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Evaluate the integral. 6x2 sin AX dx Step 1 To use the integration-by-parts formula u dy = uvV – v du, we must choose one part of 6x2 sin Tx dx to be u, with the rest becoming dv. Since the goal is to produce a simpler integral, we will choose u = 6x2. This means that sin ( Tx) dv = dx. sin (7 x) Step 2 Now, since u = 6x2, then du = 12x dx. 12x Step 3 With our choice that dv = sin TX dx, then v = sin Ax dx, which can be calculated using the substitution w = TX. In this case, we have dx = dw, and we get v = (in terms of x).

Evaluate the integral. 6x2 sin AX dx Step 1 To use the integration-by-parts formula u dy = uvV – v du, we must choose one part of 6x2 sin Tx dx to be u, with the rest becoming dv. Since the goal is to produce a simpler integral, we will choose u = 6x2. This means that sin ( Tx) dv = dx. sin (7 x) Step 2 Now, since u = 6x2, then du = 12x dx. 12x Step 3 With our choice that dv = sin TX dx, then v = sin Ax dx, which can be calculated using the substitution w = TX. In this case, we have dx = dw, and we get v = (in terms of x).

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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Evaluate the integral. 6x2 sin AX dx Step 1 To use the integration-by-parts formula u dy = uvV – v du, we must choose one part of 6x2 sin Tx dx to be u, with the rest becoming dv. Since the goal is to produce a simpler integral, we will choose u = 6x2. This means that sin ( Tx) dv = dx. sin (7 x) Step 2 Now, since u = 6x2, then du = 12x dx. 12x Step 3 With our choice that dv = sin TX dx, then v = sin Ax dx, which can be calculated using the substitution w = TX. In this case, we have dx = dw, and we get v = (in terms of x).
Transcribed Image Text:Evaluate the integral. 6x2 sin AX dx Step 1 To use the integration-by-parts formula u dy = uvV – v du, we must choose one part of 6x2 sin Tx dx to be u, with the rest becoming dv. Since the goal is to produce a simpler integral, we will choose u = 6x2. This means that sin ( Tx) dv = dx. sin (7 x) Step 2 Now, since u = 6x2, then du = 12x dx. 12x Step 3 With our choice that dv = sin TX dx, then v = sin Ax dx, which can be calculated using the substitution w = TX. In this case, we have dx = dw, and we get v = (in terms of x).
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