On-line Math 21

5.4  Techniques of integration

In general, finding integrals is a much more complicated process than finding derivatives. The straightforward algorithms of differentiation may take some time to be natural for you, but you never have a situtation where there is no algorithm. With integration, though, it is not uncommon to have to guess which method to use, to have to try several methods before you find the right one, and even to find that no method will work.

We deal here with the two most important methods of integration, substitution and integration by parts. These techniques, even in this day of computers that can integrate most functions you might need, are still useful, since they do more than provide algorithms to find antiderivatives. They are general techniques to decompose problems into rational components.

There are other techniques of integration that you will learn about in subsequent courses: partial fractions (linked here to a brief, optional introduction), trigonometric substitutions, and various tricks. These techniques combine the two methods we discuss here with algebra and trigonomety to produce refined techniques that apply in specific situations. These two techniques, however, are the basis for all the others.

5.4.1  Substitution

The point of substitution is to run the chain rule backwards. The chain rule says that, if F¢(x) = f(x) , then

(F(u(x))¢ = F¢(u(x)) u¢(x) = f(u(x)) u¢(x),

or, in Leibniz notation,

d dx (F(u(x)) = f(u(x)) du dx .

That last version is the key to substitution. Then, if we want to integrate exactly the expression:

ó õ f(u(x)) du dx dx,

the integral would be F(u(x))+C .

Now, if we think about u as a variable, u(x) = u , and we blithely use algebra on differentials, so that

du dx dx = du,

we can write the line above by:

ó õ f(u)du,

which is F(u) . But, for this last case, you forgot that x was the variable at all. u became the variable, and you just were asked for the antiderivative of f , which is just F . The trick is in the fine print: the switch of the differentials. This is the whole reason for the dx at the end. That way, all you have to do to correctly substitute is to, if you can, turn everything, including the dx , into u 's, using the formula for u(x) and the equation

du dx dx = du.

Then integrate the expression involving u , and substitute back.

Example 1

ó õ x Ö   x2+4   dx.

Solution

Example 2

ó õ x2 x3+7 dx

Solution

Exercise 1

ó õ sin2(x)cos(x)dx =

Exercise 2

ó õ e3x+1dx =

Exercise 3

ó õ x x4+1 dx =

Hint: For this one, the standard substitution of the whole denominator will not work. Look back over the standard list of integrals to see which one this can be made to resemble. That will give you the hint you need to find what substitution to make.

For definite integrals, keeping in the same spirit, you have to change the limits of integration into the u equivalents, too. Alternately, you could do the indefinite integration, substitute back in terms of x , and evaluate the definite integral using that function of x and the x -limits of integration.

Example 3

ó õ 4 0    ____ Ö3x+4    dx.

Solution

Exercise 4

ó õ 2 0  x2 x3+7 dx =

Exercise 5

ó õ 1 -1  x2sin(x3)dx =

Exercise 6

ó õ 2 -2  x Ö   4-x2   dx =

Hint: Something odd happens with this one to the limits of integration when you do the substitution.

5.4.2  Integration by parts

This technique is the most powerful in terms of applications beyond mere computing of integrals. It's absolutely vital to understanding much of differential equations.

The idea is sort of ``robbing Peter to pay Paul''. We are going to try to run the product rule backwards. However, the right-hand side has two pieces,

(u v)¢ = u¢ v+u v¢,

so it's a little hard to do that. We can, however, integrate both sides, which gives:

ó õ (u v)¢dx = ó õ u¢ vdx+ ó õ u v¢dx,

which doesn't look like much improvement. However, you can do the left-hand side, it's just u v . Also, you can solve for one of the terms on the right-hand side. We're going to pick the second one. Solving for that term,

ó õ u v¢dx = u v- ó õ u¢ vdx.

Now, that's it, really. One integral, òu v¢dx , is expressed in terms of the other, òu¢ vdx . If you have an integral you can think of as the integral of a product, and you can pick which piece you want to be u and which piece v¢, then you can trade that integral in for the integral of v u¢.

We usually write this, like we did for substitution, in differential notation, where du = u¢dx and dv = v¢dx . This then is integration by parts as you should remember it:

ó õ u dv = u v- ó õ v du.

Let's try a simple example:

Example 4

ó õ xexdx

Solution

Example 5

ó õ x2sin(x)dx.

Solution

Exercise 7

ó õ xln(x)dx =

Exercise 8

ó õ ln(x)dx =

Example 6

ó õ exsin(x)dx

Solution

If you have an integration by parts problem with limits of integration, be sure to evaluate the u v part:

ó õ b a  u dv = u v| ab- ó õ b a  v du.

5.4.3  Partial fractions

Email Address (Required to submit answers):