Problem with two improper integral problems. In the first one I am supposed to just tell if the improper integral 
$$\int\limits_1^\infty \frac{\cos^2(x)}{x^2}\,dx$$
is convergent or divergent. I am not supposed to calculate the integral.
The best I can do is show that $$\int\limits_1^\infty \frac{1}{x^2}\,dx$$ is convergent and since that $$\int\limits_1^\infty \frac{\cos^2(x)}{x^2}\,dx\lt\int\limits_1^\infty \frac{1}{x^2}\,dx$$ and is there for also convergent. But I have a feeling that I am completely wrong on this one.
The other problem is: Is it possible that $$\int\limits_{-5}^{10} \cos(\tanh(x^2))\,dx =10\pi.$$ 
The teacher then gave a hint that don't even try to calculate the integral. On this problem I have no idea were to start.
Any help is greatly appreciated.
 A: For the first one, you are right. But you should elaborate your proof.
Remember, there is a theorem that tells you that if $0<f(x)<g(x)$ for all $x>a$, then if the integral $$\int_a^\infty g(x)$$ converges, the integral $$\int_a^\infty f(x)$$ also converges.
For the second, a hint:
$\cos$ is bounded by $1$, and the length of the integral is $15$.
$10\cdot \pi$ is greater than $30$, on the other hand.
A: Note that $-1 \leq \cos(x) \leq 1$.
There is a theorem that says that if $ m \leq f(x) \leq M$ then 
$m(a - b) \leq \int_a^b f(x) \> dx \leq M(b-a) $.
Using this theorem,
$ \int_{-5}^{10} \cos(\tanh^2(x)) \> dx \leq 15 < 10 \pi$
We can conclude that it is not possible.
A: since $\cos^2 x$ fluctuates between 0 & 1 hence
$$\frac{\cos^2 x}{x^2}<\frac{1}{x^2} \ \ (1<x<\infty)$$
$$\int_1^{\infty}\frac{\cos^2 x}{x^2}<\int_1^{\infty}\frac{1}{x^2} \ \ (1<x<\infty)$$
but integral $\int_1^{\infty}\frac{1}{x^2}dx$ converges hence by comparison test, integral $\int_1^{\infty}\frac{\cos^2x}{x^2}dx$ also converges
A: Let $f$ defined by
$f(x)=\frac{cos^2(x)}{x^2}$.
$f$ is continuous at $ [1,+\infty)$
and
$(\forall x\geq 1) \;   0\leq f(x) \leq\frac{1}{x^2}$
but
$\int_1^{+\infty} \frac{1}{x^2}$
is convergent , so 
$\int_1^{+\infty}f(x)dx$
is also convergent.
