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The indefinite integral dæ can be solved using either of the three methods: 1. integration by 4+2? substitution, 2. integration by parts, and 3. integration by trigonometric substitution. The method of integration by substitution is based on the chain rule for differentiation. 1. If we applied integration by substitution, we would substitute: u = (4+x^2)^1/2 du x/(x^2+4)^1/2 da u^2-4 %3D 2. If we applied integration by parts, we would use: xp: 4+22 dv = %3D u = x^2 2x du da, v = V4 + x2. %3D 3. If we used integration by substitution, we would use a right-triangle, like the one pictured below, with: hyp opp adj hyp = V4+ x? opp = adj = x This would give: sec 0 =V4 + æ? tan 0 sec? 0 de da tan 0 = This trigonometric substitution would give: xp: V4+z? S tan 0 sec 0 do. 2.

The indefinite integral dæ can be solved using either of the three methods: 1. integration by 4+2? substitution, 2. integration by parts, and 3. integration by trigonometric substitution. The method of integration by substitution is based on the chain rule for differentiation. 1. If we applied integration by substitution, we would substitute: u = (4+x^2)^1/2 du x/(x^2+4)^1/2 da u^2-4 %3D 2. If we applied integration by parts, we would use: xp: 4+22 dv = %3D u = x^2 2x du da, v = V4 + x2. %3D 3. If we used integration by substitution, we would use a right-triangle, like the one pictured below, with: hyp opp adj hyp = V4+ x? opp = adj = x This would give: sec 0 =V4 + æ? tan 0 sec? 0 de da tan 0 = This trigonometric substitution would give: xp: V4+z? S tan 0 sec 0 do. 2.

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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The indefinite integral dæ can be solved using either of the three methods: 1. integration by 4+2? substitution, 2. integration by parts, and 3. integration by trigonometric substitution. The method of integration by substitution is based on the chain rule for differentiation. 1. If we applied integration by substitution, we would substitute: u = (4+x^2)^1/2 du x/(x^2+4)^1/2 da u^2-4 %3D 2. If we applied integration by parts, we would use: xp: 4+22 dv = %3D u = x^2 2x du da, v = V4 + x2. %3D
Transcribed Image Text:The indefinite integral dæ can be solved using either of the three methods: 1. integration by 4+2? substitution, 2. integration by parts, and 3. integration by trigonometric substitution. The method of integration by substitution is based on the chain rule for differentiation. 1. If we applied integration by substitution, we would substitute: u = (4+x^2)^1/2 du x/(x^2+4)^1/2 da u^2-4 %3D 2. If we applied integration by parts, we would use: xp: 4+22 dv = %3D u = x^2 2x du da, v = V4 + x2. %3D
3. If we used integration by substitution, we would use a right-triangle, like the one pictured below, with: hyp opp adj hyp = V4+ x? opp = adj = x This would give: sec 0 =V4 + æ? tan 0 sec? 0 de da tan 0 = This trigonometric substitution would give: xp: V4+z? S tan 0 sec 0 do. 2.
Transcribed Image Text:3. If we used integration by substitution, we would use a right-triangle, like the one pictured below, with: hyp opp adj hyp = V4+ x? opp = adj = x This would give: sec 0 =V4 + æ? tan 0 sec? 0 de da tan 0 = This trigonometric substitution would give: xp: V4+z? S tan 0 sec 0 do. 2.
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