Showing the divergence of $\int_0^{\infty} \frac{1}{1+\sqrt{t}\sin(t)^2} dt$

How can I show the divergence of

$$\int_0^x \frac{1}{1+\sqrt{t}\sin(t)^2} dt$$

as $x\rightarrow\infty?$

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After word from the chat, it appears using \mathrm is fine. I just had not seen it used often (and I find it ugly!). Sorry I removed the mathrm in your post, feel free to add it back. – Joe Jun 1 '12 at 21:57

For $t \gt 0$:

$$1 + t \ge 1 + \sqrt{t}\sin^2t$$

Or:

$$\frac{1}{1 + t} \le \frac{1}{1 + \sqrt{t}\sin^2t}$$

Now consider:

$$\int_0^x \frac{dt}{1 + t} \le \int_0^x \frac{dt}{1 + \sqrt{t}\sin^2t}$$

The LHS diverges as $x \to +\infty$, so the RHS does too.

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Thank you for your answers! I am realizing that my question is trivial, why couldn't I find this simple inequality alone! – Chon Jun 2 '12 at 10:51

Use $1+\sqrt{t}\sin^2(t)\leqslant1+\sqrt{t}$ uniformly over $t$.

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