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.
For the first one, you are right. But you should elaborate your proof.

Remember, there is a theorem that tells you that if $0&lt;f(x)&lt;g(x)$ for all $x&gt;a$, 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.
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.