The equation of a curve is $y = \frac{1 - \sin x}{\cos x}$.

1. Apply the quotient rule to differentiate the function.

Let $u = 1 - \sin x$ and $v = \cos x$.
Then $\frac{du}{dx} = -\cos x$ and $\frac{dv}{dx} = -\sin x$.

Using the quotient rule $\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$:
$\frac{dy}{dx} = \frac{\cos x (-\cos x) - (1 - \sin x)(-\sin x)}{\cos^2 x}$

2. Expand and simplify the numerator using the Pythagorean identity.

$\frac{dy}{dx} = \frac{-\cos^2 x + \sin x - \sin^2 x}{\cos^2 x}$
$= \frac{\sin x - (\sin^2 x + \cos^2 x)}{\cos^2 x}$
Since $\sin^2 x + \cos^2 x = 1$:
$= \frac{\sin x - 1}{\cos^2 x}$

3. Split the fraction and convert to secant and tangent to reach the required form.

$\frac{dy}{dx} = \frac{\sin x}{\cos^2 x} - \frac{1}{\cos^2 x}$
$= \left(\frac{\sin x}{\cos x}\right)\left(\frac{1}{\cos x}\right) - \left(\frac{1}{\cos x}\right)^2$
$= \tan x \sec x - \sec^2 x \quad \text{(Shown)}$

1. Express the integral of the derivative from part (a) using the original function.

From (a), $\frac{d}{dx}\left(\frac{1 - \sin x}{\cos x}\right) = \tan x \sec x - \sec^2 x$.

Integrating both sides with respect to $x$ from $0$ to $\frac{\pi}{4}$:
$\int_{0}^{\frac{\pi}{4}} (\tan x \sec x - \sec^2 x) \, dx = \left[ \frac{1 - \sin x}{\cos x} \right]_{0}^{\frac{\pi}{4}}$

2. Split the integral and rearrange to make the required integral the subject.

$\int_{0}^{\frac{\pi}{4}} \tan x \sec x \, dx - \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx = \left[ \frac{1 - \sin x}{\cos x} \right]_{0}^{\frac{\pi}{4}}$
$\int_{0}^{\frac{\pi}{4}} \tan x \sec x \, dx = \left[ \frac{1 - \sin x}{\cos x} \right]_{0}^{\frac{\pi}{4}} + \int_{0}^{\frac{\pi}{4}} \sec^2 x \, dx$

3. Integrate $\sec^2 x$ and combine the terms.

Since $\int \sec^2 x \, dx = \tan x$, we have:
$\int_{0}^{\frac{\pi}{4}} \tan x \sec x \, dx = \left[ \frac{1 - \sin x}{\cos x} + \tan x \right]_{0}^{\frac{\pi}{4}}$

4. Evaluate the expression at the upper and lower limits.

Substitute $x = \frac{\pi}{4}$:
$\frac{1 - \sin(\frac{\pi}{4})}{\cos(\frac{\pi}{4})} + \tan\left(\frac{\pi}{4}\right) = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} + 1 = (\sqrt{2} - 1) + 1 = \sqrt{2}$

Substitute $x = 0$:
$\frac{1 - \sin(0)}{\cos(0)} + \tan(0) = \frac{1 - 0}{1} + 0 = 1$

Subtracting the lower limit value from the upper limit value:
$\int_{0}^{\frac{\pi}{4}} \tan x \sec x \, dx = \sqrt{2} - 1$

1. Express the integrand in partial fractions (as would have been done in the missing part (i)).

Let $\frac{-14x^2 + 14x - 3}{x(2x - 1)^2} = \frac{A}{x} + \frac{B}{2x - 1} + \frac{C}{(2x - 1)^2}$.
$-14x^2 + 14x - 3 = A(2x - 1)^2 + Bx(2x - 1) + Cx$

Substitute $x = 0$:
$-3 = A(-1)^2 \implies A = -3$

Substitute $x = \frac{1}{2}$:
$-14(\frac{1}{4}) + 14(\frac{1}{2}) - 3 = C(\frac{1}{2})$
$-\frac{7}{2} + 7 - 3 = \frac{1}{2}C \implies \frac{1}{2} = \frac{1}{2}C \implies C = 1$

Compare coefficients of $x^2$:
$-14 = 4A + 2B$
$-14 = 4(-3) + 2B \implies 2B = -2 \implies B = -1$

Thus, $\frac{-14x^2 + 14x - 3}{x(2x - 1)^2} = -\frac{3}{x} - \frac{1}{2x - 1} + \frac{1}{(2x - 1)^2}$.

2. Integrate the partial fractions.

$\int \frac{-14x^2 + 14x - 3}{x(2x - 1)^2} \, dx = \int \left( -\frac{3}{x} - \frac{1}{2x - 1} + (2x - 1)^{-2} \right) \, dx$
$= -3 \ln |x| - \frac{1}{2} \ln |2x - 1| + \frac{(2x - 1)^{-1}}{(-1)(2)} + C$
$= -3 \ln |x| - \frac{1}{2} \ln |2x - 1| - \frac{1}{2(2x - 1)} + C$