9–61. Trigonometric integrals Evaluate the following integrals.
40. ∫[0 to π/6] tan⁵(2x) sec(2x) dx

Verified step by step guidance

1. Step 1: Recognize that the integral involves a combination of trigonometric functions, specifically tan⁵(2x) and sec(2x). To simplify, recall that the derivative of tan(u) is sec²(u), which suggests a substitution method might be effective.
2. Step 2: Let u = tan(2x). Then, the derivative of u with respect to x is du/dx = 2 sec²(2x). Rearrange this to express dx in terms of du: dx = du / (2 sec²(2x)).
3. Step 3: Rewrite the integral using the substitution u = tan(2x). Since tan⁵(2x) = u⁵ and sec(2x) remains in the integral, substitute dx as well: ∫[0 to π/6] tan⁵(2x) sec(2x) dx = ∫[0 to π/6] u⁵ sec(2x) * (du / (2 sec²(2x))).
4. Step 4: Simplify the integral by canceling one sec(2x) term in the numerator with one in the denominator, leaving: ∫ u⁵ * (1 / 2 sec(2x)) du. Note that sec²(2x) is related to tan²(2x) via the identity sec²(θ) = 1 + tan²(θ), which can be used to express sec²(2x) in terms of u.
5. Step 5: Adjust the limits of integration to match the substitution. When x = 0, u = tan(2 * 0) = 0. When x = π/6, u = tan(2 * π/6) = tan(π/3). The integral now becomes ∫[0 to tan(π/3)] u⁵ / 2 du, which can be solved by applying the power rule for integration.

Key Concepts

Trigonometric Functions

Trigonometric functions, such as tangent (tan) and secant (sec), are fundamental in calculus, particularly in integration. The tangent function is defined as the ratio of the sine and cosine functions, while the secant function is the reciprocal of the cosine function. Understanding their properties and relationships is crucial for evaluating integrals involving these functions.

Integration Techniques

Integration techniques, such as substitution and integration by parts, are essential for solving complex integrals. In this case, recognizing that the derivative of sec(2x) is tan(2x) sec(2x) allows for a substitution that simplifies the integral. Mastery of these techniques enables students to tackle a variety of integral forms effectively.

Definite Integrals

Definite integrals calculate the area under a curve between two specified limits, in this case, from 0 to π/6. The evaluation of definite integrals involves finding the antiderivative of the integrand and then applying the Fundamental Theorem of Calculus. Understanding how to compute definite integrals is crucial for solving problems in calculus, especially those involving trigonometric functions.