How to find g(x) when applying this inequality rule? We are given the following inequality rule:

If $f(x) \le g(x)$, for $x \in [a, b]$, then
$$\int_a^b f \le \int_a^bg$$

Now we want to apply this to prove the following:
$$
\int_0^1 \frac{x^3}{2-\sin^4x} dx \le \frac{1}{4} \log(2)
$$
The solution in the book:
$$
|\sin x| \le |x| \\
\implies \sin^4 x \le x^4 \\
\implies 2-\sin^4 x\ge 2 - x^4 > 0 \\
\implies \frac{x^3}{2-\sin^4 x} \le \frac{x^3}{2 - x^4}
$$
for $x\in[0, 1]$. Then applying the given inequality rule is straightforward.
However, I'm pretty much stuck at step 1: given that we have an integral $\int_a^b f$ on the LHS and a constant $\frac{1}{4}\log(2)$ on the RHS, how do I get started to find $g(x)$ and hence proceed to set up the prerequisite $f(x) \le g(x)$?
I wouldn't even have thought about using $|\sin x| \le |x|$, from the given inequality. Is there a general principle that applies for solving a problem like this?
 A: Let
$$
f(x) = {x^3 \over 2 - \sin^4(x) }, \ \ \mbox{where} \ \ 0 \leq x \leq 1
$$
As finding an exact value for
$$
I = \int\limits_0^1 \ f(x) \ dx
$$
is a tough task, we try to find an upper bound bound for the value of $I$.
(Obviously, $I \geq 0$.)
In order for finding a suitable $g(x)$ such that
$$
f(x) \leq g(x), \ \ x \in [0, 1]
$$
we proceed as follows.
We know that for $x \in [0, 1]$,
$$
| \sin x | \leq | x | 
$$
which gives
$$
\sin^4 x \leq x^4
$$
Immediately,
$$
2 - \sin^4 x \geq 2 - x^4
$$
Hence, we have
$$
f(x) = {x^3 \over 2 - \sin^4 x} \leq g(x) = {x^3 \over 2 - x^4} \tag{1}
$$
Integrating both sides of (1) from $0$ to $1$, we get
$$
I = \int\limits_0^1 \ f(x) \ dx \leq J = \int\limits_0^1 {x^3 \over 2 - x^4} \ dx \tag{2}
$$
For evaluating $J$, we use the substitution
$$
u = 2 - x^4 \ \ \mbox{which gives} \ \ du = - 4 x^3 \ dx
$$
or
$$x^3 dx = -{1 \over 4} \ du
$$
Thus,
$$
J = - {1 \over 4} \ \int\limits_{2}^1 \ {du \over u} = 
= {1 \over 4} \ \int\limits_1^2 \ {du \over u}
$$
Integrating,
$$
J = {1 \over 4} \ \left[ \ln|u| \right]_1^2 = {1 \over 4} \ln 2
$$
Hence, we conclude that
$$
\int\limits_0^1 \ f(x) \ dx \leq {1 \over 4} \ \ln 2
$$
