Integration by Parts - Substitution

In summary, the indefinite integral of (sin(ln16x))/xdx can be evaluated by substituting u = ln16x and du = 1/x, and then using the trigonometric identity for the integral of sin(u). The final solution will include a constant of integration. Alternative methods, such as integration by parts, are not necessary for this problem.
  • #1
sugarxsweet
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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  • #2
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 won't 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
sugarxsweet said:

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
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
encorelui2 said:
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.
 

1. What is the difference between Integration by Parts and Substitution?

Integration by Parts and Substitution are two different techniques used to evaluate integrals. Integration by Parts involves rewriting an integral in a different form in order to make it easier to solve, while Substitution involves replacing a variable in an integral with a new variable in order to simplify it.

2. When should I use Integration by Parts?

Integration by Parts is typically used when the integral involves a product of functions, such as a polynomial and an exponential function. It is also useful when the integral involves a trigonometric function raised to a power.

3. How do I choose which function to substitute in Substitution?

In Substitution, the goal is to choose a new variable that will simplify the integral. This is typically done by choosing a function that appears in the integral and substituting it with a new variable. It may also be helpful to choose a function that has a simple derivative.

4. Can I use both Integration by Parts and Substitution in the same integral?

Yes, it is possible to use both Integration by Parts and Substitution in the same integral. This is known as double integration by parts or double substitution, and it can be used in cases where one technique alone is not enough to solve the integral.

5. Are there any general tips for solving integrals using these techniques?

One tip for solving integrals using Integration by Parts and Substitution is to always check your answer by differentiating it. This can help catch any mistakes that may have been made during the integration process. It is also important to practice and become familiar with the various integration formulas and techniques in order to solve integrals efficiently.

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