$\cfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}$

: Let $I = \int {\cfrac{{\sqrt {\tan x} }}{{\sin x\cos x}}} dx = \int {\cfrac{{\sqrt {\tan x} }}{{\cfrac{{\sin x}}{{\cos \,x}} \cdot {{\cos }^2}x}}} dx$

$= \int {\cfrac{{\sqrt {\tan x} }}{{\tan x}}} \cdot {\sec ^2}xdx = \int {{{\left( {\tan x} \right)}^{ - 1/2}} \cdot {{\sec }^2}xdx}$

Put $\tan x = t$ $\Rightarrow$ ${\sec ^2}xdx = dt$

$\therefore$ $I = \int {{t^{ - \cfrac{1}{2}}}} dt = \cfrac{{{t^{ - \cfrac{1}{2} + 1}}}}{{ - \cfrac{1}{2} + 1}} + C = 2{t^{1/2}} + C = 2\sqrt {\tan x} + C$