$\text{ Evaluate }\int \sqrt{\tan \theta }d\theta$= $\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan \theta -1}{\sqrt{2\tan \theta }}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{A}{B}\right|+C$

$$\begin{gathered} \text{ Let }I=\int \sqrt{\tan \theta }d\theta \\ \text{ Let }\tan \theta =x^2\text{ . Then, }d(\tan \theta )=d\left(x^2\right)\text{ or }{\sec }^2\theta d\theta =2xdx \\ \text{ or }\begin{array}{l}d\theta =\frac{2x\cdot dx}{{\sec }^2\theta }=\frac{2xdx}{1+{\tan }^2\theta }=\frac{2xdx}{1+x^4}\\ I=\int \sqrt{x^2}\times \frac{2xdx}{1+x^4}=\int \frac{2x^2}{x^4+1}dx\\ =\int \frac{2}{x^2+1/x^2}dx\\ =\int \frac{1+1/x^2+1-1/x^2}{x^2+1/x^2}dx\\ =\int \frac{1+1/x^2}{x^2+1/x^2}dx+\int \frac{1-1/x^2}{x^2+1/x^2}dx\\ =\int \frac{1+1/x^2}{(x-1/x{)}^2+2}dx+\int \frac{1-1/x^2}{(x+1/x{)}^2-2}dx\\ putting x-\frac{{\displaystyle 1}}{{\displaystyle x}}=u in the first integral and x+\frac{1}{x}=v\text{ in the second integral we get}\\ \begin{array}{l}I=\int \frac{du}{u^2+(\sqrt{2}{)}^2}+\int \frac{dv}{v^2-(\sqrt{2}{)}^2}\\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{u}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{v-\sqrt{2}}{v+\sqrt{2}}\right|+C\\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-1/x}{\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{x+1/x-\sqrt{2}}{x+1/x+\sqrt{2}}\right|+C\\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{x\sqrt{2}}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{x^2-x\sqrt{2}+1}{x^2+x\sqrt{2}+1}\right|+C\\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan \theta -1}{\sqrt{2\tan \theta }}\right)\\ =\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{\tan \theta -1}{\sqrt{2\tan \theta }}\right)+\frac{1}{2\sqrt{2}}\log \left|\frac{\tan \theta -\sqrt{2\tan \theta }+1}{\tan \theta +\sqrt{2\tan \theta }+1}\right|+C\end{array}\end{array} \end{gathered}$$