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calculate-D-dxdy-1-x-2-1-y-2-with-D-x-h-R-2-0-x-1-and-0-y-x-




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}
$${calculate}\:\int\int_{{D}} \:\:\frac{{dxdy}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)} \\ $$$${with}\:{D}\:=\left\{\left({x},{h}\right)\in{R}^{\mathrm{2}} \:\:/\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\:\mathrm{0}\leqslant{y}\:\leqslant{x}\right\} \\ $$
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)
$$=\underset{\mathrm{0}} {\overset{{x}} {\int}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\left(\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\right){dy} \\ $$$$=\:\underset{\mathrm{0}} {\overset{{x}} {\int}}\:\frac{\mathrm{1}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\:.\left(\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \:\:{dy} \\ $$$$=\:\frac{\pi}{\mathrm{4}}\int_{\mathrm{0}} ^{{x}} \:\frac{\mathrm{1}}{\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\:{dy}\: \\ $$$$=\:\frac{\pi}{\mathrm{4}}\:\left\{\mathrm{tan}^{−\mathrm{1}} \left({y}\right)\:\mid_{\mathrm{0}} ^{{x}} \:\right\}\:=\frac{\pi}{\mathrm{4}}.\mathrm{tan}^{−\mathrm{1}} \left({x}\right) \\ $$

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