Today we’re going to talk about some different techniques we can use to calculate integrals.
Some integrals, like those of simple polynomial functions, are easy to calculate.
In the last video of this series we talked about how to reverse the Power Rule to calculate
the integrals of these types of functions.
But sometimes this isn't enough.
If you're not able to calculate an integral given the function's original form, you'll
need to try some other techniques to simplify your function, or break it into pieces that
are easier to integrate.
In general, you should try the following methods in the following order when you get stuck
on an integral problem: 1.
u-substitution 2.
integration by parts 3.
partial fractions As you become more comfortable with solving
integral problems, you'll get better at identifying quickly which method you should use, and you
can skip right to integration by parts, or partial fractions.
Keep in mind that there are other methods beyond these three for solving integrals.
For now, we'll review only these three in more detail, but once you master these you
should check out integration of exponential, logarithmic, and trigonometric functions.
U-Substitution is a great method to try before the others because it's usually the easiest
and fastest method.
It allows you to take a complicated piece of your integral, and call it u, thereby simplifying
the integral significantly.
To perform a u-substitution, you need to identify u, then take the derivative of u, which you'll
call du.
Remember that you have to include a dx when you calculate du.
It's usually easy to remember this because you have to solve your du equation for dx
before you can substitute into your equation, putting in u and du.
If u-substitution was the right method to choose, and if you've done your substitutions
correctly, you should now have either gotten rid of all your x's, or be able to cancel
out any that remain, leaving you with only u's in your function.
From here, you should be able to calculate the integral easily.
You know you've got a great candidate for a u-substitution problem if you can see two
terms that are off by one degree.
For example, you've got x and x^2, x^2 and x^3, x^3 and x^4, etc.
Even ln(x) and x.
The reason these are good candidates is because, if you set u equal to the higher degree term,
and then you calculate du, you'll see that by taking the derivative, du will include
the lower degree term, which means that, after substituting back into your integral, you
might be able to cancel your lower degree terms.
To use integration by parts, you're looking for a function you can easily break down into
two pieces.
These two pieces should be multiplied together in your original function.
Ideally, you want one of the pieces to have a derivative that is significantly simpler
than the original piece.
By simpler, we usually mean something of a lower degree.
For example, the derivative of x is 1, which is much simpler than x.
So, for example, if you're asked to integrate the function xsinx, you should immediately
think of integration by parts, because you have two functions that are multiplied together,
x and sinx, and one of those parts, x, becomes much simpler when you take its derivative.
In this method, you'll identify the two separate pieces of your function.
You'll be integrating one of them and taking the derivative of the other.
You should first identify the one you want to derive first, and that piece should become
simpler when you take its derivative.
In this problem, the derivative of x is 1, which is simpler than x.
The derivative of sinx on the other hand, is cosx, which isn't really any simpler than
sinx.
Therefore, you'll want to derive x, which means you'll be integrating sinx.
You'll say that the piece you're deriving, x, is equal to u, and you'll calculate its
derivative, du, and get 1.
You'll call the piece your integrating dv, remembering always to include a dx with this
part, and then take its integral, v to get -cosx.
When you integrate, the dx falls away.
Now that you have each of these four pieces, all you need to do is plug them into the integration
by parts formula, which tells you that the integral of u times dv is equal to u times
v minus the integral of v times du.
Once you plug in for u, v, v again, and du, the remaining integral should be much simpler
to calculate than the original integral.
If u-substitution and integration by parts aren't working out, you may want to try using
partial fractions.
This method can be useful whenever you have a rational function, or fraction to integrate.
In fact, if your original function is a fraction, you may want to consider trying partial fractions
before u-substitution or integration by parts.
This method will allow you to split a fraction into smaller, more manageable fractions that
are easier to integrate.
Before you can actually perform the partial fraction decomposition, you need to make sure
that the degree of the numerator is not greater than the degree of the denominator.
If it is, you need to perform polynomial long division on the fraction to reduce it.
Once you have, you want to factor the denominator as much as possible.
After it's completely factored, split each factor of the denominator into its own fraction,
where the numerator is a sequence of new variables, A, B, C, as many as you need to cover the
number of factors that you have.
The sum of these fractions should be set equal to the original function.
Keep in mind that the numerator of a linear, or first degree factor, will be the single
variable, A, B, C, etc.
If the denominator is a quadratic factor, where the degree is 2 or more, the numerator
will be Ax+B, Cx+D, etc.
If you have a repeated factor, two or more of the same factor, then you need include
a fraction for every degree of that factor lower than the factor itself.
In other words, if the original denominator factors into (x+1)^3, then you have a factor
of (x+1) that's repeated three times.
Your partial fraction decomposition should then include one fraction for x+1, another
for (x+1)^2, and another for (x+1)^3.
Once you've set up your decomposition, you'll need to multiply both sides of your equation
by the denominator from the left-hand side, then cancel terms to simplify as much as possible.
From here, all that's left to do is expand your terms and use simultaneous equations
to solve for each of your variables, A, B, C, etc.
After you've found their values, plug them back into your original decomposition.
You should now have a new set of fractions that you can integrate to get your final answer.
Next time we’ll use what we've learned today to review some applications of integration.
I’ll see you then.