TITLE: Trigonometric solution in a double integral?
QUESTION [3 upvotes]: I have to evaluate the given double integral:
$$\int_0^1 \int_0^2 \sqrt{4u^2+4v^2+1} \, du \, dv$$
It seems to me the only way to evaluate this is with a trigonometric substitution. However, even if I could figure out what kind of substitution could simplify this, I can't figure out how I would reevaluate the limits of the double integral with the substitution(s) I chose. Is it even possible to do this with double integrals?

REPLY [2 votes]: In probabilistic terms, we want to compute $2\cdot\mathbb{E}[\sqrt{16X^2+4Y^2+1}]$ where $X$ and $Y$ are independent random variables, uniformly distributed over $[0,1]$. The PDF of $X^2$ is supported on $(0,1]$ and given by $\frac{1}{2\sqrt{t}}$. By convolution, the PDF of $16X^2+4Y^2$ is supported on $[0,20]$ and given by $2\pi$ for $t\in[0,4]$, by $4\arctan\frac{2}{\sqrt{t-4}}$ for $t\in[4,16]$ and by $4\arctan\frac{2}{\sqrt{t-4}}-4\,\text{arccot}\frac{4}{\sqrt{t-16}}$ for $t\in[16,20]$.
The original integral is so given by $8(J_1+J_2+J_3)$ where:
$$ J_1 = \frac{1}{2}\int_{0}^{4}\pi\sqrt{t+1}\,dt = \frac{\pi}{3}(5\sqrt{5}-1), $$
$$ J_2 = \int_{4}^{20}\sqrt{t+1}\,\arctan\frac{2}{\sqrt{t-4}}\,dt $$
$$ J_3 = -\int_{16}^{20}\sqrt{t+1}\,\text{arccot}\frac{4}{\sqrt{t-16}}\,dt. $$
$J_2$ and $J_3$ are not so pleasant to compute, but integration by parts and suitable substitutions give that they boil down to integrals of rational functions, given by combinations of $\pi$, the square root of some natural numbers and values of $\arctan$ and $\text{arcsinh}$. The final form is:
$$\large\scriptstyle \frac{1}{24} \left(16 \sqrt{21}-2 \arctan\left(\frac{8}{\sqrt{21}}\right)-7 \log(5)+76\log(2+\sqrt{21})+14 \log\left(4+\sqrt{21}\right)-38\log(17)\right)$$
and the numerical value of it is $\approx\color{red}{5.23352}$.

REPLY [0 votes]: $u=r\cos\theta$ and $v=r\sin\theta$, and we have $r=\frac{2}{\cos\theta}$ and $r=\frac{1}{\sin\theta}$ and the determinat of matrix jacobian is $r$
$$\int_0^{\arctan\left(\frac{1}{2}\right)}\int_0^{\frac{2}{\cos\theta}}\sqrt{1-r^2} \,r \, dr \, d\theta + \int_{\arctan\left(\frac{1}{2}\right)}^{\pi/2} \int_0^{\frac{1}{\sin\theta}}\sqrt{1-r^2} \, r \, dr \, d\theta$$
we solve the second integral.
$$\int_{\arctan\left(\frac{1}{2}\right)}^{\frac{\pi}{2}}\int_0^{\frac{1}{\sin\theta}} r\sqrt{1-r^2}\,dr=\frac{2}{3} \int_{\arctan\left(\frac{1}{2}\right)}^{\frac{\pi}{2}}\left[-(1-r^2)^{\frac{3}{2}}\right|_0^{\frac{1}{\sin\theta}}\,d\theta$$
$$=\frac{2}{3}\int\limits_{\arctan\left(\frac{1}{2}\right)}^{\frac{\pi}{2}} 1-\cot^3\theta\,d\theta=\frac23\left(\frac{\pi}{2}-\arctan\left(\frac12\right)\right)-\frac{2}{3}\int\limits_{\arctan\left(\frac{1}{2}\right)}^{\frac{\pi}{2}} \cot^3\theta\,d\theta$$
now we focus our attention to solve the last integral
$$\int\cot^3\theta\,d\theta=\int\frac{\cot\theta}{\sin^2\theta}-\cot\theta\,d\theta=-\frac12\cot^2\theta-\ln(\sin\theta)$$
god bless you
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