Step 1: Understanding the Concept:
To integrate high powers of the tangent function, we factor out $\tan^2 x$ and replace it using the Pythagorean identity $\tan^2 x = \sec^2 x - 1$. This sets up integrals that can be solved using $u$-substitution.
Step 2: Key Formula or Approach:
Identity: $\tan^2 x = \sec^2 x - 1$.
Integral rule: $\int [f(x)]^n \cdot f'(x) dx = \frac{[f(x)]^{n+1}}{n+1} + C$.
Step 3: Detailed Explanation:
Let $I = \int \tan^4 x dx$.
Rewrite $\tan^4 x$ by splitting off $\tan^2 x$:
\[ I = \int \tan^2 x \cdot \tan^2 x dx \]
Use the identity on one of the terms:
\[ I = \int \tan^2 x (\sec^2 x - 1) dx \]
Distribute the multiplication:
\[ I = \int \tan^2 x \sec^2 x dx - \int \tan^2 x dx \]
For the first integral, notice that the derivative of $\tan x$ is $\sec^2 x$. Let $u = \tan x$, then $du = \sec^2 x dx$. The integral becomes $\int u^2 du = \frac{u^3}{3}$.
So, $\int \tan^2 x \sec^2 x dx = \frac{\tan^3 x}{3}$.
For the second integral, use the identity again:
\[ \int \tan^2 x dx = \int (\sec^2 x - 1) dx = \int \sec^2 x dx - \int 1 dx \]
We know the standard integrals: $\int \sec^2 x dx = \tan x$ and $\int 1 dx = x$.
So, $\int \tan^2 x dx = \tan x - x$.
Combine everything together:
\[ I = \frac{\tan^3 x}{3} - (\tan x - x) + k \]
\[ I = \frac{1}{3} \tan^3 x - 1 \cdot \tan x + 1 \cdot x + k \]
Compare this result with the given form $a \tan^3 x + b \tan x + cx + k$:
$a = \frac{1}{3}$
$b = -1$
$c = 1$
Now, calculate the required expression:
\[ a - b + c = \left(\frac{1}{3}\right) - (-1) + (1) \]
\[ a - b + c = \frac{1}{3} + 1 + 1 = \frac{1}{3} + 2 = \frac{7}{3} \]
Step 4: Final Answer:
The value of $a - b + c$ is $\frac{7}{3}$.