Integration by Parts: Get v in udv=uv-vdu

In summary, when using the integration by parts formula, the value for v is found by integrating the dv term. The general rule for finding u is given by the acronym LIPET, with logarithms being the highest priority and trigonometric functions being the lowest. If the integrand only contains exponential and trigonometric functions, either one can be chosen as u.
  • #1
alane1994
36
0
When integrating by parts the formula is

[tex]\int{u} dv=uv-\int{v} du[/tex]

I understand where to get the u, and du... but where does one get the v?
 
Physics news on Phys.org
  • #2
You get the v from:

$\displaystyle \int\,dv$
 
  • #3
alane1994 said:
When integrating by parts the formula is

[tex]\int{u} dv=uv-\int{v} du[/tex]

I understand where to get the u, and du... but where does one get the v?

For example, consider the integral

\[\int e^x\sin x\,dx\]

The general rule for finding your $u$ is given by the anagram LIPET, where it goes from highest priority to lowest priority.

L: Logarithm
I: Inverse Trigonometric
P: Polynomials/Algebra
E: Exponential
T: Trigonometric

In the case that you only have exponential and trig, either one can be your $u$.

So, for example, if your integrand contains inverse trig and exponential functions, you let $u$ be the inverse trig function.

In the particular example I gave, we only have exponential and trigonometric functions in the integrand, so we can pick either term to be our $u$. Without loss of generality, say $u=e^x$. Now, $\,dv$ is everything else; that is, $\,dv=\sin x\,dx$. Finding $\,du$ is straightforward (we just differentiate to get $\,du=e^x\,dx$). To find $v$, you need to integrate the $\,dv$ term, that is, $v=\int\,dv=\int \sin x\,dx=-\cos x$.

So, to find $v$ in general from integration by parts, you always end up integrating the value you have for $\,dv$.

I hope this makes sense!

EDIT: Ninja'd
 
  • #5
MarkFL said:
Interesting...I was taught the rule:

L: Logs
I: Inverse Trigonometric
A: Algebra/Polynomials
T: Trigonometric
E: Exponential

Integration by parts - Wikipedia, the free encyclopedia

What I posted is the practically the same thing...just using different letters and the order of the last two are reversed. :P
 
  • #6
This leads me to believe that perhaps the order of the last two may depend on the integrand being considered. I am glad to know there is a variation, and that the rule I was taught has some exceptions.(Star)
 
  • #7
I was never taught that but wish I had. After doing a bunch of these kinds of integrals you just develop an intuition about what to choose for $u$ and $dv$ but before that point it can be really tricky to do, so I think the acronym you guys mentioned is a great tool for making this topic quicker to master.
 
  • #8
Jameson said:
I was never taught that but wish I had. After doing a bunch of these kinds of integrals you just develop an intuition about what to choose for $u$ and $dv$ but before that point it can be really tricky to do, so I think the acronym you guys mentioned is a great tool for making this topic quicker to master.

I was never taught this when I took calculus in high school; I only found out about this anagram when I worked as a tutor at a community college. When I saw it, I was like...woah...this simplifies the process quite a lot. Since then, I've told my students about it and it has made doing IBP problems easier for them.
 
  • #9
I was never taught an acronym for this. I was given a general rule of thumb that went like this. Many functions get simpler when you differentiate them, most functions get more complicated when you integrate them. When integrating by parts, you want to differentiate the part that looks most likely to become simpler when differentiated, and you should integrate the part that looks least likely to get more complicated when integrated.
 
  • #10
I never learned it with the table, so I'm grateful for Chris to mention it.

In undergrad Physics we did these the same way my Advanced Mechanics instructor did: Start with answer and work it backward to see if it makes sense. If not, then vary your answer. etc. (Whew)

This didn't bother me so much...after all there are only so many integration problems that you have to deal with. But when he started doing it with differential equations I got a bit confused! (Tmi)

-Dan
 
  • #11
I'm learning Integration by Parts now in Calculus 1, and at first I was very confused in choosing [TEX]u[/TEX] and the [TEX]dv[/TEX], i learned the "L I A T " in youtube and has helped me a lot
 
  • #12
Chris L T521 said:
For example, consider the integral

\[\int e^x\sin x\,dx\]

The general rule for finding your $u$ is given by the anagram LIPET, where it goes from highest priority to lowest priority.

L: Logarithm
I: Inverse Trigonometric
P: Polynomials/Algebra
E: Exponential
T: Trigonometric

In the case that you only have exponential and trig, either one can be your $u$.

So, for example, if your integrand contains inverse trig and exponential functions, you let $u$ be the inverse trig function.

In the particular example I gave, we only have exponential and trigonometric functions in the integrand, so we can pick either term to be our $u$. Without loss of generality, say $u=e^x$. Now, $\,dv$ is everything else; that is, $\,dv=\sin x\,dx$. Finding $\,du$ is straightforward (we just differentiate to get $\,du=e^x\,dx$). To find $v$, you need to integrate the $\,dv$ term, that is, $v=\int\,dv=\int \sin x\,dx=-\cos x$.

So, to find $v$ in general from integration by parts, you always end up integrating the value you have for $\,dv$.

I hope this makes sense!

EDIT: Ninja'd

That helps alot! Thank you very much! The terminology and way that my professor explained it did not help very much at all. I appreciate all the help.
 
  • #13
Just a innocent passer-by here. I also wanted to thank you both for the acronyms...I too was just taught in Calculus II to basically develop an intuition for IBP, and in cases when my intuition fails me to just try different options and see which work and which don't. I'm excited to try LIPET and LIATE.
 

FAQ: Integration by Parts: Get v in udv=uv-vdu

What is integration by parts?

Integration by parts is a method used in calculus to find the integral of a product of two functions. It is typically used when the integral cannot be solved by basic integration techniques.

How does integration by parts work?

The integration by parts formula, uv - ∫vdu, involves breaking down the integrand into two parts, u and dv, and then using the formula to find the integral of the product. This method is based on the product rule of differentiation in calculus.

When should I use integration by parts?

Integration by parts is useful when the integral involves a product of two functions, one of which can be easily integrated while the other cannot. It is also helpful when the integral involves a function that can be differentiated multiple times.

What are the steps for solving an integral using integration by parts?

The steps for solving an integral using integration by parts are as follows: 1) Identify u and dv in the integrand, 2) Apply the integration by parts formula, uv - ∫vdu, 3) Simplify the resulting integral, 4) Solve for the remaining variable, and 5) Check the answer for accuracy.

Are there any tips for using integration by parts?

Some tips for using integration by parts include choosing u and dv wisely to make the resulting integral easier to solve, using tabular integration for integrals that require multiple applications of the integration by parts formula, and practicing various examples to become comfortable with the method.

Back
Top