Question:

The given integral is $$I = \int \frac{\tan\left(\frac{\pi}{4} - x\right)}{\cos^2 x \sqrt{\tan^3 x + \tan^2 x + \tan x}} dx$$ Using the identity $\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x} = \frac{(1 - \tan x)(1 + \tan x)}{(1 + \tan x)^2} = \frac{1 - \tan^2 x}{(1 + \tan x)^2}$, we have: $$I = \int \frac{(1 - \tan^2 x) dx}{(1 + \tan x)^2 \cos^2 x \sqrt{\tan^3 x + \tan^2 x + \tan x}}$$ Dividing the numerator and denominator by $\tan^2 x$ and using $\frac{1}{\cos^2 x} = \sec^2 x$: $$I = \int \frac{-\left(1 - \frac{1}{\tan^2 x}\right) \sec^2 x dx}{\left(\tan x + 2 + \frac{1}{\tan x}\right) \sqrt{\tan x + 1 + \frac{1}{\tan x}}}$$ Let $y = \sqrt{\tan x + 1 + \frac{1}{\tan x}}$. Then $y^2 = \tan x + 1 + \frac{1}{\tan x} \Rightarrow 2y \, dy = \left(\sec^2 x - \frac{1}{\tan^2 x} \sec^2 x\right) dx = \sec^2 x \left(1 - \frac{1}{\tan^2 x}\right) dx$. Also, $\tan x + 2 + \frac{1}{\tan x} = y^2 + 1$. Substituting these into the integral: $$I = \int \frac{-2y \, dy}{(y^2 + 1) \cdot y} = -2 \int \frac{dy}{1 + y^2}$$ $$I = -2 \tan^{-1} y + C = -2 \tan^{-1} \left(\sqrt{\tan x + 1 + \frac{1}{\tan x}}\right) + C$$