# Integration by Parts - Substitution

1. Sep 10, 2012

### sugarxsweet

1. The problem statement, all variables and given/known data
Evaluate the following indefinite integral:
∫(sin(ln16x))/xdx

2. Relevant equations

3. 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?

2. Sep 10, 2012

### sikrut

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: $$\int u\, du = uv - \int v\, du$$

3. Sep 10, 2012

### sugarxsweet

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

4. Sep 10, 2012

### ehild

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

ehild

5. Sep 10, 2012

### uart

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

6. Sep 10, 2012

### encorelui2

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. Sep 10, 2012

### SammyS

Staff Emeritus
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.