I need to solve the integral
$$\int_{0}^{1}\int_{0}^{1} \sqrt{1 + 4(x^2 + y^2)}\,dx\,dy$$
I am using polar coordinates here to get :
$$ \int_{0} ^{\pi/4}\int_{0}^{\sec \theta} \sqrt{(1 + 4r^2)} r \,dr \,d\theta + \int_{\pi/4} ^{\pi/2}\int_{0}^{\operatorname{cosec}\theta} \sqrt{(1 + 4r^2)} r \,dr \,d\theta$$
After this integral becomes too complex to solve further . For eg : the first integral gives :
$$\int_{0}^{\pi/4}\frac1{12}{((1 + 4\sec^2\theta)^{3/2} - 1)}\, d\theta$$
After this I am stuck how to proceed further, Please help.
Thank You.
It boils down to integrate
$$I= \int_0^{\pi/4}(1+4\sec^2\theta)^{3/2}\,d\theta$$ Let $\sinh t= {\frac{2}{\sqrt{5}}}\tan\theta$ to proceed
\begin{align} I &=\int_0^{\sinh^{-1}\frac{2}{\sqrt{5}}} \frac{50\cosh^4 t}{4+{5}\sinh^2t }dt =\int_0^{\sinh^{-1}\frac{2}{\sqrt{5}}} \frac{25(1+\cosh 2t)^2}{3+5\cosh2t}dt\\ &=\int_0^{\sinh^{-1}\frac{2}{\sqrt{5}}} \left(5 \cosh2t + 7 + \frac{4}{3+5\cosh2t}\right)dt\\ &= 6 + \frac72 \ln{5}+\cot^{-1}3 \end{align}