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Ch. 8 - Integration Techniques
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.3.53

9–61. Trigonometric integrals Evaluate the following integrals.
53. ∫ from 0 to π/4 of sec⁴θ dθ

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Step 1: Recognize that the integral involves sec⁴θ, which is a trigonometric function. To simplify the integration, recall the reduction formula for sec⁴θ: sec⁴θ = 1 + 2sec²θ.
Step 2: Split the integral using the reduction formula: ∫ sec⁴θ dθ = ∫ (1 + 2sec²θ) dθ. This allows us to integrate each term separately.
Step 3: For the first term, ∫ 1 dθ, the integral is straightforward and equals θ.
Step 4: For the second term, ∫ 2sec²θ dθ, recall the integral of sec²θ is tanθ. Therefore, ∫ 2sec²θ dθ = 2tanθ.
Step 5: Combine the results of the two integrals and evaluate the definite integral from 0 to π/4. Substitute the limits of integration into the combined expression θ + 2tanθ.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Trigonometric Functions

Trigonometric functions, such as secant (sec), are fundamental in calculus, representing relationships between angles and sides in right triangles. The secant function is defined as the reciprocal of the cosine function, sec(θ) = 1/cos(θ). Understanding these functions is crucial for evaluating integrals involving trigonometric expressions.
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Integration Techniques

Integration techniques are methods used to find the integral of a function. For trigonometric integrals, techniques such as substitution, integration by parts, or recognizing standard integral forms are often employed. Mastery of these techniques is essential for solving integrals like ∫ sec⁴θ dθ.
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Definite Integrals

Definite integrals calculate the area under a curve between two specified limits, in this case, from 0 to π/4. The result of a definite integral is a numerical value representing this area. Understanding how to evaluate definite integrals is key to solving problems that involve specific bounds, such as the one presented in the question.
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