My First Problem

For calculating $\mathfrak{I}_1$ do the substitution $x=\sin^2\theta$ such that $dx=\sin 2\theta d\theta$ and integral transforms to:- $\int_0^{\frac{\pi}{2}} sin^{10}\theta (1-\sin^2\theta)^5 \sin 2\theta d\theta$ $=2\int_{0}^{\pi}\sin^{11}\theta\cos^{11}\theta$ Now recalling Beta Functions $\Large{*}$ : $\mathfrak{I}_1=B(6,6)=\dfrac{5!\times 5!}{11!}=\dfrac{1}{2772}$ Hence:- $\lfloor\dfrac{\sqrt{2772}}{13}\rfloor-12=\color{#D61F06}{1}$

Write $\mathfrak{I}_2$ as:- $\lim_{n\to \infty}\dfrac{1}{n}\left(\dfrac{\left(\displaystyle\sum_{r=1}^{\infty}\left(\frac rn\right)^2\right)\left(\displaystyle\sum_{r=1}^{\infty}\left(\frac rn\right)^4\right)}{\displaystyle\sum_{r=1}^{\infty}\left(\frac rn\right)^7 }\right)$ and apply Reimann Sums to get : $\mathfrak{I}_2= \dfrac{(\int_0^1x^2 dx)(\int_0^1x^4 dx)}{(\int_0^1x^7 dx)}$

$=\dfrac{(\frac 13)(\frac 15)}{(\frac 18)}=\color{#D61F06}{\dfrac{8}{15}}$ For $\mathfrak{I_3}$ , apply $\int_{-a}^a f(x)dx=2\int_{0}^af(x)dx$ if $f(x)$ is an even function and $\int_{-a}^a f(x)dx=0$ if $f(x)$ is a odd function. $\mathfrak{I}_3=2\int_{0}^{\frac{\pi}{3}} \left(\dfrac{\pi}{2-\cos(\phi +\frac{\pi}{3})}\right)d\phi+\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left(\dfrac{1996\phi^{1997}}{2-\cos(|\phi|+\frac{\pi}{3})}\right)\, d\phi$ $=2\pi\int_{\frac{\pi}{3}}^{\frac{2\pi}{3}} \dfrac{d\phi}{2-\cos \phi}+0$ Now substitute $\tan \frac{\phi}2=x$ and use $cos ~\phi=\dfrac{1-\tan^2\frac{\phi}2}{1+\tan^2 \frac{\phi}2}$ $\implies \mathfrak{I}_3=4\pi\int_{\frac{1}{\sqrt 3}}^{\sqrt 3}\dfrac{d(\tan x)}{3\tan^2 x+1}$ $=\dfrac{4\pi}{\sqrt 3}(\tan^{-1} \sqrt 3 x)|_{\frac{1}{\sqrt 3}}^{^{\sqrt 3}}$ $\color{#D61F06}{=\dfrac{4\pi}{\sqrt 3}\tan^{-1}\frac 12}$ $\large \color{#D61F06}{\therefore \sqrt{\dfrac{1\times 8\times 15\times 4\times 3\times 1\times 2}{5}}}$ $\huge =\boxed{24}$ $\Large *$ -An alternate approach here may be Using $I.B.P$ on $\mathfrak{I}(n)=\int_0^1(\sin x)^n\, dx$

Phew!! ;-)