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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.

StubbornAtom
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night_crawler
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    While it's going to wind up the same regardless, I think, you should be able to integrate without the transformation to polar coordinates; take the inner integral as $\int\ \sqrt{(1+4y^2)+4x^2}\ dx$, evaluate this via the usual means, and then plug in the limits; you should get a complicated but manageable expression for the result that can then be integrated with respect to $y$. – Steven Stadnicki Feb 03 '21 at 19:01
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    Use the substitution $5\cosh^2t = 1+4\sec^2\theta$. For more details on why this works, see this post – Ninad Munshi Feb 03 '21 at 19:09

1 Answers1

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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}

Quanto
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