calculate-0-e-2x-1-e-4x-dx-

Question Number 38126 by maxmathsup by imad last updated on 22/Jun/18

c a l c u l a t e \int_{0}^{\infty} e^{- 2 x} \sqrt{1 + e^{- 4 x}} d x .

Commented by math khazana by abdo last updated on 26/Jun/18

l e t I = \int_{0}^{\infty} e^{- 2 x} \sqrt{1 + e^{- 4 x}} d x c h a n g e m e n t

e^{- 2 x} = t g i v e - 2 x = l n (t) a n d

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

2 I =_{t = s h (u)} \int_{0}^{l n (1 + \sqrt{2})} c h (t) c h (t) d t

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

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

= \frac{l n (1 + \sqrt{2})}{2} + \frac{1}{4} s h (2 l n (1 + \sqrt{2}))

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

Commented by math khazana by abdo last updated on 26/Jun/18

I = \frac{l n (1 + \sqrt{2})}{4} + \frac{1}{16} {{(1 + \sqrt{2})}^{2} - {(1 + \sqrt{2})}^{- 2}} .

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

t = e^{- 2 x} d t = - 2 e^{- 2 x} d x

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

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

u s e f o r m u l a \int \sqrt{x^{2} + a^{2}} d x

Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18