Integration by Parts - Substitution

  • #1

Homework Statement


Evaluate the following indefinite integral:
∫(sin(ln16x))/xdx


Homework Equations





The Attempt at a Solution


let u = ln16x
therefore du=16/16x=1/x

∫sinudu
=-cosu
=-cos(ln16x)

Why is this showing as the wrong answer?
 

Answers and Replies

  • #2
49
1
you need to substitute with two variables, u & v.

[itex] ln(16x) [/itex] has a function within itself, therefore substituting that whole part with u wont work.

keep in mind: [tex] \int u\, du = uv - \int v\, du [/tex]
 
  • #3
Sorry, stupid question - if there's both u and v, how do I choose which one goes with which?
 
  • #4
ehild
Homework Helper
15,543
1,912

Homework Statement


Evaluate the following indefinite integral:
∫(sin(ln16x))/xdx


Homework Equations





The Attempt at a Solution


let u = ln16x
therefore du=16/16x=1/x

∫sinudu
=-cosu
=-cos(ln16x)

Why is this showing as the wrong answer?

Didn't you forget the arbitrary constant? The integral is -cos(ln16x))+C.

ehild
 
  • #5
uart
Science Advisor
2,776
9
Yep, I can't see anything wrong except that you forgot the constant.:smile:
 
  • #6
Substitute for 16x and then further substitute for log(16x). You should end up with Int([e^a][sin(a)])/16

At 1st I did it just like u did, then I saw sikrut's answer... Reworked it & verified with Matlab. Turns out Matlab agrees with Sikrut.
 
  • #7
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,365
1,032
Substitute for 16x and then further substitute for log(16x). You should end up with Int([e^a][sin(a)])/16

At 1st I did it just like u did, then I saw sikrut's answer... Reworked it & verified with Matlab. Turns out Matlab agrees with Sikrut.

This problem is adequately handled by substitution, exactly as you did it --- of course, include the constant of integration.

No need for integration by parts. I don't see how that would even be helpful.
 

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