Put $x=\sqrt{\tan \theta}$ so that $d x=\frac{\sec ^2 \theta d \theta}{2 \sqrt{\tan \theta}}$ $\therefore I=\frac{1}{2} \int \frac{\sec ^2 \theta d \theta}{\sin ^{3 / 2} \theta\left(\sec ^2 \theta\right)^{3 / 2}}=\frac{1}{2} \int \frac{\cos \theta d \theta}{\sin ^{3 / 2} \theta}$
Put $\sin \theta=\mathrm{t}, \cos \theta \mathrm{d} \theta=\mathrm{dt}$
$\begin{aligned}
&\mathrm{I}=\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}^{3 / 2}}=\frac{1}{2} \int \mathrm{t}^{-3 / 2} \mathrm{dt}=\frac{1}{2}(-2) \mathrm{t}^{-1 / 2}+\mathrm{c} \\
&\mathrm{I}=-\frac{1}{\sqrt{\mathrm{t}}}+\mathrm{c}=-\frac{1}{\sqrt{\sin \theta}}+\mathrm{c} \quad \ldots \text { (i) }
\end{aligned}$
we have $x=\sqrt{\tan \theta} \quad \therefore \quad \tan \theta=x^2$
$\therefore \quad \sin \theta=\frac{x^2}{\sqrt{1+x^4}}, \quad \sqrt{\sin \theta}=\frac{x}{\left(1+x^4\right)^{1 / 4}}$
$\therefore I=\frac{-1}{\sqrt{\sin \theta}}+c=-\frac{\left(1+x^4\right)^{1 / 4}}{x}+c$