关于Ahmed积分变体及类似积分的替代推导方法问询

我最近推导出了四个与Ahmed积分形式相近的定积分，具体结果如下：

$$
\begin{aligned}
&I_1=\int_{0}^{1}
\frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} }
\arcsin\left ( \sqrt{\frac{2-y^2}{4} } \right )
\text{d}y = \frac{11\pi^2}{288},\
&I_2=\int_{0}^{1}
\frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} }
\arcsin\left ( \sqrt{\frac{1-y^2}{3} } \right )
\text{d}y = \frac{\pi^2}{8}-\frac{\pi}{2} \arcsin\left ( \frac{1}{\sqrt{3}} \right ),\
&I_3=\int_{0}^{1}
\frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} }
\arctan\left ( \frac{\sqrt{(1-y^2 )\left ( 2+y^2 \right ) }} {2}\right )
\text{d}y
=\pi\arctan\left ( \frac{1}{\sqrt{2} } \right ) -\frac{\pi^2}{6},\
&I_4=\int_{0}^{1}
\frac{1}{\left ( 1+y^2 \right )\sqrt{2+y^2} }
\arctan\left ( {\sqrt{\frac{4-y^4}5}}\right )
\text{d}y
=\frac\pi2\arctan\left ( \sqrt{\frac35} \right ) -\frac{\pi^2}{15}.\
\end{aligned}
$$

这些结果我是通过一些复杂的技巧推导得到的，不过我还想到了两种替代推导思路，……

备注：内容来源于Stack Exchange，提问作者蜜柑しぇっり