- Thread starter
- #1

#### alane1994

##### Active member

- Oct 16, 2012

- 126

[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?

- Thread starter alane1994
- Start date

- Thread starter
- #1

- Oct 16, 2012

- 126

[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?

- Admin
- #2

- Moderator
- #3

- Jan 26, 2012

- 995

For example, consider the integral

[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?

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

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

- Admin
- #4

L: Logs

I: Inverse Trigonometric

A: Algebra/Polynomials

T: Trigonometric

E: Exponential

Integration by parts - Wikipedia, the free encyclopedia

- Moderator
- #5

- Jan 26, 2012

- 995

What I posted is the practically the same thing...just using different letters and the order of the last two are reversed.

L: Logs

I: Inverse Trigonometric

A: Algebra/Polynomials

T: Trigonometric

E: Exponential

Integration by parts - Wikipedia, the free encyclopedia

- Admin
- #6

- Admin
- #7

- Jan 26, 2012

- 4,052

- Moderator
- #8

- Jan 26, 2012

- 995

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.

- Moderator
- #9

- Feb 7, 2012

- 2,715

- Aug 30, 2012

- 1,135

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.

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!

-Dan

- Feb 14, 2012

- 79

- Thread starter
- #12

- Oct 16, 2012

- 126

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.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

- Oct 3, 2012

- 114