Answer
$$ - \frac{2}{{\sqrt {\tan x} }} + C $$
Work Step by Step
$$\eqalign{
& \int {{{\left( {\tan x} \right)}^{ - 3/2}}{{\sec }^2}x} dx \cr
& {\text{use the substitution method}}{\text{: }} \cr
& u = \tan x,\,\,\,\,du = {\sec ^2}xdx,\,\,\,\,\,dx = \frac{{du}}{{{{\sec }^2}x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {{{\left( {\tan x} \right)}^{ - 3/2}}{{\sec }^2}x} dx = \int {{u^{ - 3/2}}{{\sec }^2}x} \left( {\frac{{du}}{{{{\sec }^2}x}}} \right) \cr
& = \int {{u^{ - 3/2}}du} \cr
& {\text{integrating by the power rule gives}} \cr
& = \frac{{{u^{ - 1/2}}}}{{ - 1/2}} + C \cr
& = - 2{u^{ - 1/2}} + C \cr
& = - \frac{2}{{\sqrt u }} + C \cr
& {\text{write in terms of }}x;{\text{ replace }}u = \tan x \cr
& = - \frac{2}{{\sqrt {\tan x} }} + C \cr} $$
