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
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Please help me with these questions. I have completed steps 1-6. Just need help with steps 7. Thank you
Transcribed Image Text: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
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F88e366cf-af4b-482a-858f-e216b02790ea%2Fe8deb57d-c2d0-4f77-8b6f-a2e59b7d1485%2Fk8ar2po_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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
=
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