Let $I = \int_2^\infty \frac{sin^2(x)}{x^\alpha(x^\alpha + \sin(x))}dx$. Then $I$ is divergent iff $\exists \epsilon > 0: \forall A(\epsilon) = A\ge 2 : \exists A_1, A_2 > A: |\int_{A_1}^{A_2} \frac{sin^2(x)}{x^\alpha(x^\alpha + \sin(x))}dx| \ge \epsilon$. Let $A_1 = n\pi$ and $A_2 = 2n\pi$ for some large enough $n\in\mathbb{N}$. Since we're working on a positive interval and $x \ge\sin(x)$ for $x\ge 2$ we can drop the absolute value and we get the following chain of inequalities:
$\int_{n\pi}^{2n\pi} \frac{sin^2(x)}{x^\alpha(x^\alpha + \sin(x))}dx \ge \int_{n\pi}^{2n\pi} \frac{sin^2(x)}{x(x + \sin(x))}dx \ge \int_{n\pi}^{2n\pi} \frac{sin^2(x)}{x(x + 1)}dx \ge \int_{n\pi}^{2n\pi} \frac{sin^2(x)}{2n\pi(2n\pi + 1)}dx \ge \frac{n\pi}{2} \frac 1 {2n\pi(2n\pi + 1)} = \frac 1{4(2n\pi + 1)}$.
My question is if I set $\epsilon = \frac 1{4(2n\pi + 1)}$ would the argument be valid or not? I'm pretty sure it wouldn't be but I can't construct an integral in which the $n$ cancels out completely.
EDIT:
$\int_2^\infty \frac{sin^2(x)}{x^\alpha(x^\alpha + \sin(x))}dx \ge \int_2^\infty \frac{sin^2(x)}{x^\alpha(x^\alpha + 1)}dx \ge \int_2^\infty \frac{sin^2(x)}{x^\alpha(x^\alpha + 1)}dx = \int_2^\infty \frac{sin^2(x)}{x^{2\alpha} + x^\alpha}dx \ge \int_2^\infty \frac{sin^2(x)}{2x^{2\alpha}}dx$
So now we apply the original statement to the final integral:
$\exists\epsilon>0: \forall A(\epsilon)=A\ge2:\exists A_1,A_2>A: |\int_{A_1}^{A_2} \frac{sin^2(x)}{2x^{2\alpha}}dx| \ge\epsilon$. Let $A_1 = n\pi$ and $A_2 = 2n\pi$ for a large enough $n\in\mathbb{N}$. Then:
$|\int_{n\pi}^{2n\pi} \frac{sin^2(x)}{2x^{2\alpha}}dx|\ge\int_{n\pi}^{2n\pi} \frac{sin^2(x)}{2x^{2\alpha}}dx \ge \int_{n\pi}^{2n\pi} \frac{sin^2(x)}{2x}dx \ge \int_{n\pi}^{2n\pi} \frac{sin^2(x)}{4n\pi}dx = \frac{\frac{n\pi}{2}}{4n\pi} = \frac 1 8 \implies \epsilon = \frac 1 8$. Therefore the integral is divergent which means our original integral is divergent as well.
For $a>1/2$: $\int_2^\infty \frac{sin^2(x)}{x^\alpha(x^\alpha + \sin(x))}dx \le \int_2^\infty \frac{1}{x^\alpha(x^\alpha-1)}dx = \int_2^\infty \frac{1}{x^{2\alpha} - x^\alpha}dx \le \int_2^\infty \frac{1}{x^{2\alpha} - 2x^{2\alpha}}dx = \int_2^\infty \frac{1}{-x^{2\alpha}}dx$ and this integral is convergent when $2\alpha > 1$ i.e. when $\alpha > 1/2$.