Evaluating the Integral \(\int \frac{\tan x}{\ln(\cos x)} \, dx\)

Evaluating the Integral \(\int \frac{\tan x}{\ln(\cos x)} \, dx\)

Step-by-Step Solution

Step 1: Make a substitution:

\[ \text{Let } u = \ln(\cos x) \]

Differentiate both sides:

\[ \frac{du}{dx} = \frac{d}{dx}[\ln(\cos x)] = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x \] \[ \Rightarrow du = -\tan x \, dx \] \[ \Rightarrow -du = \tan x \, dx \]

Step 2: Rewrite the integral:

\[ \int \frac{\tan x}{\ln(\cos x)} dx = \int \frac{-du}{u} = -\int \frac{1}{u} du \]

Step 3: Integrate:

\[ -\ln|u| + C = -\ln|\ln(\cos x)| + C \]

\[ \boxed{-\ln|\ln(\cos x)| + C} \]

Verification by Differentiation

To verify, we can differentiate our result:

\[ \frac{d}{dx} \left[-\ln|\ln(\cos x)|\right] = -\frac{1}{\ln(\cos x)} \cdot \frac{d}{dx}[\ln(\cos x)] \] \[ = -\frac{1}{\ln(\cos x)} \cdot \left(\frac{1}{\cos x} \cdot (-\sin x)\right) \] \[ = \frac{\tan x}{\ln(\cos x)} \]

Which matches our original integrand, confirming the solution is correct.