Answer
Converges
Work Step by Step
Given $$\int_{0}^{\pi} \frac{d t}{\sqrt{t}+\sin t} .$$
Since for $0 \leq t \leq \pi, $ $$0 \leq \frac{1}{\sqrt{t+\sin t}} \leq \frac{1}{\sqrt{t}}$$ and
\begin{align*}
\int_{0}^{\pi} \frac{d t}{\sqrt{t}} &= \sqrt{t} \bigg|\int_{0}^{\pi}\\
&=2\sqrt{\pi}
\end{align*}
converges, then by the Direct Comparison Test,
$$\int_{0}^{\pi} \frac{d t}{\sqrt{t}+\sin t} .$$
also converges.