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Question Number 68043 by mhmd last updated on 03/Sep/19

\int_{π / 2}^{π} e^{c o s x} \sqrt{1 - e^{c o s x}} s i n x d x

Commented by mathmax by abdo last updated on 03/Sep/19

l e t I = \int_{\frac{π}{2}}^{π} e^{c o s x} \sqrt{1 - e^{c o s x}} s i n x d x v h a 7 g e m e n t c o s x = - t g i v e

I = \int_{0}^{1} e^{- t} \sqrt{1 - e^{- t}} (d t) = \int_{0}^{1} e^{- t} \sqrt{1 - e^{- t}} d t a f t e r w e d o t h e

c h a n g e m e n t \sqrt{1 - e_{}^{- t}} = u \Rightarrow 1 - e^{- t} = u^{2} \Rightarrow e^{- t} = 1 - u^{2} \Rightarrow

- t = l n (1 - u^{2}) \Rightarrow t = - l n (1 - u^{2}) \Rightarrow

I = \int_{0}^{\sqrt{1 - e^{- 1}}} (1 - u^{2}) u \times \frac{2 u}{1 - u^{2}} d u = 2 \int_{0}^{\sqrt{1 - e^{- 1}}} u^{2} d u

= \frac{2}{3} {[u^{3}]}_{0}^{\sqrt{1 - e^{- 1}}} = \frac{2}{3} {\sqrt{1 - e^{- 1}}}^{3} = \frac{2}{3} (1 - e^{- 1}) \sqrt{1 - \frac{1}{e}}

= \frac{2}{3} (1 - \frac{1}{e}) \frac{\sqrt{e - 1}}{\sqrt{e}} = \frac{2}{3} (\frac{e - 1}{e}) \frac{\sqrt{e - 1}}{\sqrt{e}}

\Rightarrow I = \frac{2 (e - 1) \sqrt{e - 1}}{3 e \sqrt{e}} .

Answered by mr W last updated on 03/Sep/19

= - \int_{π / 2}^{π} e^{c o s x} \sqrt{1 - e^{c o s x}} d \cos x

= \int_{- 1}^{0} e^{t} \sqrt{1 - e^{t}} d t

= \int_{- 1}^{0} \sqrt{1 - e^{t}} {d e}^{t}

= \int_{\frac{1}{e}}^{1} \sqrt{1 - u} d u

= - \int_{\frac{1}{e}}^{1} \sqrt{1 - u} d (1 - u)

= - \frac{2}{3} {[{(1 - u)}^{\frac{3}{2}}]}_{\frac{1}{e}}^{1}

= \frac{2}{3} {(1 - \frac{1}{e})}^{\frac{3}{2}}

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