Answer
Converges
Work Step by Step
Given: $\int_{0}^{\pi} \frac{d t}{\sqrt{t}+\sin t}$
Since, $ 0 \leq \dfrac{1}{\sqrt{t+\sin t}} \leq \dfrac{1}{\sqrt{t}}$; $0 \leq t \leq \pi$
and $\int_{0}^{\pi} \dfrac{d t}{\sqrt{t}}=[2\sqrt{t} ]_{0}^{\pi}=2\sqrt{\pi}$
Thus the integral converges.
Then by the Direct Comparison Test,
$\int_{0}^{\pi} \dfrac{d t}{\sqrt{t}+\sin t} $
also converges.