All in "One"!

We note that, for $x \in [7,9]$ , $\lfloor x \{x\}\rfloor$ are integers having the range of $[0,\lfloor x \rfloor]$ (see figure above) and that $\sin^2 (\lfloor x \{x\}\rfloor) \ge 0$ , therefore, $\text{sgn}\left(\sin^2 (\lfloor x \{x\}\rfloor)\right) = \begin{cases} 0, & \lfloor x \{x\}\rfloor = 0 \\ 1, & \lfloor x \{x\}\rfloor > 0 \end{cases}$ .

We note that for $x \in [k,k+1)$ , where $k = 7, 8$ , there is an $x_{k1}$ such $\lfloor x \{x\}\rfloor \ge 1$ for $x \ge x_{k1}$ . Then the integral $\displaystyle I_k = \int_k^{k+1} \text{sgn}\left(\sin^2 (\lfloor x \{x\}\rfloor)\right) dx$ is the area when $\lfloor x \{x\}\rfloor \ge 1$ (the blue region in figure) or $I_k = 1 \times (k+1-x_{k1}) = \lfloor x \rfloor + 1 - x_{k1} = 1-\left \{x_{k1} \right \}$ . Now $\left \{x_{k1} \right\}$ is given by:

$\begin{aligned} x \{x\} & \ge 1 \\ (\lfloor x \rfloor + \{x\}) \{x\} & \ge 1 \\ (k + \{x\}) \{x\} & \ge 1 \\ \{x\}^2 + k\{x\} - 1 & \ge 0 \\ \implies \{x_{k1}\} & = \frac {\sqrt{(k^2+4)}-k}2 & \small \color{#3D99F6} \text{Note that } 0 \le \{x_{k1}\} < 1 \\ \{x_{71}\} & = \frac {\sqrt{53}-7}2 \\ \{x_{81}\} & = \frac {\sqrt{68}-8}2 \end{aligned}$ )

Now, we have:

$\begin{aligned} I & = \int_7^9 \text{sgn}\left(\sin^2 (\lfloor x \{x\}\rfloor)\right) dx \\ & = I_7 + I_8 \\ & = 1- \{x_{71}\} + 1- \{x_{81}\} \\ & = 2 - \frac {\sqrt{53}-7}2 - \frac {\sqrt{68}-8}2 \\ & = \frac {19 - (\sqrt{53} + \sqrt{68})}2 \end{aligned}$

$\implies a \times (b+c-30) = \boxed{1729}$

Even answer of this problem is framed beautifully