Show that $\frac{d}{dx} \ln(e^x \cos x) = 1 - \tan x$. Using part (i), find $\int \tan x \, dx$. Hence show that $\int_{0}^{\frac{\pi}{3}} \tan x \, dx = \ln 2$.

1. Simplify the expression using laws of logarithms.

$\ln(e^x \cos x) = \ln(e^x) + \ln(\cos x)$
$= x + \ln(\cos x)$

2. Differentiate the simplified expression with respect to $x$.

$\frac{d}{dx} [x + \ln(\cos x)] = 1 + \frac{1}{\cos x} \cdot (-\sin x)$

3. Simplify the derivative to obtain the required result.

$= 1 - \frac{\sin x}{\cos x}$
$= 1 - \tan x$

1. Use the result from part (i) to set up an integral equation

Since $\frac{d}{dx} \ln(e^x \cos x) = 1 - \tan x$, integrating both sides gives:
$\int (1 - \tan x) \, dx = \ln(e^x \cos x) + C$

2. Rearrange to find the integral of $\tan x$

$\int 1 \, dx - \int \tan x \, dx = \ln(e^x \cos x) + C$
$x - \int \tan x \, dx = \ln(e^x \cos x) + C$
$\int \tan x \, dx = x - \ln(e^x \cos x) + C'$

(Optional simplification)
Since $\ln(e^x \cos x) = \ln(e^x) + \ln(\cos x) = x + \ln(\cos x)$,
$\int \tan x \, dx = x - (x + \ln(\cos x)) + C' = -\ln(\cos x) + C'$

1. State the definite integral using the antiderivative found in part (ii)

From (ii), $\int \tan x \, dx = x - \ln(e^x \cos x)$.

$\int_{0}^{\frac{\pi}{3}} \tan x \, dx = \left[ x - \ln(e^x \cos x) \right]_{0}^{\frac{\pi}{3}}$

2. Substitute the limits and simplify the expression

$= \left( \frac{\pi}{3} - \ln\left(e^{\frac{\pi}{3}} \cos \frac{\pi}{3}\right) \right) - \left( 0 - \ln(e^0 \cos 0) \right)$

$= \frac{\pi}{3} - \ln\left(e^{\frac{\pi}{3}} \cdot \frac{1}{2}\right) - \left( - \ln(1 \cdot 1) \right)$

3. Apply logarithm laws to show the final result

$= \frac{\pi}{3} - \left( \ln e^{\frac{\pi}{3}} + \ln \frac{1}{2} \right) + 0$

$= \frac{\pi}{3} - \frac{\pi}{3} - \ln(2^{-1})$

$= -(-\ln 2) = \ln 2$ (Shown)