# My First Problem

Calculus Level 5

$\mathfrak{I}_1=\int_0^1x^5(1-x)^5\, dx=\dfrac{1}{a}$ 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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