calculate ∫∫_D ((dxdy)/((1+x^2 )(1+y^2 ))) with D ={(x,h)∈R^2 /0≤x≤1 and 0≤y ≤x}


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Question Number 78701 by abdomathmax last updated on 20/Jan/20
$calculate ∫∫_D ((dxdy)/((1+x^2 )(1+y^2 ))) with D ={(x,h)∈R^2 /0≤x≤1 and 0≤y ≤x}$
$c a l c u l a t e \int \int_{D} \frac{d x d y}{(1 + x^{2}) (1 + y^{2})}$ $w i t h D = {(x, h) \in R^{2} / 0 ⩽ x ⩽ 1 a n d 0 ⩽ y ⩽ x}$
Commented by john santu last updated on 20/Jan/20
$=∫_0 ^x ∫_0 ^1 ((dx/((1+x^2 )(1+y^2 ))))dy = ∫_0 ^x (1/((1+y^2 ))) .(tan^(−1) (x))∣_0 ^1 dy = (π/4)∫_0 ^x (1/((1+y^2 ))) dy = (π/4) {tan^(−1) (y) ∣_0 ^x } =(π/4).tan^(−1) (x)$
$= \underset{0}{\int^{x}} \underset{0}{\int^{1}} (\frac{d x}{(1 + x^{2}) (1 + y^{2})}) d y$ $= \underset{0}{\int^{x}} \frac{1}{(1 + y^{2})} . (\tan^{- 1} (x)) ∣_{0}^{1} d y$ $= \frac{π}{4} \int_{0}^{x} \frac{1}{(1 + y^{2})} d y$ $= \frac{π}{4} {\tan^{- 1} (y) ∣_{0}^{x}} = \frac{π}{4} . \tan^{- 1} (x)$
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