The value of $\iint_{A}\frac{r dr d\theta}{({r^{2}+a^{2})}^{\frac{1}{2}}}$,where $A$ is a loop of $r^{2}=a^{2}cos(2\theta)$ is to be calculated .
$\int_{\theta=\frac{-\pi}{4}}^{\theta=\frac{\pi}{4}}\int_{r=0}^{(a^{2}cos(2\theta))^{\frac{1}{2}}}\frac{r dr d\theta}{({r^{2}+a^{2})}^{\frac{1}{2}}}+ \int_{\theta=\frac{3\pi}{4}}^{\theta=\frac{5\pi}{4}}\int_{r=0}^{(a^{2}cos(2\theta))^{\frac{1}{2}}}\frac{r dr d\theta}{({r^{2}+a^{2})}^{\frac{1}{2}}}$
$=\frac{1}{2}\int_{\theta=\frac{-\pi}{4}}^{\theta=\frac{\pi}{4}}\int_{r=0}^{(a^{2}cos(2\theta))^{\frac{1}{2}}}\frac{r dr d\theta}{({r^{2}+a^{2})}^{\frac{1}{2}}}+ \frac{1}{2}\int_{\theta=\frac{3\pi}{4}}^{\theta=\frac{5\pi}{4}}\int_{r=0}^{(a^{2}cos(2\theta))^{\frac{1}{2}}}\frac{r dr d\theta}{({r^{2}+a^{2})}^{\frac{1}{2}}}$
$=\int_{\theta=\frac{-\pi}{4}}^{\theta=\frac{\pi}{4}} [({a^{2}cos(2\theta)+a^{2})}^{\frac{1}{2}}-a] +\int_{\theta=\frac{3\pi}{4}}^{\theta=\frac{5\pi}{4}} [({a^{2}cos(2\theta)+a^{2})}^{\frac{1}{2}}-a]$
$=a\int_{\theta=\frac{-\pi}{4}}^{\theta=\frac{\pi}{4}} [({cos(2\theta)+1)}^{\frac{1}{2}}-1] +a\int_{\theta=\frac{3\pi}{4}}^{\theta=\frac{5\pi}{4}} [({cos(2\theta)+1)}^{\frac{1}{2}}-1]$
$=a\int_{\theta=\frac{-\pi}{4}}^{\theta=\frac{\pi}{4}} [(2^{\frac{1}{2}}cos(\theta)-1]+ a\int_{\theta=\frac{3\pi}{4}}^{\theta=\frac{5\pi}{4}} [(2^{\frac{1}{2}}cos(\theta)-1]$
$=[2a]+[-2\sqrt{2}a-\dfrac{\pi}{2}]$
Is it correct now.
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