To evaluate the integral \[\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx\], substitution and trigonometric identities are employed.
First, it is noted that \(\tan x = \frac{\sin x}{\cos x}\), implying \(\sqrt{\tan x} = \frac{\sqrt{\sin x}}{\sqrt{\cos x}}\).
The integral is then rewritten as:
\(\int \frac{\sqrt{\tan x}}{\sin x \cos x} \, dx = \int \frac{\sqrt{\sin x}/\sqrt{\cos x}}{\sin x \cos x} \, dx = \int \frac{1}{\sqrt{\sin x} \, \cos x \, \sqrt{\cos x}} \, dx\)
Let \(\sqrt{\tan x} = t\). This implies \(\tan x = t^2\). Differentiating both sides yields:
\(d(\tan x) = d(t^2) \Rightarrow \sec^2 x \, dx = 2t \, dt\)
Given \(\tan x = \frac{\sin x}{\cos x}\), we have:
\(\sin x = t^2 \cos x\). Differentiating this equation gives \(d(\sin x) = d(t^2 \cos x) = 2t \cos x \, dt + t^2 (-\sin x) \, dx\)
Rearrangement of the terms leads to:
\(\sec^2 x \, dx = \frac{2t \, dt}{(1 + t^2)}\)
The integral transformation is then:
\(\int \frac{1}{t^2(1+t^2)} \cdot \frac{2t}{1+t^2} \, dt = \int \frac{2}{t^3(1+t^2)^2} \, dt\)
Each substitution simplifies the expression, facilitating a smooth path to the antiderivative.
The evaluated integral, utilizing the secant property, simplifies to:
\(\frac{2}{\cos^2 x}\), which matches the provided correct solution.