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 $x\rightarrow\infty?$

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.
