# Improper Integral inequality

Explain why the following inequality is true

$$0\le \int_1^\infty \frac{\sin^2(x)}{\sqrt{x^3+x}} dx \le 2$$

Any help will be greatly appreciated!

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Hint: The inequality $0\le{ (\sin x)^2\over \sqrt{x^3+x}}\le {1\over x^{3/2}}$ is valid for $x\ge1$.
Given that $$0<\sin^2 x\leq1$$ and $$\frac{1}{\sqrt{x^3+x}}<\frac 1 {x^{3/2} }\text{ for }x>1$$ you have that
$$0<\frac{{{{\sin }^2}x}}{{\sqrt {{x^3} + x} }} \leq \frac{1}{{{x^{3/2}}}}$$
But then $$\int\limits_1^\infty\frac{dx}{x^{3/2}}=2$$