\[\Large\displaystyle \int_0^{1} \frac{\arccos^4 \left(x^2\right)}{\sqrt{1-x^2}}\,\mathrm{d}x\]

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\(\Large\mathbf{Solution:}\)
Let \(I\) denote the integral. Using the substitution \(x=\sqrt{\cos t}\), we get
\[I=\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{t^4\sin(t)}{\sqrt{\cos t-\cos^2t}}\mathrm{d}t=\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\frac{t^4\cos\left(\dfrac{t}{2} \right)}{\sqrt{1-2\sin^2\left(\dfrac{t}{2} \right)}}\mathrm{d}t\]
Next, substitute \(\displaystyle 2\sin^2\left(\frac{t}{2}\right)=\sin^2\theta\) to get
\[I=16\int_0^{\frac{\pi}{2}}\left[\sin^{-1}\left(\frac{\sin\theta}{\sqrt{2}} \right) \right]^4\; \mathrm{d}\theta\]
Now, plug in the series \(\displaystyle (\sin^{-1}z)^4=\sum_{k=1}^\infty \frac{H_{k-1}^{(2)}(2z)^{2k}}{k^2\dbinom{2k}{k}}\) and integrate termwise:
\[\begin{align*} I &= 24\int_0^{\frac{\pi}{2}}\left(\sum_{k=1}^\infty \frac{H_{k-1}^{(2)}2^k}{k^2\dbinom{2k}{k}}\sin^{2k}(\theta) \right)\; \mathrm{d}\theta= 24\sum_{k=1}^\infty \frac{H_{k-1}^{(2)}2^k}{k^2\dbinom{2k}{k}}\int_0^{\frac{\pi}{2}}\sin^{2k}(\theta)\mathrm{d}\theta \\ &= 12\pi \sum_{k=1}^\infty \frac{H_{k-1}^{(2)}2^k}{k^2\dbinom{2k}{k}}\left(\frac{\dbinom{2k}{k}}{2^{2k}} \right)= 12\pi\sum_{k=1}^\infty \frac{H_{k-1}^{(2)}}{k^2 2^k}= 12\pi\sum_{k=1}^\infty \frac{\zeta(2)-\psi_1(k)}{k^2 2^k} \\ &= 12\pi\zeta(2)\text{Li}_2\left(\frac{1}{2}\right)-12\pi\sum_{n=1}^\infty\frac{\psi_1(k)}{k^2 2^k} \end{align*}\]

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For the last series
\[\begin{align*} \sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2} &= -\sum_{n=1}^\infty\psi_1(n)\left(\frac{\ln(2)}{2^n n}+\int_0^{\frac{1}{2}} x^{n-1}\ln(x)\mathrm{d}x\right) \\ &= -\ln(2)\sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}-\int_0^{1\over 2}\frac{\ln(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)\mathrm{d}x \tag{1} \end{align*}\]
1.Evaluation of\(\displaystyle \int_0^{1\over 2}\frac{\ln(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)\mathrm{d}x\)
Using the identity \(\text{Li}_2(x)+\text{Li}_2(1-x)=\zeta(2)-\ln(x)\ln(1-x)\), we have
\[\begin{align*} \int_0^{1\over 2}\frac{\ln(x)}{1-x}\left(\zeta(2)-\text{Li}_2(x) \right)\mathrm{d}x &= \color{red}{\int_0^{1\over 2}\frac{\ln^2(x)\ln(1-x)}{1-x}\mathrm{d}x}+\int_0^{1\over 2}\frac{\ln(x)\text{Li}_2(1-x)}{1-x}\mathrm{d}x \\ &={-\frac{\pi^4}{360}-\frac{\ln^4(2)}{4}}+{\left(\frac{1}{2}\text{Li}_2^2(1-x) \right)\Bigg|_0^{1\over 2}} \\ &= -\frac{19\pi^4}{1440}-\frac{\pi^2}{24}\ln^2(2)-\frac{\ln^4(2)}{8}\tag{2} \end{align*}\]
The red integral was evaluated using equation 12 on page 310 of Leonard Lewin - Polylogarithms and associated functions. The proof is discussed on page 203 and 204.
2.Evaluation of\(\displaystyle \sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}\)
\[\begin{align*} \sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n}& = \sum_{n=1}^\infty \psi_1(n)\int_0^{1\over 2}x^{n-1}\mathrm{d}x \\
&= \int_0^{\frac{1}{2}}\frac{\zeta(2)-\text{Li}_2(x)}{1-x}\mathrm{d}x\\
&= \color{Purple}{\int_0^{1\over 2}\frac{\ln(x)\ln(1-x)}{1-x}\mathrm{d}x}+\int_0^{1\over 2}\frac{\text{Li}_2(1-x)}{1-x}\mathrm{d}x\\
&= {\frac{\ln^3(2)}{2}+\frac{1}{2}\int_0^{\frac{1}{2}}\frac{\ln^2(1-x)}{x}\mathrm{d}x }+{\zeta(3)-\text{Li}_3\left(\frac{1}{2}\right)} \\ &= {\frac{\ln^3(2)}{2}+\frac{1}{2}\left( -\ln^3(2)-2\ln(2)\text{Li}_2\left(\frac{1}{2}\right)+2\zeta(3)-2\text{Li}_3\left( \frac{1}{2}\right)\right) }\\
&~~~+{\zeta(3)-\text{Li}_3\left(\frac{1}{2}\right)} \\ &= \frac{\pi^2}{12}\ln(2)+\frac{\ln^3(2)}{6}+\frac{\zeta(3)}{4}\tag{3} \end{align*}\]
The purple integral was evaluated using the generalized result found in this thread.
3.The Final Answer
Substitute (2) and (3) into (1) to get
\[\large\boxed{\displaystyle \sum_{n=1}^\infty\frac{\psi_1(n)}{2^n n^2}=\color{Teal}{\frac{19\pi^4}{1440}-\frac{\pi^2}{24}\ln^2(2)-\frac{\ln^4(2)}{24}-\frac{\zeta(3) \ln(2)}{4}}}\]

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Hence The final result of the initial integral follows from the above answer
\[\Large\boxed{\displaystyle \int_0^{1} \frac{\arccos^4 \left(x^2\right)}{\sqrt{1-x^2}}\,\mathrm{d}x=\color{Blue}{ \frac{\pi^5}{120}+\frac{\pi }{2}\ln^4 2 + 3\pi \zeta(3)\ln 2 - \frac{\pi^{3} }{2}\ln^2 2}}\]