# Thread: How do I know what the inner and outer function is?

1. ## How do I know what the inner and outer function is?

Some equations are easy to figure out because you can add brackets around them, but sometimes it's really difficult!

For example this:

$\displaystyle e^{x^{2}}$

How do you find out what the inner and outer function is in order to apply the chain rule?

And what about integration by parts, if you want to integrate this:

$\displaystyle \int x\cdot \cos x dx$

How do you find out, which is the u and which is the v?

2. Originally Posted by No Logic Sense
Some equations are easy to figure out because you can add brackets around them, but sometimes it's really difficult!

For example this:

$\displaystyle e^{x^{2}}$

How do you find out what the inner and outer function is in order to apply the chain rule?

And what about integration by parts, if you want to integrate this:

$\displaystyle \int x\cdot \cos x dx$

How do you find out, which is the u and which is the v?
$\displaystyle f(x) = e^x$

$\displaystyle g(x) = x^2$

now, is $\displaystyle e^{x^2}$ = $\displaystyle f[g(x)]$ or $\displaystyle g[f(x)]$ ?

for integration by parts, someone came up with a rule of thumb called the ILATE rule which gives what may be the best order for choosing which function to make u ...

I ... Inverse Trig functions
L ... Log functions
A ... Algebraic functions
T ... trig functions
E ... exponential functions

so, for $\displaystyle \int x \cos{x} \, dx$ you have an Algebraic function, $\displaystyle x$ , and a Trig function, $\displaystyle \cos{x}$ . The ILATE rule says to let $\displaystyle u$ be the Algebraic function, $\displaystyle x$ , because A comes before T in the word ILATE.

3. But I thought that in $\displaystyle e^{x^{2}}$ the $\displaystyle e^{x}$ would be one function and then the $\displaystyle ^{2}$ would be the other. Can x be used two times?

That ILATE rule seems really useful! Am I all set if I just use that rule when integrating by parts?

By the word, I searched on Wikipedia and they wrote it LIATE, where the logarithm is before the inverse functions. So now there are two types of the rule, is it ILATE or LIATE?

Integration by parts - Wikipedia, the free encyclopedia

I tried using the rule and integrate by parts, where:

$\displaystyle u=x$
$\displaystyle v=\cos x$

$\displaystyle \int x \cos{x} \, dx$

$\displaystyle \int \frac{x^{2}}{2} \cdot \cos x - \int \frac{x^{2}}{2} \sin x \, dx$

4. Are you sure that is correct?

I made a little typo where I should have written -sin x, but still, I'm not sure if the ILATE rule works here.

5. But I thought that in the would be one function and then the would be the other.
that is incorrect ... how does one justify the lone exponent "2" to be a function?

By the word, I searched on Wikipedia and they wrote it LIATE, where the logarithm is before the inverse functions. So now there are two types of the rule, is it ILATE or LIATE?
I've seen both LIATE and ILATE ... it most likely will not affect you since it is rare to integrate the product of a log and inverse trig function.
Remember that it is just a rule of thumb ... something to get you started.

I tried using the rule and integrate by parts, where:

I have no idea how you arrived at what you show above.

this is a straight forward use of integration by parts ...

$\displaystyle \int x\cos{x} \, dx$

$\displaystyle u = x$ ... $\displaystyle dv = \cos{x} \, dx$

$\displaystyle du = dx$ ... $\displaystyle v = \sin{x}$

$\displaystyle \int u \cdot dv = u \cdot v - \int v \cdot du$

$\displaystyle \int x\cos{x} \, dx = x\sin{x} - \int \sin{x} \, dx$

$\displaystyle \int x\cos{x} \, dx = x\sin{x} + \cos{x} + C$

6. Well, that's weird. It seems like my math book has a different formula for integrating by parts.

In my math book, they wrote this:

$\displaystyle \int u(x) \cdot v(x) \, dx = U(x) \cdot v(x) - \int U(x) \cdot v'(x) \, dx$

I have no idea why they have written this in my math book. Is your formula an easier method?

Also, why is v in your example sin x? Shouldn't it be cos x?

7. The two formulas are exactly the same with only notational differences.

$\displaystyle \int u(x) \cdot v(x) \, dx = U(x) \cdot v(x) - \int U(x) \cdot v'(x) \, dx$
"U(x)" is the anti-derivative of "u(x)" while in
$\displaystyle \int u \cdot dv = u \cdot v - \int v \cdot du$
has "u" and "v" reversed from your formula and "dv" is the differential of v(x): dv/dx= v' so dv= v' dx.

If, in your formula, you replace U(x) by v(x), v'(x) dx by du, u(x)dx by v(x)dx, you will get the other formula.

8. Originally Posted by HallsofIvy
The two formulas are exactly the same with only notational differences.

$\displaystyle \int u(x) \cdot v(x) \, dx = U(x) \cdot v(x) - \int U(x) \cdot v'(x) \, dx$
"U(x)" is the anti-derivative of "u(x)" while in
$\displaystyle \int u \cdot dv = u \cdot v - \int v \cdot du$
has "u" and "v" reversed from your formula and "dv" is the differential of v(x): dv/dx= v' so dv= v' dx.

If, in your formula, you replace U(x) by v(x), v'(x) dx by du, u(x)dx by v(x)dx, you will get the other formula.
Oh, okay. So I guess the one that they wrote in my math book makes it more complicated compared to the other one, right? Seeing as the ILATE rule doesn't apply there?

9. Originally Posted by No Logic Sense
Well, that's weird. It seems like my math book has a different formula for integrating by parts.

In my math book, they wrote this:

$\displaystyle \int u(x) \cdot v(x) \, dx = U(x) \cdot v(x) - \int U(x) \cdot v'(x) \, dx$

I have no idea why they have written this in my math book. Is your formula an easier method?

Also, why is v in your example sin x? Shouldn't it be cos x?
the formula you quoted is incorrect.

the standard formula for integration by parts is ...

$\displaystyle \int u \cdot dv = u \cdot v - \int v \cdot du$

As far as the integral I solved ... my work is correct. Check it yourself by taking the derivative of my solution.

10. Originally Posted by skeeter
the formula you quoted is incorrect.

the standard formula for integration by parts is ...

$\displaystyle \int u \cdot dv = u \cdot v - \int v \cdot du$

As far as the integral I solved ... my work is correct. Check it yourself by taking the derivative of my solution.
well, the formula I wrote is mentioned in my math book so I just assumed it was correct. but the one that you wrote seems to be used in Wikipedia and a lot of other places so I guess it's easier to use.

I'm 100% that the answer you gave is correct as well, because in my math book, the guy tried to integrate by parts in an example, but instead of having u = x and v = cos x, he switched around so he had v = x and u = cos x and then he came to the same answer as you did, he just used the other formula.

But the formula in my math book seems really long and unnecessary compared to the other one. Can I forget about the formula in my math book?