$$\int\int_D{\sqrt y\over \sqrt x(1+xy)}dxdy$$ $$1\le x\le 3 $$ $$0\le y\le 1 $$ so I just inversed the heads, and I'm stuck at integral $$\int_1^3{1 \over \sqrt x}\int_0^1{\sqrt y \over(1+xy)}dydx$$ and then I substitute in the 2nd integral , as $$2\int_0^1{t^2 \over 1+xt^2}dt$$any ideas?
double integral square root
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$\begingroup$
integration
2 Answers
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Integrals of the form $\int_{0}^{1}\frac{q(z^2)}{1+z^2}\,dz$ with $q$ being a polynomial are straightforward to compute, hence:
$$\begin{eqnarray*}\iint_{(1,3)\times(0,1)}\frac{\sqrt{y}}{\sqrt{x}(1+xy)}\,dx\,dy &\stackrel{(x,y)\mapsto(u^2,v^2)}{=}& \iint_{(1,\sqrt{3})\times(0,1)}\frac{4v^2}{1+u^2 v^2}\,du\,dv\\&=& \int_{0}^{1}4v\left(\arctan(v\sqrt{3})-\arctan(v)\right)\,dv\\&\stackrel{IBP}{=}&\frac{\pi}{6}-\int_{0}^{1}2v^2\left(\frac{\sqrt{3}}{1+3v^2}-\frac{1}{1+v^2}\right)\,dv\\&=&\color{red}{2-\frac{2}{\sqrt{3}}-\frac{\pi}{9}}.\end{eqnarray*}$$
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Hint: $$2\int_0^1{t^2 \over 1+xt^2}dt=2\int_0^1{(1/x)(1+xt^2-1) \over 1+xt^2}dt$$