Answer
$$
\int_{\pi / 4}^{\pi / 3} \frac{\sqrt{\tan \theta}}{\sin 2 \theta} d \theta =\sqrt[4]{3}-1
$$
Work Step by Step
$$
\begin{split}
\int_{\pi / 4}^{\pi / 3} \frac{\sqrt{\tan \theta}}{\sin 2 \theta} d \theta
& = \int_{\pi / 4}^{\pi / 3} \frac{\sqrt{\frac{\sin \theta}{\cos \theta}}}{2 \sin \theta \cos \theta} d \theta \\
& = \int_{\pi / 4}^{\pi / 3} \frac{1}{2}(\sin \theta)^{-1 / 2}(\cos \theta)^{-3 / 2} d \theta \\
& = \int_{\pi / 4}^{\pi / 3} \frac{1}{2} \left(\frac{\sin \theta}{\cos \theta}\right)^{-1 / 2}(\cos \theta)^{-2} d \theta \\
& =\int_{\pi / 4}^{\pi / 3} \frac{1}{2}(\tan \theta)^{-1 / 2} \sec ^{2} \theta d \theta \\
& =\int_{\pi / 4}^{\pi / 3} \frac{1}{2}(\tan \theta)^{-1 / 2} (d \tan \theta)\\
& = [\sqrt{\tan \theta}]_{\pi / 4}^{\pi / 3} \\
& =\sqrt[4]{3}-1 .
\end{split}
$$
So
$$
\int_{\pi / 4}^{\pi / 3} \frac{\sqrt{\tan \theta}}{\sin 2 \theta} d \theta =\sqrt[4]{3}-1
$$
