2}}$$$$=\frac{\pi }{4}-\frac{1}{2}arcsin\left(e^{-1}\right)-\frac{1}{4}sin\left(2arcsin\left(e^{-1}\right)\right).$$

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