\int _0^{\frac{\pi }{4}}tan^4\theta \:sec^4\theta \:d\theta  calc 2

We are tasked with evaluating the following integral:

\[
I = \int_0^{\frac{\pi}{4}} \tan^4(\theta) \sec^4(\theta) \, d\theta
\]

### Step 1: Rewrite \(\sec^4(\theta)\) in terms of \(\tan(\theta)\)
We can express \(\sec^2(\theta)\) in terms of \(\tan(\theta)\) using the identity:

\[
\sec^2(\theta) = 1 + \tan^2(\theta)
\]

Therefore, \(\sec^4(\theta)\) becomes:

\[
\sec^4(\theta) = (\sec^2(\theta))^2 = (1 + \tan^2(\theta))^2
\]

Now we substitute this into the integral:

\[
I = \int_0^{\frac{\pi}{4}} \tan^4(\theta) (1 + \tan^2(\theta))^2 \, d\theta
\]

### Step 2: Expand \((1 + \tan^2(\theta))^2\)

\[
(1 + \tan^2(\theta))^2 = 1 + 2\tan^2(\theta) + \tan^4(\theta)
\]

Now the integral becomes:

\[
I = \int_0^{\frac{\pi}{4}} \tan^4(\theta) \left( 1 + 2\tan^2(\theta) + \tan^4(\theta) \right) \, d\theta
\]

### Step 3: Distribute \(\tan^4(\theta)\)

\[
I = \int_0^{\frac{\pi}{4}} \left( \tan^4(\theta) + 2\tan^6(\theta) + \tan^8(\theta) \right) \, d\theta
\]

### Step 4: Integrate term by term

Now we evaluate each term individually:
\[
I_1 = \int_0^{\frac{\pi}{4}} \tan^4(\theta) \, d\theta
\]
\[
I_2 = \int_0^{\frac{\pi}{4}} 2\tan^6(\theta) \, d\theta
\]
\[
I_3 = \int_0^{\frac{\pi}{4}} \tan^8(\theta) \, d\theta
\]

To evaluate these integrals, we use reduction formulas for powers of tangent, or we can solve them using standard techniques such as substitution. I'll calculate these integrals now.

The individual integrals and the total result are as follows:

\[
I_1 = \int_0^{\frac{\pi}{4}} \tan^4(\theta) \, d\theta = \frac{\pi}{4} - \frac{2}{3}
\]
\[
I_2 = \int_0^{\frac{\pi}{4}} 2\tan^6(\theta) \, d\theta = \frac{26}{15} - \frac{\pi}{2}
\]
\[
I_3 = \int_0^{\frac{\pi}{4}} \tan^8(\theta) \, d\theta = \frac{\pi}{4} - \frac{76}{105}
\]

Adding these together:

\[
I = \frac{12}{35}
\]

Thus, the value of the integral is \( \frac{12}{35} \).

Would you like further details on any step or a deeper explanation?

Here are 5 follow-up questions for more practice:
1. How do you apply reduction formulas to integrals involving trigonometric powers?
2. What is the general approach to solving integrals with products of powers of trigonometric functions?
3. How would you compute \( \int_0^{\frac{\pi}{4}} \sec^6(\theta) \, d\theta \)?
4. What substitution could simplify \( \int \tan^n(\theta) \sec^m(\theta) \, d\theta \)?
5. How does the identity \( \sec^2(\theta) = 1 + \tan^2(\theta) \) simplify integration?

**Tip**: For higher powers of trigonometric functions, substitution with identities or recurrence relations can significantly simplify the process.

Evaluate the integral: \( \int_0^{\frac{\pi}{4}} \tan^4(\theta) \sec^4(\theta) \, d\theta \)

This page provides a detailed solution to the integral \( \int_0^{\frac{\pi}{4}} \tan^4(\theta) \sec^4(\theta) \, d\theta \). The solution involves using trigonometric identities and integration techniques, with a final result of \( \frac{12}{35} \). Suitable for advanced high school or undergraduate calculus students.

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