I hope I created a quite clear Problem

Solving the definite integral in the denominator:

$\begin{aligned} \displaystyle \int_0^{{\pi}/2} \sin^9 \theta \,d \theta &= \dfrac 12 \cdot 2 \int_0^{{\pi}/2} \sin^9 \theta \,d \theta \\ &= \dfrac 12 \cdot B \left( 5, \dfrac 12 \right) \\ &= \dfrac 12 \cdot \dfrac{ \Gamma (5) \Gamma \left( \frac 12 \right) }{\Gamma \left( \frac{11}{2} \right)} & \small{\color{#3D99F6} \text{Using } B(m,n) = \dfrac{\Gamma(m) \Gamma(n)}{\Gamma(m+n)}} \\ &= \dfrac 12 \cdot \dfrac{256}{315} \\ &= \dfrac{128}{315} \end{aligned}$

Thus $f(x) = \sqrt{\dfrac{ (4-|x|) (|x|+2) }{ (10-|x|) \left( \frac{128}{315} - |x| \right) }}$ (note that $\sqrt{x^2} = |x|$ ).

Hence for $f(x)$ to be defined over $\mathbb{R}$ we must have

$\dfrac{ (4-|x|) (|x|+2) }{ (10-|x|) \left( \frac{128}{315} - |x| \right) } \ge 0$

or equivalently

$(4-|x|) (|x|+2) (10-|x|) \left( \frac{128}{315} - |x| \right) \ge 0 \qquad \qquad \small \color{#3D99F6} \left( \text{excluding } \left\{ \dfrac{128}{315} , 10 \right\} \right)$

Assume $|x| = t$ , then we have

$\begin{aligned} & (4-t) (t+2) (10-t) \left( \frac{128}{315} - t \right) \ge 0 \\ \implies & (t-4) (t+2) (t-10) \left( t - \frac{128}{315} \right) \le 0 & \small \color{#3D99F6} \left( \text{excluding } \left\{ \dfrac{128}{315} , 10 \right\} \right) \end{aligned}$

From above, we get

$t \in \left[ -2 , \dfrac{128}{315} \right) \cup [4,10)$

Since $t = |x| \ge 0$ , so we have

$t \in \left[ 0 , \dfrac{128}{315} \right) \cup [4,10) \implies |x| \in \left[ 0 , \dfrac{128}{315} \right) \cup [4,10)$

Hence

$x \in (-10,-4] \cup \left( -\dfrac{128}{315} , \dfrac{128}{315} \right) \cup [4,10)$

Thus, we have

$\left\{ x : x \in \mathbb{Z} \land x \in (-10,-4] \cup \left( -\dfrac{128}{315} , \dfrac{128}{315} \right) \cup [4,10) \right\} \implies x \in \{-9,-8,-7,-6,-5,-4,0,4,5,6,7,8,9\}$

Giving us a total of $\boxed{13}$ solutions for integer values of $x$ .

For the function $f(x)$ presented here we require the radicand to be strictly non-negative. The definite integral in the denominator computes to $\frac{128}{315}$ (try it yourselves), and $\sqrt{x^2} = |x|$ in each factored term. This gives us:

$f(x) = \sqrt{ \frac{(4 - |x|)(2 + |x|)}{(10 - |x|)(128/315 - |x|)}}.$

The radicand will be non-negative for all $x$ in the domain $(-10, -4] \cup (-128/315, 128/315) \cup [4, 10)$ . The integral values included in this domain are:

$x = {-9, -8, -7, -6, -5, -4, 0 , 4, 5, 6, 7, 8, 9}$ ,

or $\boxed{13}$ in total.