$\int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} \mathrm{~d} x=$

The integral is evaluated using substitution after rewriting the integrand in terms of $\tan x$.

Recall that $\sin x \cdot \cos x = \tan x \cdot \cos^2 x$. Substituting into the integral gives:

$\int \frac{\sqrt{\tan x}}{\tan x \cdot \cos^2 x} dx = \int \frac{1}{\sqrt{\tan x}} \cdot \frac{1}{\cos^2 x} dx = \int \frac{\sec^2 x}{\sqrt{\tan x}} dx$

Let $u = \tan x$, so $du = \sec^2 x dx$. The integral simplifies to:

$\int u^{-1/2} du = 2u^{1/2} + C$

Substituting back yields the final result:

$2\sqrt{\tan x} + C$

This corresponds to option B.