Working on integrating something by parts can be a very tedious and often confusing line to approach certain integrals. Although in many cases, it is necessary!

Previously, we have derived the formula to allow us to integrate by parts:

\int \! u \, \mathrm{d}v = uv - \int \! v \,\mathrm{d}u

We can see that for any mixed integral that we come across, we would know what u and \mathrm{d}v is by substitution, and from then, we would have to determine du and v is to solve the Right hand side.
In other words, we would have to integrate \mathrm{d}v to obtain v and differentiate u  to obtain \mathrm{d}u .

But given a particular integral, say \int \! e^x \ln x \, \mathrm{d}x for example, we come across our first hurdle. Which function shall we take as u and \mathrm{d}v ? In otherwords we can translate our problem to being:

 “Which is easier to differentiate?”
“Which is easier to integrate?”

To determine this, try and answer the following question:
Which is easier to determine?

\int \! \ln x \, \mathrm{d} x  or \frac {\mathrm{d}(\ln x)}{\mathrm{d} x}

Even with a basic understanding of derivatives, one of the first things you end up learning is that \frac {\mathrm{d}(\ln x)}{\mathrm{d} x} = \frac {1}{x}  and the evaluation of \int \! ln x \, \mathrm{d} x  is not as obvious (certainly not as simple to say the least).

Now consider the same question but this time for e^x
Which is easier to determine?

\int \! e^x \, \mathrm{d} x  or \frac {\mathrm{d}(e^ x)}{\mathrm{d} x}

In this case, we are lucky that the derivative of e^x is itself! Also, since integration is essentially the inverse of differentiation, then we can equivalently say that the integral of  e^x is itself too. This makes exponentials, functions which are both easy to differentiate and to integrate!

Going back to our choices of u and \mathrm{d}v , since we know that \ln x is by far easier to differentiate e^x (because \ln x is certainly not easier to integrate!) then we take  u = \ln x and \mathrm{d}v=e^x \mathrm{d}x

We are not very much concerned in what \int e^x \ln x evaluates to, but rather how \int e^x \ln x is evaluated.

What if we had an integral involving \ln x and sin(x) ?
Or e^x and x^2 ?
Which function would we choose to take as u and \mathrm{d}v ?

It would be great if we could create a system whereby we can determine which functions to take as u and \mathrm{d}v .
So we ask the following question:

Can we create an easy reference system, such that the choice of functions for u and \mathrm{d}v will be simplified?

and surprisingly the answer is

YES! We can!

There are 4 types of functions you will normally be dealing with when doing integration.
These are:

  • Algebraic (ALG)
  • Exponentials (EXP)
  • Logarithmic (LOG)
  • Trigonometric (TRIG)

We will create our reference system focused on the choice of u , since the other function will automatically be taken to be \mathrm{d}v .

We shall be ranking the types of functions from left to right, the left most function that we have being the one to take as u , (or the right most function as \mathrm{d}v ).

We have already seen that LOG type functions are not easy to integrate, so if we have a  LOG in our integral, ideally we would always like to take that as u so that we may differentiate first. Thus, LOG will come first in our acronym.

L _ _ _

EXP functions are both easy to integrate and differentiate. Since integration usually is the headache when it comes to evaluation (differentiation is normally not very difficult in most elementary cases), if we have EXP in our integral, we would rather not pick it to be differentiated but rather save it for the more difficult option of integration! Because of this, it comes last in our acronym.

L _ _ E

Now ALG functions are polynomials such as x, x^2, x^n and \frac {1}{x} . While TRIG functions include trigonometric functions such as sin(x), cos(x), cosec(x) and arcsin(x) (which is sin^{-1}(x) ). In this case, we shall choose ALG over TRIG for differentiation, not because trigonometric functions are incredibly difficult to differentiate, but rather because we might have x which when differentiated, differentiates to 1 . In this sense, ALG type functions are simpler.


And there we go! Now you need not worry when you see a product of 2 different functions to integrate since you can use LATE to pick which functions are what, and simply substitute into the by parts formula!


\int e^x x^2 \mathrm{d}x \Rightarrow (u = x^2) \wedge (\mathrm{d}v = e^x \mathrm{d}x)  (ALG comes before EXP)

\int e^x sin(x) \mathrm{d}x \Rightarrow (u = sin(x)) \wedge (\mathrm{d}v = e^x \mathrm{d}x)  (TRIG comes before EXP)

\int x^2 sin(x) \mathrm{d}x \Rightarrow (u = x^2) \wedge (\mathrm{d}v = sin(x) \mathrm{d}x)  (ALG comes before TRIG)

Thanks for reading!



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