show using the comparison theorem that the function is divergent or convergent. $$\int_{3}^{\infty}\frac{1}{\sqrt{x}-\sin x}\mathrm{d}x$$
I used $$\ 1/\sin x$$ as a function to compare but when doing its integral as $x$ reaches infinity the integral would become undefined. what should I do?
 A: I explained in my comment on how you can show
$$ \frac{1}{\sqrt{x}+1} \leq \frac{1}{\sqrt{x}-\sin(x)} \leq \frac{1}{\sqrt{x}-1}.$$
Let $u = \sqrt{x}$, $du = \frac{dx}{2\sqrt{x}}$, so we can rewrite the lower bound as
$$ \int_3^\infty \frac{dx}{\sqrt{x}+1} = 2\int_\sqrt{3}^\infty \frac{u}{u+1}du.$$
Note
$$ 1 - \frac{1}{u+1} = \frac{u}{u+1},$$
so rewrite the above as
$$2\int_\sqrt{3}^\infty \frac{u}{u+1}du = 2\int_\sqrt{3}^\infty \left( 1 - \frac{1}{u+1}\right)du  = 2 \lim_{s \rightarrow \infty} \int_{\sqrt{3}}^s\left( 1 - \frac{1}{u+1}\right)du  \\ = 2 \lim_{s \rightarrow \infty} \left[ u - \log(u+1) \bigg|_{u=\sqrt{3}}^s \right]  = 2 \lim_{s \rightarrow \infty} \left[ s - \log(s+1) - \sqrt{3} + \log(\sqrt{3} +1)\right] \\ = \infty.$$
Note I'm being a little reckless and assumming you know $\lim_{s \rightarrow \infty} (s - \log(s+1)) = \infty.$
In case you don't know this, notice that
$$s - \log(s+1) = \log(\exp(s - \log(s+1))).$$
Examining the inside, we see
$$ \exp(s - \log(s+1)) = \frac{\exp(s)}{\exp(\log(s+1))} = \frac{\exp(s)}{s+1}.$$
Use L'Hospital to get
$$\lim_{s \rightarrow \infty} \frac{\exp(s)}{s+1} = \lim_{s \rightarrow \infty}\frac{\exp(s)}{1} = \infty.$$
Moreover $\log$ is continuous on its domain so
$$\lim_{s \rightarrow \infty} (s - \log(s+1)) = \lim_{s \rightarrow \infty} \log\left( \frac{\exp(s)}{s+1}\right) \\ = \log \left( \lim_{s \rightarrow \infty} \frac{\exp(s)}{s+1}\right) = \infty.$$
A: Hint
As pointed just above, we have
$$(\forall x\ge 3)\; \sqrt{x}\ge 1$$
and
$$0<\sqrt{3}-1\le \sqrt{x}-\sin(x)\le \sqrt{x}+1\le 2\sqrt{x}$$
thus
$$\frac{1}{2\sqrt{x}}\le \frac{1}{\sqrt{x}-\sin(x)}$$
but
$$\int_3^{+\infty}\frac{dx}{\sqrt{x}}\; diverges, \; so$$
