calculate-1-2-0-x-1-x-2-y-2-3-2-dydx-

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

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 .