Solve Improper Integral: $\int_{1}^{\infty} 1/(x^2+ 3 \ |sin x| +2) dx$

[Archimedes II]

Homework Statement

$\displaystyle \int_{1}^{\infty} 1/(x^2+ 3 \ |sin x| +2) dx$

Homework Equations

N/A

The Attempt at a Solution

$\displaystyle \int_{1}^{\infty} 1/(x^2+ 3 \ |sin x| +2) dx =$

$\displaystyle lim_{t\rightarrow \infty} \int_{1}^{t} 1/(x^2+ 3 \ |sin x| +2) dx$

Side Work

$\displaystyle \int 1/(x^2+ 3 \ |sin x| +2) dx$

I have now clue how to solve this integral. It can't be simplified. U substitution doesn't work nor does a trigonometric substitution. Once I can solve the indefinite itegral I can solve the rest on my own.

Thanks in advance.

 

[Dick]

Homework Statement

$\displaystyle \int_{1}^{\infty} 1/(x^2+ 3 \ |sin x| +2) dx$

Homework Equations

N/A

The Attempt at a Solution

$\displaystyle \int_{1}^{\infty} 1/(x^2+ 3 \ |sin x| +2) dx =$

$\displaystyle lim_{t\rightarrow \infty} \int_{1}^{t} 1/(x^2+ 3 \ |sin x| +2) dx$

Side Work

$\displaystyle \int 1/(x^2+ 3 \ |sin x| +2) dx$

I have now clue how to solve this integral. It can't be simplified. U substitution doesn't work nor does a trigonometric substitution. Once I can solve the indefinite itegral I can solve the rest on my own.

Thanks in advance.

You can't solve the indefinite integral. You just want to prove the improper integral exists. Try a comparison test.

 

[Archimedes II]

You can't solve the indefinite integral. You just want to prove the improper integral exists. Try a comparison test.

Oh ok thanks that makes since now.