Does this integral converge or diverge? Can you help me to integrate this?

Question

Does this integral converge or diverge? $$\int_1^{\infty}\frac{\cos x}{\sqrt{x}+\cos x}\,dx$$ I tried to simplify it using trigonometric functions, but I got lost.

Answer 1

Hint for the convergence/divergence. Note that by integration by parts, $$\begin{align} \int_1^{\infty}\frac{\cos (x)}{\sqrt{x}+\cos (x)}\,dx&= \left[\frac{\sin (x)}{\sqrt{x}+\cos (x)}\right]_1^{\infty}- \int_1^{\infty}\sin(x)\cdot\frac{\sin(x)-\frac{1}{2\sqrt{x}}}{(\sqrt{x}+\cos x)^2}\,dx\\ &= \left[\frac{\sin (x)}{\sqrt{x}+\cos (x)}\right]_1^{\infty}- \int_1^{\infty}\frac{\sin^2(x)}{(\sqrt{x}+\cos x)^2}\,dx+\int_1^{\infty}\frac{1}{2\sqrt{x}(\sqrt{x}+\cos x)^2}\,dx \end{align}$$ Now the last integral is convergent, whereas $$\int_1^{\infty}\frac{\sin^2(x)}{(\sqrt{x}+\cos x)^2}\,dx\geq \int_1^{\infty}\frac{\sin^2(x)}{(\sqrt{x}+1)^2}\,dx\geq \int_1^{\infty}\frac{\sin^2(x)}{4x}\,dx.$$ It remains to show that the integral on the right is divergent which implies that the given integral diverges to $-\infty$.