Textbook Question
7–64. Integration review Evaluate the following integrals.
53. ∫ eˣ sec(eˣ + 1) dx
Ch. 8 - Integration Techniques
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 8, Problem 8.2.48
48. Integral of sec³x Use integration by parts to show that:
∫ sec³x dx = (1/2) secx tanx + (1/2) ∫ secx dx
Verified step by step guidance1
Start with the integral \( \int \sec^3 x \, dx \). Rewrite \( \sec^3 x \) as \( \sec x \cdot \sec^2 x \) to prepare for integration by parts.
Choose \( u = \sec x \) and \( dv = \sec^2 x \, dx \). Then compute \( du = \sec x \tan x \, dx \) and \( v = \tan x \) since the derivative of \( \tan x \) is \( \sec^2 x \).
Apply the integration by parts formula: \( \int u \, dv = uv - \int v \, du \). Substitute the chosen \( u \), \( v \), \( du \), and \( dv \) to get \( \int \sec^3 x \, dx = \sec x \tan x - \int \tan x \cdot \sec x \tan x \, dx \).
Simplify the integral \( \int \tan x \cdot \sec x \tan x \, dx = \int \sec x \tan^2 x \, dx \). Use the identity \( \tan^2 x = \sec^2 x - 1 \) to rewrite the integral as \( \int \sec x (\sec^2 x - 1) \, dx = \int \sec^3 x \, dx - \int \sec x \, dx \).
Substitute back into the equation and solve for \( \int \sec^3 x \, dx \) to isolate it on one side. This will lead to the expression \( \int \sec^3 x \, dx = \frac{1}{2} \sec x \tan x + \frac{1}{2} \int \sec x \, dx \), as required.
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6mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Integration by Parts
Integration by parts is a technique derived from the product rule of differentiation. It transforms the integral of a product of functions into simpler integrals, using the formula ∫u dv = uv - ∫v du. Choosing appropriate u and dv is crucial to simplify the integral effectively.
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Integration by Parts for Definite Integrals
Trigonometric Identities
Trigonometric identities relate different trigonometric functions and are essential for simplifying integrals involving trig functions. For example, knowing that d/dx (sec x) = sec x tan x helps in recognizing parts of the integral and applying integration by parts correctly.
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Integral of sec x
The integral of sec x, ∫ sec x dx, is a standard integral that equals ln |sec x + tan x| + C. Recognizing this integral within the problem allows for expressing the solution in terms of known functions, completing the integration process.
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