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4. 16x – 2x2 – 23 dx -

4. 16x – 2x2 – 23 dx -

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
Trigonometric Substitution There are certain integrals that contain no trigonometry that can be solved with the aid of trigonometry. Consider integrands involving va? -x, a? +x, or vx - a (where a is constant). These can be evaluated by making the following substitutions. Va? + x x= asin 6 x= atan 0 X= asec 0 dx = acos ede dx = a sec*o de dx = asec Otan ede For each integral, figure out which trigonometric substitution should be used. The goal is to rewrite the integrand so that there is no radical. After integrating, draw and label a right triangle (as shown below). This will allow you to rewrite your integrated expression back in terms of x. For this last part of the problem, you may need one or more trigonometric identities. For example, the triangle that corresponds with exercise 1 looks like this. If we let x = 2sin 0, it follows that sin 6 = x/2. By expressing the side length opposite the given angle as x and the hypotenuse as 2, the side length adjacent the given angle works out conveniently to 4 -x, which is the integrand. !3!
Transcribed Image Text:Trigonometric Substitution There are certain integrals that contain no trigonometry that can be solved with the aid of trigonometry. Consider integrands involving va? -x, a? +x, or vx - a (where a is constant). These can be evaluated by making the following substitutions. Va? + x x= asin 6 x= atan 0 X= asec 0 dx = acos ede dx = a sec*o de dx = asec Otan ede For each integral, figure out which trigonometric substitution should be used. The goal is to rewrite the integrand so that there is no radical. After integrating, draw and label a right triangle (as shown below). This will allow you to rewrite your integrated expression back in terms of x. For this last part of the problem, you may need one or more trigonometric identities. For example, the triangle that corresponds with exercise 1 looks like this. If we let x = 2sin 0, it follows that sin 6 = x/2. By expressing the side length opposite the given angle as x and the hypotenuse as 2, the side length adjacent the given angle works out conveniently to 4 -x, which is the integrand. !3!
4. V16x- 2x2-23 dx
Transcribed Image Text:4. V16x- 2x2-23 dx
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