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Evaluate the indefinite integral. | x34 sin(x35) dx

Evaluate the indefinite integral. | x34 sin(x35) 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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I just would like some help answering this question

If we choose 

u = x35,

 then 

du = 35x34 dx.



If 

u = x35

 is substituted into 

  x34 sin(x35) dx
 
 
,

 then we have 

  x34 sin u dx
 
 
 = 
  sin u(x34 dx)
 
 
.



We must also convert 

x34 dx

 into an expression involving u.

We know that 

du = 35x34 dx,

 and so 

x34 dx = 
WA МАТH 151 — 5.5 HW —MJ - МА X b My Questions | bartleby -> A webassign.net/web/Student/Assignment-Responses/submit?pos=3&dep=28475968&tags=autosave#question2632218_3 TIIJ qutJLION TiA5 JTVCI ai pai t5 triaL mUst Dt CompicLTu styuTIICially. IT you 5Kip a pait UI CTe qutStION, you WII TIUL TEttivt ally poiTCS TOI CITIC SKippeu part, aiTu you vwIII TIVC DE able to come back to the skipped part. Tutorial Exercise Evaluate the indefinite integral. |x34 sin(x35) dx Step 1 We must decide what to choose for u. If u = f(x), then du = f '(x) dx, and so it is helpful to look for some expression in x34 sin(x35) dx for which the derivative is also present, though perhaps missing a constant factor. For example, x35 is part of this integral, and the derivative of x35 is 35x34 which is also present except for a constant. 3534 Step 2 If we choose u = x35, then du = 35x34 dx. If u = x35 is substituted into x34 sin(x35) dx, then we have x34 sin u dx = sin u(x34 dx). We must also convert x34 dx into an expression involving u. We know that du = 35×34 dx. and so x34 dx =. 35 du +
Transcribed Image Text:WA МАТH 151 — 5.5 HW —MJ - МА X b My Questions | bartleby -> A webassign.net/web/Student/Assignment-Responses/submit?pos=3&dep=28475968&tags=autosave#question2632218_3 TIIJ qutJLION TiA5 JTVCI ai pai t5 triaL mUst Dt CompicLTu styuTIICially. IT you 5Kip a pait UI CTe qutStION, you WII TIUL TEttivt ally poiTCS TOI CITIC SKippeu part, aiTu you vwIII TIVC DE able to come back to the skipped part. Tutorial Exercise Evaluate the indefinite integral. |x34 sin(x35) dx Step 1 We must decide what to choose for u. If u = f(x), then du = f '(x) dx, and so it is helpful to look for some expression in x34 sin(x35) dx for which the derivative is also present, though perhaps missing a constant factor. For example, x35 is part of this integral, and the derivative of x35 is 35x34 which is also present except for a constant. 3534 Step 2 If we choose u = x35, then du = 35x34 dx. If u = x35 is substituted into x34 sin(x35) dx, then we have x34 sin u dx = sin u(x34 dx). We must also convert x34 dx into an expression involving u. We know that du = 35×34 dx. and so x34 dx =. 35 du +
WA МАТH 151 — 5.5 HW —MJ - МА X b My Questions | bartleby x + -> A webassign.net/web/Student/Assignment-Responses/submit?pos=3&dep=28475968&tags=autosave#question2632218_3 If u = f(x), then du = f '(x) dx, and so it is helpful to look for some expression in x34 sin(x35) dx for which the derivative is also present, though perhaps missing a constant factor. For example, x35 is part of this integral, and the derivative of x35 is 35x34 which is also present except for a constant. 3534 Step 2 If we choose u = x35, then du = 35x34 dx. If u = x35 is substituted into x34 sin(x35) dx, then we have x34 sin u dx = sin u(x34 dx). We must also convert x34 dx into an expression involving u. We know that du = 35x34 dx, and so x34 dx =| 35 du. Submit Skip (you cannot come back) Need Help? Read It Submit Answer
Transcribed Image Text:WA МАТH 151 — 5.5 HW —MJ - МА X b My Questions | bartleby x + -> A webassign.net/web/Student/Assignment-Responses/submit?pos=3&dep=28475968&tags=autosave#question2632218_3 If u = f(x), then du = f '(x) dx, and so it is helpful to look for some expression in x34 sin(x35) dx for which the derivative is also present, though perhaps missing a constant factor. For example, x35 is part of this integral, and the derivative of x35 is 35x34 which is also present except for a constant. 3534 Step 2 If we choose u = x35, then du = 35x34 dx. If u = x35 is substituted into x34 sin(x35) dx, then we have x34 sin u dx = sin u(x34 dx). We must also convert x34 dx into an expression involving u. We know that du = 35x34 dx, and so x34 dx =| 35 du. Submit Skip (you cannot come back) Need Help? Read It Submit Answer
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