calculate-0-e-x-1-e-2x-dx-

Question Number 38127 by maxmathsup by imad last updated on 22/Jun/18

c a l c u l a t e \int_{0}^{\infty} e^{- x} \sqrt{1 + e^{- 2 x}} d x

Commented by abdo mathsup 649 cc last updated on 23/Jun/18

l e t I = \int_{0}^{\infty} e^{- x} \sqrt{1 + e^{- 2 x}} d x c h a n g e m e n t e^{- x} = t

g i v e - x = l n (t) \Rightarrow x = - l n (t) a n d

I = - \int_{0}^{1} t \sqrt{1 + t^{2}} \frac{- d t}{t}

= \int_{0}^{1} \sqrt{1 + t^{2}} d t a l s o c h s n g . t = s h u g i v e

I = \int_{0}^{a r g s h (1)} c h (u) c h (u) d u

= \int_{0}^{l n (1 + \sqrt{2})} \frac{1 + c h (2 u)}{2} d u

= \frac{1}{2} l n (1 + \sqrt{2}) + \frac{1}{4} {[s h (2 u)]}_{0}^{l n (1 + \sqrt{2})}

= \frac{1}{2} l n (1 + \sqrt{2}) + \frac{1}{4} {[\frac{e^{2 u} - e^{- 2 u}}{2}]}_{0}^{l n (1 + \sqrt{2})}

= \frac{1}{2} l n (1 + \sqrt{2}) + \frac{1}{8} {{(1 + \sqrt{2})}^{2} - {(1 + \sqrt{2})}^{- 2}}