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!
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!
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 $\int\limits_1^\infty\frac{dx}{x^{3/2}}=2$