Cyclic Functions and Integration by Parts: Where Did I Go Wrong?

In summary: I tried it with sin on the left instead of g and it worked out fine. So, it seems like the order of integration matters.
  • #1
pivoxa15
2,255
1
Why is it that when I do integration by parts on cyclic functions such as (sinx)e^(inx), I get a trivial answer like C=C, C is a constant

Have I done something wrong or are there other methods of doing those integrals?
 
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  • #2
I don't understand your question, could you give a better example of what you mean? Integrals like this usually work out nicely by using parts twice and then collecting like terms of the original integral on one side of teh equation (I'm not sure how much sense that made if any).
 
  • #3
As soon as you get the integral of your original function, stop using parts. Merge it with the original function on the left and divide by the new constant. Using parts again just proves that it is equal (c=c)
 
  • #4
I did the integral originally posted with a different order, exponetial on the left and it worked out nicely. The first time I did it, I had sin on the left and it turned into the trivial C=C (following all the usual rules). Looks like the order of integration matters.
 
  • #5
If, integrating something like [itex]\int f(x)g(x)dx[/itex], you do an integration by parts, letting u= f(x), dv= g(x)dx, you will, of course, get another integral [itex]\int vdu[/itex]. If you do another integration by parts, this time reversing the roles (letting u= the old v and dv= the old u dx) you will just reverse what you did in the first integration and, yes, everything will cancel.
 
  • #6
HallsofIvy said:
If, integrating something like [itex]\int f(x)g(x)dx[/itex], you do an integration by parts, letting u= f(x), dv= g(x)dx, you will, of course, get another integral [itex]\int vdu[/itex]. If you do another integration by parts, this time reversing the roles (letting u= the old v and dv= the old u dx) you will just reverse what you did in the first integration and, yes, everything will cancel.

I see what you are getting at here, I realized that I made the exact mistake you pointed out here.
 

1. What is trivial integration by parts?

Trivial integration by parts is a technique used in calculus to simplify the integration process by breaking down a complicated integral into smaller, more manageable parts. It involves using the product rule of differentiation in reverse to find the antiderivative of a function.

2. When should I use trivial integration by parts?

Trivial integration by parts is typically used when the integral involves a product of two functions, where one function is easier to integrate than the other. It can also be used to solve integrals involving trigonometric functions, logarithmic functions, and exponential functions.

3. How do I perform trivial integration by parts?

To perform trivial integration by parts, you need to identify the two functions in the integral and assign one as u and the other as dv. Then, use the formula ∫u dv = uv - ∫v du to find the antiderivative of the integral.

4. What are the benefits of using trivial integration by parts?

Trivial integration by parts can simplify a complex integral into smaller, more manageable parts, making it easier to solve. It can also be used to solve a wide range of integrals involving different types of functions.

5. Are there any limitations to trivial integration by parts?

Trivial integration by parts may not always work for every integral, especially if the integral involves functions that do not have an antiderivative or if the integral is too complex. In some cases, it may be more efficient to use other integration techniques.

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