Integration by parts difficulties

In summary, the conversation discusses a specific problem in Acheson's book "From Calculus to Chaos" involving integration by parts. The problem involves a step from equation 1 to equation 2, where the first term is integrated by parts using u = \dot{y}_A and dv = \dot{\eta}dt. The person asking for help has been stuck on it and is looking for suggestions.
  • #1
falcao
9
0
Just working through a problem Acheson's book (From Calculus to Chaos) if anyone knows it.. eq 8.6 in this book..

As he's working through the problem he makes the step of this:


[tex]
m\int_{t_{1} }^{t_{2}} \left( \dot{y}_{\scriptscriptstyle A }\dot{\eta} - g\eta \right)\,dt
[/tex] (1)

to

[tex]m\left[\dot{y}_{\scriptscriptstyle A} \eta\right]_{t_{1}}^{t_{2}} -
m\int_{t_{1}}^{t_{2}}\left(\ddot{y}_{\scriptscriptstyle A} + g\right)\eta\,dt
[/tex] (2)

... using integration by parts.

Now, I've been stuck on it for a little while and I can't really figure it out.. I tried 'taking out' the [tex]\eta[/tex] but had no luck, if anyone had any suggestions of what to let == dv and what to let == u in the formula it would probably sort me out!
[tex]\int u\,dv = uv - \int v\,du
[/tex]
Thanks in advance! (ps. I can post further workings if necessary)
 
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  • #2
He only did integration by parts on the first term in the integral, the [itex]\dot{y}_A\dot{\eta}[/itex] term. The [itex]g\eta[/itex] term was not integrated by parts. [itex]u = \dot{y}_A[/itex], [itex]dv = \dot{\eta}dt[/itex].
 
  • #3
Mute said:
He only did integration by parts on the first term in the integral, the [itex]\dot{y}_A\dot{\eta}[/itex] term. The [itex]g\eta[/itex] term was not integrated by parts. [itex]u = \dot{y}_A[/itex], [itex]dv = \dot{\eta}dt[/itex].
sorry, my bad, although I probably wouldn't have realized if I looked at it for a full day! :redface:thanks for your help
 

1. What is integration by parts and why is it difficult?

Integration by parts is a mathematical technique used to evaluate integrals that involve products of functions. It involves using the product rule from calculus to rewrite the integral in a different form. It can be difficult because it requires knowledge of both the product rule and integration techniques.

2. What are some common challenges when using integration by parts?

Some common challenges when using integration by parts include determining the correct choice for the two functions involved, calculating the integral of the chosen function, and dealing with complicated integrands that require multiple applications of the integration by parts formula.

3. How can I determine which function to choose as u and which to choose as dv?

There is no specific rule for choosing which function to use as u and which to use as dv in integration by parts. A common approach is to choose u as the function that becomes simpler or easier to integrate after multiple applications of the product rule, and dv as the more complicated function. Practice and experience can also help in determining the best choices.

4. Are there any tips or tricks for making integration by parts easier?

Some tips for making integration by parts easier include choosing u and dv carefully, using integration tables or software to check your work, and breaking down complicated integrals into smaller parts. It can also be helpful to review and practice the product rule and integration techniques to improve your understanding and speed with integration by parts.

5. Can integration by parts be used for all integrals?

No, integration by parts can only be used for integrals that can be rewritten in the form ∫u dv. Some integrals may require other techniques such as substitution or partial fractions. It is important to have a variety of integration techniques in your toolkit to tackle different types of integrals.

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