<s>[INST] Does $\int_0^\infty\frac{\cos^2x}{x^2+5x+11}dx$ converge or diverge? When I'm learning convergence, my teacher just show me about condition to convergence or not. But I haven't meet a function that contain both trigonometric and normal polynomial. When I asked one of my friends, he tell me that using Taylor to developed $\cos x$, but I'm afraid that is not the good solution.
$$\int_0^\infty\frac{\cos^2(x)}{x^2+5x+11}dx$$
Thanks :)
 [/INST] Let $$I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx$$ The integrand is non-negative and since $\cos^2(x) \in [0,1]$, $\forall x \in \mathbb{R}$, we get that $$0 \leq I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac1{x^2 + 5x+11} dx = \int_0^{\infty} \frac{dx}{(x+5/2)^2 + 19/4}$$
Now make use of the identity $$\int \frac{dx}{x^2 + a^2} = \frac{\arctan (x/a)}a$$ and bound the integral $I$. Hence, we get that $$0 \leq I \leq \left. \frac{2}{\sqrt{19}} \arctan \left(\frac{2x+5}{\sqrt{19}} \right) \right \rvert_{0}^{\infty} = \frac{2}{\sqrt{19}} \left( \frac{\pi}{2} - \arctan \left(\frac{5}{\sqrt{19}}\right) \right) = \frac{\pi - 2 \arctan \left( \dfrac{5}{\sqrt{19}} \right)}{\sqrt{19}} \approx 0.328984$$
EDIT
Another quick argument is along the lines of what mixedmath has done but there is no need to split the integral over two regions $$I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac1{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac{dx}{(x+5/2)^2} = \left. \left(-\frac1{x+5/2} \right) \right \rvert_{0}^{\infty} = \frac25$$
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