Welcome to our community

Be a part of something great, join today!

Integration by Parts

alane1994

Active member
Oct 16, 2012
126
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?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You get the v from:

$\displaystyle \int\,dv$
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
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
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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)
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
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.
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,715
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.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,135
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
 

Chipset3600

Member
Feb 14, 2012
79
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
 

alane1994

Active member
Oct 16, 2012
126
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.
 

skatenerd

Active member
Oct 3, 2012
114
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.