calculate ∫_1 ^2 ∫_0 ^x (1/((x^2 +y^2 )^(3/2) ))dydx


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Question Number 64904 by mathmax by abdo last updated on 23/Jul/19

$c a l c u l a t e \int_{1}^{2} \int_{0}^{x} \frac{1}{{(x^{2} + y^{2})}^{\frac{3}{2}}} d y d x$
Commented by ~ À ® @ 237 ~ last updated on 23/Jul/19
$le domaine d integration D ={(x . y) / 1<x<2 0<y<x} en posant x=u et y=utanv on a J(x y)=utanv(1+tan^2 v) D={(u. utanv)/ 1<u<2 et 0<tanv<1} ={(u. utanv)/ 1<u<2 et 0<v<(π/4) } on obtient alors I= ∫_1 ^2 ∫_0 ^(π/4) ((utanv(1+tan^2 v)dudv)/(((√((u^2 +u^2 tan^2 v))))^3 )) =∫_1 ^2 (1/u^2 )du ∫_0 ^(π/4) ((tanv(1+tan^2 v)dv)/((1+tan^2 v)^(3/2) )) =[−(1/u)]_1 ^2 [(1/2) ((−1)/(((3/2)−1)(1+tan^2 v)^(((3/2)−1)) ))]_0 ^(π/4) = = (((2−(√2)))/4)$
$l e d o m a i n e d i n t e g r a t i o n D = {(x . y) / 1 < x < 2 0 < y < x}$ $e n p o s a n t x = u e t y = u t a n v o n a J (x y) = u t a n v (1 + {t a n}^{2} v)$ $D = {(u . u t a n v) / 1 < u < 2 e t 0 < t a n v < 1}$ $= {(u . u t a n v) / 1 < u < 2 e t 0 < v < \frac{π}{4}}$ $o n o b t i e n t a l o r s$ $I = \int_{1}^{2} \int_{0}^{\frac{π}{4}} \frac{u t a n v (1 + {t a n}^{2} v) d u d v}{{(\sqrt{(u^{2} + u^{2} {t a n}^{2} v)})}^{3}}$ $= \int_{1}^{2} \frac{1}{u^{2}} d u \int_{0}^{\frac{π}{4}} \frac{t a n v (1 + {t a n}^{2} v) d v}{{(1 + {t a n}^{2} v)}^{\frac{3}{2}}}$ $Missing \left or extra \right$ $=$ $= \frac{(2 - \sqrt{2})}{4}$
Commented by mathmax by abdo last updated on 23/Jul/19

$t h a n k y o u s i r .$
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