- #1

- 12

- 0

## 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?

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- Thread starter sugarxsweet
- Start date

- #1

- 12

- 0

Evaluate the following indefinite integral:

∫(sin(ln16x))/xdx

let u = ln16x

therefore du=16/16x=1/x

∫sinudu

=-cosu

=-cos(ln16x)

Why is this showing as the wrong answer?

- #2

- 49

- 1

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

keep in mind: [tex] \int u\, du = uv - \int v\, du [/tex]

- #3

- 12

- 0

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.

- #6

- 7

- 0

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

- 11,368

- 1,035

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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