Answer
$$\int(\tan x)^{-3/2}\sec^2 xdx=-\frac{2}{\sqrt{\tan x}}+C$$
Work Step by Step
$$A=\int(\tan x)^{-3/2}\sec^2 xdx$$
We set $a=\tan x$, then $$da=\sec^2 xdx$$
Therefore, $$A=\int a^{-3/2}da$$ $$A=\frac{a^{-1/2}}{-\frac{1}{2}}+C$$ $$A=-2a^{-1/2}+C$$ $$A=-2(\tan x)^{-1/2}+C$$ $$A=-\frac{2}{\sqrt{\tan x}}+C$$
