Ch. 8 - Integration Techniques

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 8 - Integration Techniques

Problem 8.3.7

Chapter 8, Problem 8.3.7

7. How would you evaluate ∫ tan¹⁰x sec²x dx?

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Recognize that the integral involves a trigonometric function raised to a power, specifically tan¹⁰(x), and sec²(x). Recall that the derivative of tan(x) is sec²(x), which suggests a substitution method might be appropriate.

Let u = tan(x). Then, the derivative of u with respect to x is du/dx = sec²(x), or equivalently, du = sec²(x) dx. This substitution simplifies the integral.

Rewrite the integral in terms of u. Since tan(x) = u and sec²(x) dx = du, the integral becomes ∫ u¹⁰ du.

Apply the power rule for integration, which states that ∫ uⁿ du = (uⁿ⁺¹)/(n+1) + C, where n ≠ -1. Here, n = 10, so the integral becomes (u¹¹)/11 + C.

Finally, substitute back u = tan(x) to express the result in terms of x. The final expression is (tan¹¹(x))/11 + C.

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Key Concepts

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

Integration Techniques

Integration techniques are methods used to find the integral of a function. In this case, recognizing that the integral involves the function tan(x) and its derivative sec²(x) is crucial. This allows for the use of substitution, where we can let u = tan(x), simplifying the integral significantly.

Substitution Method

The substitution method is a powerful technique in calculus that simplifies the process of integration. By substituting a part of the integral with a new variable, we can transform the integral into a more manageable form. For the integral ∫ tan¹⁰x sec²x dx, substituting u = tan(x) leads to a straightforward integration of u¹⁰ du.

Definite and Indefinite Integrals

Understanding the difference between definite and indefinite integrals is essential in calculus. An indefinite integral, like ∫ tan¹⁰x sec²x dx, represents a family of functions and includes a constant of integration. In contrast, a definite integral computes the area under the curve between two limits. Recognizing this distinction helps in correctly interpreting the results of integration.

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