The question is basically how does the curve behave when x is close to $0$.

A rough estimate would be to use cos x = 1 and sin x = x when x is close to $0$.

Then $\frac{2 + \cos x}{3\sqrt x – x^2 \sin x} \approx

\frac{3}{3\sqrt x – x^3} \approx \frac{1}{\sqrt x}$

when $x$ is close to $0$ and

$\int_0^\epsilon \frac{dx}{\sqrt x}$ is finite.

This is evocative, but it is not proof.

So let’s look at

$$\frac{1}{\sqrt x} – \frac{2 + \cos x}{3\sqrt x – x^2 \sin x} =

\frac{1}{\sqrt x} \left(1 – \frac{2 + \cos x}{3 – x^{3/2} \sin x}\right)$$

Pick some small positive $\epsilon$, say $\epsilon = \frac{1}{10}$ if you insists on a number. For $0 \le x \le \epsilon$,

- $0 \le x^{3/2}\sin x < \epsilon$
- $1 \ge \cos x > 1 – \epsilon^2$

It follows that

- $ 2 + \cos x > 3 – \epsilon^2$
- $ 3 – x^{3/2} \sin x \le 3$

Which gives us

$$1 – \frac{2 + \cos x}{3 – x^{3/2} \sin x} < 1 \text{ when } 0 \le x \le \epsilon$$

So

$$

0 < \int_0^{\epsilon} \frac{2 + \cos x}{3\sqrt x – x^2 \sin x}dx

< \int_0^{\epsilon} \frac{dx}{\sqrt x}

$$

Which is enough to prove that the integral exists.