Determine whether the integral is convergent or divergent:

[Mampac]

Hello, could you correct my steps?

I'm dealing with




1) for convenience, I'm splitting the integral to two integrals from 0 to 1 and from 0 to +inf.

First of all, is there any way to simplify [MATH]arctan(sinx)[/MATH]? I searched all the internet but there seems to be no way of doing that.

2) I'm assuming it's convergent, because [MATH]arctan(sinx)[/MATH] is bounded and [MATH]x^2[/MATH]'s power is bigger than 1.

3) For Direct Comparison Test, I have to find a function bigger that's convergent. I make the numerator bigger by replacing [MATH]arctan(sinx)[/MATH] with [MATH]{\dfrac{π}{2}}[/MATH] (which I'll take out of the integral later) and make the denominator smaller by dropping [MATH]3cosx + 4[/MATH].

4) I prove that the integral from 1 to +inf of [MATH]{\dfrac{1}{x^2}}[/MATH]is convergent since x's power is greater than 1.

However, I do have many doubts about it! Is this enough? I look at the definition, but it seems like something's missing (((

I tried looking at the Limit Comparison Test, but I'm failing to solve the limit:
5) I see I can pull out [MATH]x^2[/MATH] in the denominator and get rid of the cosine and 4, but what do I do with the arctangent? if i just ignore it I get again "something divided by [MATH]x^2[/MATH]. Is this gonna be enough to prove the convergence?

Thank you in advance <3

 

[skeeter]

[MATH]-\frac{\pi}{4} \le \arctan(\sin{x}) \le \frac{\pi}{4} \text{ for all } x \in [0,\infty)[/MATH]
[MATH]x^2+1 \le x^2 + 3\cos{x} + 4 \le x^2+7 \text{ for all } x \in [0,\infty)[/MATH]

 

[Mampac]

[MATH]-\frac{\pi}{4} \le \arctan(\sin{x}) \le \frac{\pi}{4} \text{ for all } x \in [0,\infty)[/MATH]
[MATH]x^2+1 \le x^2 + 3\cos{x} + 4 \le x^2+7 \text{ for all } x \in [0,\infty)[/MATH]

Does that mean I should replace the numerator with [MATH]{\dfrac{π}{4}}[/MATH], and the denominator with [MATH]x^2 + 1[/MATH]
Then show that the resulting fraction's limit tends to 0 thus it's convergent, thus by DCT the original integral is convergent?

 

[skeeter]

Does that mean I should replace the numerator with [MATH]{\dfrac{π}{4}}[/MATH], and the denominator with [MATH]x^2 + 1[/MATH]

... wouldn't that be strictly less than the original integrand in the improper integral for all x ?

if so, wouldn't the following be true?

[MATH]\frac{\arctan(\sin{x})}{x^2+3\cos{x}+4} < \dfrac{\pi}{4} \cdot \dfrac{1}{x^2+1} \implies \int_0^\infty \frac{\arctan(\sin{x})}{x^2+3\cos{x}+4} \, dx < \frac{\pi}{4} \int_0^\infty \frac{1}{x^2+1} \, dx[/MATH]
I believe one can determine the exact value of that last integral ...