We are tasked with finding the indefinite integral of the following expression:

\[
\int \tan^6(x) \sec^4(x) \, dx
\]

### Step 1: Simplify the integrand using trigonometric identities.

We know the following relationships between \(\tan(x)\) and \(\sec(x)\):
\[
\sec^2(x) = 1 + \tan^2(x)
\]
Thus, we can try to express the integrand in terms of \(\tan(x)\) only.

Start by rewriting the integrand as:
\[
\int \tan^6(x) \sec^4(x) \, dx = \int \tan^6(x) \sec^2(x) \cdot \sec^2(x) \, dx
\]

Since \(\sec^2(x) = 1 + \tan^2(x)\), we can write:
\[
\int \tan^6(x) \sec^2(x) \cdot \sec^2(x) \, dx = \int \tan^6(x) \sec^2(x) \cdot (1 + \tan^2(x)) \, dx
\]

### Step 2: Substitute \( u = \tan(x) \).
Let:
\[
u = \tan(x) \quad \text{so that} \quad du = \sec^2(x) \, dx
\]

Thus, the integral becomes:
\[
\int u^6 (1 + u^2) \, du
\]

### Step 3: Expand and integrate.
Now expand the integrand:
\[
\int u^6 (1 + u^2) \, du = \int (u^6 + u^8) \, du
\]

Now integrate term by term:
\[
\int u^6 \, du = \frac{u^7}{7} \quad \text{and} \quad \int u^8 \, du = \frac{u^9}{9}
\]

Thus, the result is:
\[
\frac{u^7}{7} + \frac{u^9}{9} + C
\]

### Step 4: Substitute \( u = \tan(x) \) back.
Substitute back \( u = \tan(x) \):
\[
\frac{\tan^7(x)}{7} + \frac{\tan^9(x)}{9} + C
\]

### Final answer:
\[
\int \tan^6(x) \sec^4(x) \, dx = \frac{\tan^7(x)}{7} + \frac{\tan^9(x)}{9} + C
\]

Would you like more details on any step?

Here are 5 related questions you can explore:
1. How do you integrate powers of \(\sec(x)\) and \(\tan(x)\) together?
2. What is the integral of \(\sec^4(x)\)?
3. Can substitution methods always be used for trigonometric integrals?
4. How can you solve integrals involving both \(\sin(x)\) and \(\cos(x)\)?
5. What is the general form of an integral involving powers of trigonometric functions?

**Tip:** When dealing with powers of trigonometric functions, use identities to simplify expressions before integrating.

Indefinite integral of tan^6(x)×sec^4(x)

Learn how to find the indefinite integral of tan^6(x) × sec^4(x) using trigonometric identities and substitution methods. This step-by-step solution is perfect for students in Grades 11-12 looking to master advanced calculus and trigonometry techniques.

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