Sum, floor and inverse

We note that $S_6 = \displaystyle \sum_{x=1}^6 x! = 873 \equiv 5 \text{ (mod 7)}$ and that for $n > 6$ , $S_n - S_6 = \displaystyle \sum_{\color{#D61F06} x=7}^6 x! \equiv 0 \text{ (mod 7)}$ . Therefore, for $n \ge 6$ , $S_n \equiv 5 \text{ (mod 7)}$ , $\implies S_n = 7 \left \lfloor \dfrac {S_n}7 \right \rfloor + 5$ and $\color{#3D99F6}S_n - 7 \left \lfloor \dfrac {S_n}7 \right \rfloor = 5$ .

Then, we have:

$\begin{aligned} T & = \arcsin \left(\sin \left({\color{#3D99F6}S_n - 7 \left \lfloor \frac {S_n}7 \right \rfloor}\right) \right) \\ & = \arcsin \left(\sin \left({\color{#3D99F6}5}\right) \right) \\ & = \color{#3D99F6}5 - 2\pi & \small \color{#3D99F6} \text{as } \arcsin \theta \text{ is defined within } -\frac \pi 2 \le \theta \le \frac \pi 2 \end{aligned}$

And

$\begin{aligned} I & = \int_0^1 \frac T{\sqrt{1-x^2}} dx & \small \color{#3D99F6} \text{Let }x = \sin \theta, \ dx = \cos \theta \ d\theta \\ & = \int_0^\frac \pi 2 \frac {(5-2\pi)\cos \theta}{\cos \theta} d\theta \\ & = (5-2\pi) \theta \ \bigg|_0^\frac \pi 2 \\ & = \frac 52\pi - \pi^2 \end{aligned}$

$\implies \dfrac bc + a = \dfrac 22 + 5 = \boxed{6}$

How much are you scoring in today's test @Tapas Mazumdar ?