To evaluate the integral of \(\tan^7(x) \sec^4(x) \, dx\), we can use a combination of trigonometric identities and substitution methods. 

### Step 1: Simplify using trigonometric identities
We know that:
\[
\sec^2(x) = 1 + \tan^2(x)
\]

This allows us to write:
\[
\sec^4(x) = (\sec^2(x))^2 = (1 + \tan^2(x))^2
\]

Thus, the integral becomes:
\[
\int \tan^7(x) \sec^4(x) \, dx = \int \tan^7(x) (1 + \tan^2(x))^2 \, dx
\]

### Step 2: Use substitution
Let:
\[
u = \tan(x)
\]
Then:
\[
\frac{du}{dx} = \sec^2(x)
\]
or equivalently:
\[
du = \sec^2(x) \, dx
\]
The integral becomes:
\[
\int u^7 (1 + u^2)^2 \, du
\]

### Step 3: Expand and integrate
Now, expand the integrand:
\[
(1 + u^2)^2 = 1 + 2u^2 + u^4
\]
Thus:
\[
\int u^7 (1 + 2u^2 + u^4) \, du = \int (u^7 + 2u^9 + u^{11}) \, du
\]

### Step 4: Integrate term by term
Now, integrate each term:
\[
\int u^7 \, du = \frac{u^8}{8}
\]
\[
\int 2u^9 \, du = \frac{2u^{10}}{10} = \frac{u^{10}}{5}
\]
\[
\int u^{11} \, du = \frac{u^{12}}{12}
\]

So the integral becomes:
\[
\frac{u^8}{8} + \frac{u^{10}}{5} + \frac{u^{12}}{12} + C
\]

### Step 5: Substitute back \(u = \tan(x)\)
Substitute back \(u = \tan(x)\) to get the final answer:
\[
\frac{\tan^8(x)}{8} + \frac{\tan^{10}(x)}{5} + \frac{\tan^{12}(x)}{12} + C
\]

This is the evaluated integral.

Would you like further details or have any questions?

Here are some related questions:
1. How do trigonometric identities simplify integration?
2. What are some common substitutions used in trigonometric integrals?
3. How can you verify the result of an integral?
4. What are other methods to evaluate \(\int \tan^7(x) \sec^4(x) \, dx\)?
5. How do you approach integrals involving higher powers of trigonometric functions?

**Tip:** When integrating trigonometric functions, always check if a substitution involving \(\tan(x)\) or \(\sec(x)\) simplifies the integral.

This detailed guide demonstrates how to evaluate the integral of tan^7(x) sec^4(x) dx by leveraging trigonometric identities and the substitution method. Ideal for advanced high school students in Grades 11-12, the solution walks through simplifying the integral, applying substitution, and integrating each term.

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