\( \int \tan x [ \ln ( \cos x ) ] d x \)

To solve the integral \( \int \tan x [ \ln ( \cos x ) ] d x \), we can use integration by parts. Let's set \( u = \ln(\cos x) \) and \( dv = \tan x \, dx \). Then, we need to find \( du \) and \( v \). The derivative of \( u \) gives us: \[ du = \frac{-\sin x}{\cos x} \, dx = -\tan x \, dx. \] For \( dv = \tan x \, dx \), the integral is: \[ v = -\ln(\cos x). \] Now we can apply the integration by parts formula, which states: \[ \int u \, dv = uv - \int v \, du. \] Substituting our findings into the formula, we get: \[ \int \tan x [ \ln ( \cos x ) ] d x = \ln(\cos x)(-\ln(\cos x)) - \int -\ln(\cos x)(-\tan x) \, dx. \] This simplifies to: \[ -\ln^2(\cos x) + \int \ln(\cos x) \tan x \, dx. \] At this point, we've found that \( \int \tan x [ \ln ( \cos x ) ] d x \) leads us back to the original integral, suggesting we might need additional methods or revisiting our substitutions to fully solve it. The key takeaway is understanding the interplay of logarithmic and trigonometric functions in integrals!