In my homework we are supposed to show that the integral $\displaystyle\int_1^{+\infty} \frac{\sin(x)}{\sqrt{x}+\cos(x)} \mathrm dx$ converges, using Abel's criterion. There should not be any deep result or manipulation needed.
Of course the obvious idea of writing the integrand as $\sin(x)\times \frac{1}{\sqrt{x}+\cos(x)}$ does not work since $\frac{1}{\sqrt{x}+\cos(x)}$ does not decrease towards $0$.
Instead I thought that I could write it as $\frac{\sin(x)}{1+\frac{\cos(x)}{\sqrt{x}}}\times \frac{1}{\sqrt{x}}$ and show that $\displaystyle\int_1^{+\infty}\frac{\sin(x)}{1+\frac{\cos(x)}{\sqrt{x}}}\mathrm dx$ is bounded. It seemed reasonable since $\frac{1}{1+\frac{\cos(x)}{\sqrt{x}}}$ converges to $1$ as $x\to \infty$, so it should not add too much perturbation to the boundedness of the integral of $\sin(x)$. But this is like saying that the equivalence of functions preserves the boundedness of the integral, which is only true for positive functions. And indeed if I try to prove it by hand, I realize that the fact that $\sin(x)$ changes sign prevents me from writing a good estimation to show boundedness.
Am I missing something? Should I consider a different application of Abel's criterion?
We will use the Dirichlet test which implies the convergence of the integrals $$\int\limits_1^\infty{\sin(ax)\over x^b}\,dx, \quad b>0$$ We have $${\sin x\over \sqrt{x}+\cos x}-{\sin x \over \sqrt{x}}={\cos x\sin x \over (\sqrt{x}+\cos x)\sqrt{x}}\\ ={1\over 2}{\sin(2x)\over x+\sqrt{x}\cos x}$$ It suffices to show that the integral of the last expression is convergent. We have $${\sin(2x)\over x+\sqrt{x}\cos x}-{\sin(2x)\over x}={\cos x\sin(2x)\over x(\sqrt{x}+\cos x)} $$ The obtained function is absolutely integrable. Hence the original integral is convergent.