find the integral of tan power5 theta sec theta

We are tasked with finding the integral of \( \tan^5(\theta) \sec(\theta) \, d\theta \). Let's break this down step by step.

### Problem:
\[
I = \int \tan^5(\theta) \sec(\theta) \, d\theta
\]

### Solution:
We can use a substitution approach. First, recall that the derivative of \( \tan(\theta) \) is \( \sec^2(\theta) \), which suggests that we may want to work with \( \tan(\theta) \).

Let's rewrite the integrand as follows:
\[
\tan^5(\theta) \sec(\theta) = \tan^4(\theta) \cdot \tan(\theta) \cdot \sec(\theta)
\]

Now, we express \( \tan^4(\theta) \) as \( (\tan^2(\theta))^2 \):
\[
= (\tan^2(\theta))^2 \cdot \tan(\theta) \cdot \sec(\theta)
\]

Using the identity \( \tan^2(\theta) = \sec^2(\theta) - 1 \), we substitute:
\[
= (\sec^2(\theta) - 1)^2 \cdot \tan(\theta) \cdot \sec(\theta)
\]

At this point, let \( u = \tan(\theta) \), hence \( du = \sec^2(\theta) d\theta \). The integral becomes:

\[
I = \int u^5 \, du
\]

The solution to this integral is:
\[
I = \frac{u^6}{6} + C = \frac{\tan^6(\theta)}{6} + C
\]

Thus, the integral of \( \tan^5(\theta) \sec(\theta) \, d\theta \) is:

\[
\boxed{ \frac{\tan^6(\theta)}{6} + C }
\]

### Do you want details or have any questions? 

### Related Questions:
1. How would the integral change if we had \( \tan^4(\theta) \sec(\theta) \)?
2. Can we solve \( \int \tan^3(\theta) \sec(\theta) \, d\theta \)?
3. What happens if we have \( \int \tan^6(\theta) \sec(\theta) \, d\theta \)?
4. How can substitution help in solving trigonometric integrals?
5. What is the general rule for integrating powers of tangent and secant?

### Tip:  
For integrals involving powers of trigonometric functions, use trigonometric identities and substitution to simplify the expression before integrating.

Learn how to compute the integral of tan^5(θ) sec(θ) using trigonometric identities and substitution. This detailed step-by-step solution explores the key concepts of trigonometric integration, perfect for pre-calculus and calculus students.

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