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Question Number 64559 by aliesam last updated on 19/Jul/19

Commented by mathmax by abdo last updated on 20/Jul/19
$let I =∫_(1/e) ^e (√((1−(lnx)^2 )/x))dx ⇒I =∫_(1/e) ^e (√(1−(lnx)^2 ))(dx/(√x)) changement (√x)=t give I = ∫_(1/(√e)) ^(√e) (√(1−(2lnt)^2 )) ((2tdt)/t) =2 ∫_(1/(√e)) ^(√e) (√(1−4ln^2 t)) dt changement t =(1/u) give I =2 ∫_(√e) ^(1/(√e)) (√(1−4ln^2 u))(−(du/u^2 )) =2 ∫_(1/(√e)) ^(√e) (1/u^2 )(√(1−4ln^2 u))du by psrts f′=(1/u^2 ) and g =(√(1−4ln^2 u)) ⇒ I =2{ [−(1/u)(√(1−4ln^2 u))]_(√e) ^(1/(√e)) −∫_(1/(√e)) ^(√e) (−(1/u)) ((−8lnu)/(2u(√(1−4ln^2 u)))) du} =2{....−4 ∫_(1/(√e)) ^(√e) ((lnu)/(u^2 (√(1−4ln^2 u)))) du}....be continued...$
$l e t I = \int_{\frac{1}{e}}^{e} \sqrt{\frac{1 - {(l n x)}^{2}}{x}} d x \Rightarrow I = \int_{\frac{1}{e}}^{e} \sqrt{1 - {(l n x)}^{2}} \frac{d x}{\sqrt{x}}$ $c h a n g e m e n t \sqrt{x} = t g i v e I = \int_{\frac{1}{\sqrt{e}}}^{\sqrt{e}} \sqrt{1 - {(2 l n t)}^{2}} \frac{2 t d t}{t}$ $= 2 \int_{\frac{1}{\sqrt{e}}}^{\sqrt{e}} \sqrt{1 - 4 {l n}^{2} t} d t c h a n g e m e n t t = \frac{1}{u} g i v e$ $I = 2 \int_{\sqrt{e}}^{\frac{1}{\sqrt{e}}} \sqrt{1 - 4 {l n}^{2} u} (- \frac{d u}{u^{2}}) = 2 \int_{\frac{1}{\sqrt{e}}}^{\sqrt{e}} \frac{1}{u^{2}} \sqrt{1 - 4 {l n}^{2} u} d u$ $b y p s r t s f^{'} = \frac{1}{u^{2}} a n d g = \sqrt{1 - 4 {l n}^{2} u} \Rightarrow$ $I = 2 {{[- \frac{1}{u} \sqrt{1 - 4 {l n}^{2} u}]}_{\sqrt{e}}^{\frac{1}{\sqrt{e}}} - \int_{\frac{1}{\sqrt{e}}}^{\sqrt{e}} (- \frac{1}{u}) \frac{- 8 l n u}{2 u \sqrt{1 - 4 {l n}^{2} u}} d u}$ $= 2 {. . . . - 4 \int_{\frac{1}{\sqrt{e}}}^{\sqrt{e}} \frac{l n u}{u^{2} \sqrt{1 - 4 {l n}^{2} u}} d u} . . . . b e c o n t i n u e d . . .$
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