Differentiate $\text{e}^{\tan x}$ with respect to $x$. Hence find $\int \text{e}^{\tan x} \sec^4 x \text{ d}x$.

1. Apply the chain rule to differentiate the exponential function

$\frac{\text{d}}{\text{d}x}(\text{e}^{\tan x}) = \text{e}^{\tan x} \cdot \frac{\text{d}}{\text{d}x}(\tan x)$

2. Evaluate the derivative of $\tan x$ to obtain the final expression

$= \text{e}^{\tan x} (\sec^2 x)$
$= \sec^2 x \text{ e}^{\tan x}$

1. Recall the derivative from part (a) to set up integration by parts.

From part (a), we have:
$\frac{\text{d}}{\text{d}x} (\text{e}^{\tan x}) = \text{e}^{\tan x} \sec^2 x$

We rewrite the integral to extract this exact expression:
$\int \text{e}^{\tan x} \sec^4 x \text{ d}x = \int \sec^2 x (\text{e}^{\tan x} \sec^2 x) \text{ d}x$

2. Apply integration by parts for the first time.

Let $u = \sec^2 x$ and $\frac{\text{d}v}{\text{d}x} = \text{e}^{\tan x} \sec^2 x$.
Then $\frac{\text{d}u}{\text{d}x} = 2 \sec x (\sec x \tan x) = 2 \sec^2 x \tan x$, and $v = \text{e}^{\tan x}$.

Using the integration by parts formula $\int u \frac{\text{d}v}{\text{d}x} \text{ d}x = uv - \int v \frac{\text{d}u}{\text{d}x} \text{ d}x$:
$\int \sec^2 x (\text{e}^{\tan x} \sec^2 x) \text{ d}x = \sec^2 x \text{e}^{\tan x} - \int \text{e}^{\tan x} (2 \sec^2 x \tan x) \text{ d}x$
$= \sec^2 x \text{e}^{\tan x} - \int 2 \tan x (\text{e}^{\tan x} \sec^2 x) \text{ d}x$

3. Apply integration by parts a second time to the remaining integral.

We evaluate $\int 2 \tan x (\text{e}^{\tan x} \sec^2 x) \text{ d}x$:
Let $u_2 = 2 \tan x$ and $\frac{\text{d}v_2}{\text{d}x} = \text{e}^{\tan x} \sec^2 x$.
Then $\frac{\text{d}u_2}{\text{d}x} = 2 \sec^2 x$ and $v_2 = \text{e}^{\tan x}$.

Applying integration by parts again:
$\int 2 \tan x (\text{e}^{\tan x} \sec^2 x) \text{ d}x = 2 \tan x \text{e}^{\tan x} - \int 2 \sec^2 x \text{e}^{\tan x} \text{ d}x$

4. Integrate the final term directly and combine all components to obtain the answer.

Using the derivative result from part (a) again:
$\int 2 \sec^2 x \text{e}^{\tan x} \text{ d}x = 2 \text{e}^{\tan x}$

Substitute this back into the overall expression:
$\int \text{e}^{\tan x} \sec^4 x \text{ d}x = \sec^2 x \text{e}^{\tan x} - (2 \tan x \text{e}^{\tan x} - 2 \text{e}^{\tan x}) + C$
$= \text{e}^{\tan x} (\sec^2 x - 2 \tan x + 2) + C$
(Note: Equivalently, using $\sec^2 x = 1 + \tan^2 x$, the answer can also be written as $\text{e}^{\tan x} (\tan^2 x - 2 \tan x + 3) + C$)