How can I show the divergence of
$$ \int_0^x \frac{1}{1+\sqrt{t}\sin(t)^2} dt$$
as $x\rightarrow\infty?$
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.
Use $1+\sqrt{t}\sin^2(t)\leqslant1+\sqrt{t}$ uniformly over $t$.