Evaluate the integral. e50 sin(60) do Step 1 We will begin by letting u = sin(60) and dv = e50 de. Then du = 6 cos(60) 6 cos (60) de and v = сл e50 5 50 Step 2 After integration by parts we have Step 3 5 50 sin(60) de = 50 sin (60) 5 1 - 50 1.2✔ 6/5 5 130 e* sin(60) Je We'll now apply the integration by parts procedure to the new integral Then du = -6 sin(60) dU 50 -6 sin(60) de and V = e 5 Step 4 After integration by parts we have 50 e50 cos(60) de. e50 co cos(60) de, letting U = cos(60) and dV = esg 35 /e50 e 50 cos(60) 1 cos(60) de = -1.2 -6/5 50 5 5' -ecos (60) e50 si sin(60) de. e50 de. Step 5 Notice that the original integral Se 50 I = =Se 1 50 sin(60) do, we have I 50 = e 5 sin(60) de is once again present. In fact, putting together all our work so far, if we let sin(60) - [1e50 cos(60) + 131] + C₁. 55 5 This can be re-arranged to give us I = 150 5 sin(60) -0.24✔ 6/25 e50 cos(60) - 1.44 ✔ Step 6 We can move all the I terms to one side and get Step 7 2.44 ✔ 1 6 50 61/25 I sin(60) - - e50 cos(60) + C₁. 5 25 36/25 I+C1. And now we finally have Jes e50 sin(60) de = I = 25 + C, where C = C₁. 61 Putting all of this together and incorporating the constant of integration, C, we have e50 sin(60) de =

Evaluate the integral. e50 sin(60) do Step 1 We will begin by letting u = sin(60) and dv = e50 de. Then du = 6 cos(60) 6 cos (60) de and v = сл e50 5 50 Step 2 After integration by parts we have Step 3 5 50 sin(60) de = 50 sin (60) 5 1 - 50 1.2✔ 6/5 5 130 e* sin(60) Je We'll now apply the integration by parts procedure to the new integral Then du = -6 sin(60) dU 50 -6 sin(60) de and V = e 5 Step 4 After integration by parts we have 50 e50 cos(60) de. e50 co cos(60) de, letting U = cos(60) and dV = esg 35 /e50 e 50 cos(60) 1 cos(60) de = -1.2 -6/5 50 5 5' -ecos (60) e50 si sin(60) de. e50 de. Step 5 Notice that the original integral Se 50 I = =Se 1 50 sin(60) do, we have I 50 = e 5 sin(60) de is once again present. In fact, putting together all our work so far, if we let sin(60) - [1e50 cos(60) + 131] + C₁. 55 5 This can be re-arranged to give us I = 150 5 sin(60) -0.24✔ 6/25 e50 cos(60) - 1.44 ✔ Step 6 We can move all the I terms to one side and get Step 7 2.44 ✔ 1 6 50 61/25 I sin(60) - - e50 cos(60) + C₁. 5 25 36/25 I+C1. And now we finally have Jes e50 sin(60) de = I = 25 + C, where C = C₁. 61 Putting all of this together and incorporating the constant of integration, C, we have e50 sin(60) de =

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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