Determining whether $\int_{1}^{+\infty}\frac{\sin^3 \left(x\right)}{\sqrt {x^2}}\,\mathrm{d}x $ converges or diverges.

Question

I's struggling with an integral, and not sure wich method I should use to determining whether it converges or diverges.

I know, from a software, that it should converge.

The integral is: $$ \displaystyle\int_{1}^{+\infty}\dfrac{\sin^3 \left(x\right)}{\sqrt {x^2}}\,\mathrm{d}x $$

I've been looking all over the forum for some clues with no success. Maybe I'm not using the right function to compare.

It's the square root (of a square number) in the denominator that's puzzling me. Why did the text book made this writting? Maybe it's reffering to the modulus function?

Thank you for your attention.

Answer 1

Hint: $\sin 3x = 3 \sin x - 4 \sin^3 x \implies$

$$\int_1^c \frac{\sin^3 x}{x} \, dx = \frac{3}{4}\int_1^c \frac{\sin x}{x} \, dx - \frac{1}{4}\int_1^c \frac{\sin 3x}{x} \, dx$$