# Integration by Parts - Substitution

## Homework Statement

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

## 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?

you need to substitute with two variables, u & v.

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

keep in mind: $$\int u\, du = uv - \int v\, du$$

Sorry, stupid question - if there's both u and v, how do I choose which one goes with which?

ehild
Homework Helper

## Homework Statement

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

## 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

uart
Yep, I can't see anything wrong except that you forgot the constant.

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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.