<s>[INST] How can I evaluate $\int_0^{\pi/2}\frac{x\cos{x}}{3\sin^2x+1}dx$ and $\int_0^{\pi/2}\frac{x\cos{x}}{\sin^2x+3}dx$? I do not find the closed form of the following integrals$$\int_0^{\pi/2}\frac{x\cos{x}}{3\sin^2x+1}\mathrm dx$$
$$\int_0^{\pi/2}\frac{x\cos{x}}{\sin^2x+3}\mathrm dx$$
On the other side, I find
$$\int_0^{\pi/2}\frac{x(1+\sin^2x)\cos{x}}{(3\sin^2x+1)(\sin^2x+3)}\mathrm dx=-\frac{\sqrt3}{24}\ln(3)\ln{(2-\sqrt3)}$$
 [/INST] Integrating by parts, we get
\begin{align*}
\int_{0}^{\pi/2} \frac{x\cos x}{3\sin^{2}x + 1} \, dx
&= \left[ \frac{x}{\sqrt{3}} \arctan(\sqrt{3}\sin x) \right]_{0}^{\pi/2} - \frac{1}{\sqrt{3}} \int_{0}^{\pi/2} \arctan(\sqrt{3}\sin x) \, dx \\
&= \frac{\pi^{2}}{6\sqrt{3}} - \frac{1}{\sqrt{3}} \int_{0}^{\pi/2} \arctan(\sqrt{3}\sin x) \, dx.
\end{align*}
Now by noting the identity
$$ \int_{0}^{\pi/2} \arctan(r\sin x) \, dx = 2\chi_{2}\left(\frac{\sqrt{1+r^{2}} - 1}{r} \right), $$
where $\chi_{2}$ is the Legendre chi function of order 2, it follows that
$$ \int_{0}^{\pi/2} \frac{x\cos x}{3\sin^{2}x + 1} \, dx
= \frac{\pi^{2}}{6\sqrt{3}} - \frac{2}{\sqrt{3}} \chi_{2}\left( \frac{1}{\sqrt{3}} \right). $$
There are only a handful of cases where the exact value of $\chi_{2}(z)$ are known. And unfortunately $z = 3^{-1/2}$ is not the case. Similarly,
$$ \int_{0}^{\pi/2} \frac{x\cos x}{\sin^{2}x + 3} \, dx
= \frac{\pi^{2}}{12\sqrt{3}} - \frac{2}{\sqrt{3}} \chi_{2}(2-\sqrt{3}). $$
</s>