Evaluate the integrals in Exercises 33–52.
∫ from -π/4 to π/4 of 6 tan⁴(x) dx

Verified step by step guidance

1. Recognize that the integral is \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 6 \tan^{4}(x) \, dx \). The constant 6 can be factored out of the integral, so rewrite it as \( 6 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^{4}(x) \, dx \).
2. Note that \( \tan^{4}(x) \) is an even function because \( \tan(-x) = -\tan(x) \), so \( \tan^{4}(-x) = \tan^{4}(x) \). This means the integral from \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \) can be simplified to twice the integral from 0 to \( \frac{\pi}{4} \): \( 6 \times 2 \int_0^{\frac{\pi}{4}} \tan^{4}(x) \, dx = 12 \int_0^{\frac{\pi}{4}} \tan^{4}(x) \, dx \).
3. Express \( \tan^{4}(x) \) in terms of \( \sec(x) \) and \( \tan(x) \) using the identity \( \tan^{2}(x) = \sec^{2}(x) - 1 \). So, \( \tan^{4}(x) = (\tan^{2}(x))^{2} = (\sec^{2}(x) - 1)^{2} = \sec^{4}(x) - 2\sec^{2}(x) + 1 \).
4. Rewrite the integral as \( 12 \int_0^{\frac{\pi}{4}} (\sec^{4}(x) - 2\sec^{2}(x) + 1) \, dx = 12 \left( \int_0^{\frac{\pi}{4}} \sec^{4}(x) \, dx - 2 \int_0^{\frac{\pi}{4}} \sec^{2}(x) \, dx + \int_0^{\frac{\pi}{4}} 1 \, dx \right) \).
5. Evaluate each integral separately: use reduction formulas or known integrals for \( \int \sec^{2}(x) \, dx \) and \( \int \sec^{4}(x) \, dx \), and the integral of 1 is straightforward. Then combine the results and multiply by 12 to express the final integral.

Key Concepts

Definite Integrals

A definite integral calculates the net area under a curve between two specific limits. It involves evaluating the integral function at the upper and lower bounds and subtracting these values. This concept is essential for finding the exact value of the integral from -π/4 to π/4.

Integration of Trigonometric Functions

Integrating powers of trigonometric functions like tan⁴(x) often requires using identities or substitution methods. Recognizing how to rewrite or simplify the integrand using trigonometric identities helps in evaluating the integral more easily.

Symmetry in Definite Integrals

When integrating an even or odd function over symmetric limits (like -a to a), symmetry properties can simplify the calculation. For example, the integral of an even function over symmetric limits is twice the integral from 0 to a, while the integral of an odd function is zero.