Integration by Parts - Substitution

[sugarxsweet]

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?

 

[sikrut]

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

ln(16x) has a function within itself, therefore substituting that whole part with u won't work.

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

 

[sugarxsweet]

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

 

[ehild]

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

 

[uart]

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

 

[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.

 

[SammyS]

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.