I answered the volume question Find the volume between the regions $x^2 + y^2+ z^2 = 4$ and $x = 4-y^2$.
But , then I faced two nasty integrals. One of them is $$I=\int_{\sqrt{3}}^{2}\sqrt{(y^2-3)(4-y^2)}dy.$$
Let us substitute $y=2\sin\theta$. Then,
$$I=\int_{0}^{\large\frac{\pi}{6}}4\sin^2\theta\sqrt{1-4\sin^2\theta}d\theta$$
WolframAlpha found it : https://www.wolframalpha.com/input?key=&i=int_0%5E%28pi%2F6%29sin%5E2tsqrt%281-4sin%5E2t%29dt But unfortunately I can not find it that easily although I can do the complete integral by a formula. I saw the formula here: https://arxiv.org/pdf/2102.09459.pdf
Any help or positive comment is appreciated. Thanks.
(Only subgenius friends can take a joke. I am writing this for friend Em.)
Notation for (complete) elliptic integrals: $$ K(k):=\int_0^1\frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}dt, \qquad E(k):=\int_0^1\frac{\sqrt{1-k^2t^2}}{\sqrt{1-t^2}}dt. $$ By simple algebra we also have $$ \int_0^1\frac{t^2}{\sqrt{(1-t^2)(1-k^2t^2)}}dt=\frac{K(k)-E(k)}{k^2}. $$ Integrating by parts we get $$ I:=\int_{\sqrt 3}^2\sqrt{(y^2-3)(4-y^2)}dy=\int_{\sqrt 3}^2\frac{2y^4-7y^2}{\sqrt{(y^2-3)(4-y^2)}}dy. $$ We have $$ 2y^4-7y^2=-2(y^2-3)(4-y^2)+7y^2-24 $$ hence $$ I=-2I+\int_{\sqrt 3}^2\frac{7y^2-24}{\sqrt{(y^2-3)(4-y^2)}}dy, $$ i.e., $$ I=\frac 13\int_{\sqrt 3}^2\frac{7y^2-24}{\sqrt{(y^2-3)(4-y^2)}}dy. $$ It is convenient to change integration variable by $y=2t$ to obtain \begin{align*} I&=\frac{4i}{3\sqrt 3}\int_{\sqrt 3/2}^1\frac{7t^2-6}{\sqrt{(1-t^2)(1-\frac 43t^2)}} \\ &=\Re\left(\frac{4i}{3\sqrt 3}\int_{0}^1\frac{7t^2-6}{\sqrt{(1-t^2)(1-\frac 43t^2)}}\right) \\ &=\Re\left(\frac{4i}{3\sqrt 3}\left(7\frac{K(\tfrac 2{\sqrt3})-E(\tfrac 2{\sqrt3})}{4/3}-6K(\tfrac 2{\sqrt3})\right)\right) \\ &=\Im\left(\frac{K(\tfrac 2{\sqrt3})+7E(\tfrac 2{\sqrt3})}{\sqrt 3}\right). \end{align*} One can further simplify the result by splitting real and imaginary parts (see also Separate elliptic integrals into real and imaginary parts). In particular we use DLMF 19.7.3 \begin{align*} K(k^{-1})&=k\left(K(k)-i K(k')\right),\\ E(k^{-1})&=k^{-1}\left(E(k)+iE(k')-{k'}^2K(k)-i k^2K(k')\right), \end{align*} where $k':=\sqrt{1-k^2}$ with $k=\sqrt 3/2$, $k'=1/2$ to obtain $$ \Im(K(2/\sqrt 3))=-\frac{\sqrt 3} 2K(1/2),\qquad \Im(E(2/\sqrt 3))=\frac 2{\sqrt 3}E(1/2)-\frac{\sqrt 3}2K(1/2) $$ so that finally $$ I=\Im\left(\frac{K(\tfrac 2{\sqrt3})+7E(\tfrac 2{\sqrt3})}{\sqrt 3}\right)=\frac {14}3E(1/2)-4K(1/2). $$ This agrees with the linked computation in Wolfram Alpha (up to the factor 4 omitted in the integral and up to the different convention in the notation for elliptic integrals, which in WA are expressed as functions of $k^2$ rather than $k$, explaining the argument $1/4$ rather than $1/2$ in the elliptic integrals).