The integral ${\displaystyle \underset{}{\overset{}{\int }}\frac{{\sec }^{3/2}\theta -{\sec }^{1/2}\theta }{2+{\tan }^2\theta }\tan \theta d\theta }$ is equal to

The given integral is $I={\displaystyle \underset{}{\overset{}{\int }}\frac{\left(\sec \theta -1\right)\sqrt{\sec \theta }}{1+{\sec }^2\theta }}\tan \theta d\theta$

                $={\displaystyle \underset{}{\overset{}{\int }}\frac{\left(\sec \theta -1\right)\sec \theta \tan \theta }{\left(1+{\sec }^2\theta \right)\sqrt{\sec \theta }}}d\theta$

Put $\sqrt{\sec \theta }=t\Rightarrow \frac{1}{2\sqrt{\sec \theta }}\sec \theta \tan \theta d\theta =dt$

$\therefore I={\displaystyle \underset{}{\overset{}{\int }}\frac{\left(t^2-1\right)}{1+t^4}.2dt}$

$=2{\displaystyle \underset{}{\overset{}{\int }}\frac{1-\frac{1}{t^2}}{t^2+\frac{1}{t^2}}dt=2{\displaystyle \underset{}{\overset{}{\int }}\frac{\left(1-\frac{1}{t^2}\right)dt}{{\left(1+\frac{1}{t}\right)}^2-2}}}$

Put $t+\frac{1}{t}=u\Rightarrow \left(1-\frac{1}{t^2}\right)dt=du$

$\therefore I=2{\displaystyle \underset{}{\overset{}{\int }}\frac{du}{u^2-2}=\frac{2}{2\sqrt{2}}{\log }_e\left|\frac{u-\sqrt{2}}{u+\sqrt{2}}\right|+C}$

$=\frac{1}{\sqrt{2}}{\log }_e\left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|+C$

$=\frac{1}{\sqrt{2}}{\log }_e\left|\frac{t^2-\sqrt{2t}+1}{t^2+\sqrt{2}t+1}\right|+C$

$=\frac{1}{\sqrt{2}}{\log }_e\left|\frac{\sec \theta -\sqrt{2\sec \theta }+1}{\sec \theta +\sqrt{2\sec \theta }+1}\right|+C$