Uploaded by TheIntegralCALC on 26.06.2012

Transcript:

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.

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.