Answer
\[3^{\frac{1}{4}}-1\]
Work Step by Step
Let \[I=\int_{\frac{π}{4}}^{\frac{π}{3}}\frac{\sqrt{\tan\theta}}{\sin 2\theta}\;d\theta\;\;\;\;\ldots (1)\]
Consider \[I_1=\int\frac{\sqrt{\tan\theta}}{\sin 2\theta}\;d\theta\]
\[I_1=\int\frac{\sqrt{ \frac{\sin\theta}{\cos\theta} }}{2\sin \theta\cos\theta}\;d\theta\]
\[I_1=\int\frac{1}{2\sin^{\frac{1}{2}}\theta\cos^{\frac{3}{2}}\theta}\;d\theta\]
\[I_1=\int\frac{\sec^2\theta}{2\sqrt{\tan\theta}}\;d\theta\]
Substitute $\;t=\tan\theta$
\[\Rightarrow dt=\sec^2\theta d\theta\]
\[I_1=\int\frac{dt}{2\sqrt{t}}=\frac{1}{2}\int t^{-\frac{1}{2}}dt\]
\[I_1=\frac{1}{2}(2\sqrt{t})=\sqrt t\]
\[I_1=\sqrt{\tan\theta}+C\;\;\;\ldots (2)\]
Where $C$ is constant of integration
Using (2) in (1)
\[I=\left[\sqrt{\tan\theta}\right]_{\frac{π}{4}}^{\frac{π}{3}}\]
\[I=\sqrt{\tan \frac{π}{3} }-\sqrt{\tan \frac{π}{4} }\]
\[I=\sqrt{\sqrt{3}}-1\]
\[I=3^{\frac{1}{4}}-1\]
Hence \[I=3^{\frac{1}{4}}-1\;\;.\]
