Checking $\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}} dx}$ for convergence 
Given $\displaystyle\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}$, prove that it converges.

So of course, I said:

We have to calculate $\displaystyle \lim_{b \to \infty} {\int_{0}^{\infty} {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}}$. And in order to do that we have to calculate the integral $\displaystyle \int {\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}  dx}$.

I got stuck calculating this integral, if anyone can give me a direction with this question I'll appreciate it!
 A: You have the inequalities
$$
0\le\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}\le
\frac{1}{\sqrt[3]{x^7 + 1}}\le
\frac{1}{x^{7/3}}
$$
so your integral will converge if
$$
\int_1^{\infty}\frac{1}{x^{7/3}}\,dx
$$
does. The change of the integration bounds is irrelevant, because
$$
\int_0^1\frac{\sin^2x}{\sqrt[3]{x^7 + 1}}\,dx
$$
poses no problem.
Now
$$
\int_1^{a}\frac{1}{x^{7/3}}\,dx=\left[-\frac{3}{4}x^{-4/3}\right]_1^a
=-\frac{3}{4}+\frac{3}{4}a^{-4/3}
$$
and
$$
\lim_{a\to\infty}a^{-4/3}=0
$$
It's quite improbable that you can find the "exact" value of the proposed integral, but the question was only to check for convergence.
A: You're not meant to calculate the integral, you're meant to check it converges.
Since the integrand is finite and continuous everywhere, the only possible thing which can go wrong is that it does not converge at infinity.
The behaviour at large $x$ is asymptotically $\sin^2 x^{-7/3}$. Now this is bounded by $x^{-7/3}$ which has a finite integral $\propto x^{-4/3}$ which converges at infinity; the result is that the integral does indeed converge. 
