Your classmate, Yinhe, pointed out that the part b problem actually diverges.
This is the Improper integral in question:
$latex \int_{0}^{\infty} \frac {1}{2+sin^2 x} dx$
We can use a simple comparison test.
\[sin^2 x ≤ 1\]
\[
2 +sin^2 x ≤ 3
\] \[
\frac {1}{2+sin^2 x} ≥ 1/3
\]
$latex \int \frac {1}{2+sin^2 x} dx ≥ \int \frac {1}{3} dx = \frac {x}{3} + C $
Since $latex \lim_{x \to \infty} \frac {x}{3} } = \infty$
$latex \int_{0}^{\infty} \frac {1}{3} dx$ diverges
And so $latex \int_{0}^{\infty} \frac {1}{2+sin^2 x} dx$ also diverges.
