\[\mathfrak{I}_1=\int_0^1x^5(1-x)^5\, dx=\dfrac{1}{a}\] Then \(\dfrac{\left\lfloor \sqrt{a}\right\rfloor}{4}-12=\color{red}{R}\)

\[\mathfrak{I}_2 =\lim_{n\rightarrow\infty}\dfrac{(1^2+2^2+\cdots+n^2)(1^4+2^4+\cdots+n^4)}{(1^7+2^7+\cdots+n^7)}=\color{red}{\dfrac{I}{S}}\]

\[\mathfrak{I}_3=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left(\dfrac{\pi+ 1996\phi^{1997}}{2-\cos(|\phi|+\frac{\pi}{3})}\right)\, d\phi\] such that \(\mathfrak{I}_3\) can be represented as \(\color{red}{\dfrac{H_1\pi}{\sqrt A}\tan^{-1} \left(\dfrac{B}{H_2}\right)}\)

Find:- \[\color{red}{\sqrt{\dfrac{R\times I\times S\times H_1 \times A\times B\times H_2}{5}}}\]

\[a,R,I,S,H_1,A,B,H_2~\in \mathbb Z\]
\(B\) and \(H_2\); \(I\) and \(S\) are coprime and \(A\) is square free.

\(\lfloor x\rfloor\) represents floor function while \(|x|\) represents modulus function.

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