In the preceding section, we have seen that, given an appropriate calculating device, evaluating definite integrals is not difficult - and doesn't require the Fundamental Theorem. Our success notwithstanding, there are times and places at which we must have some ability to calculate integrals (definite or indefinite) without numerical or symbolic help from a computer or calculator.

• First, even with a computer at hand, there are some integration problems that are at least as easy to do with pencil and paper - including some that look hard. The way we learn to tell which ones we would just as soon do ourselves and which we really want to turn over to the computer is by practice - enough practice to discover the patterns that signal "hard" or "easy" integrals.
  
  
• Second, for some problems - differential equations, for example - the preferred form of a solution involves explicit antidifferentiation. That is, indefinite integration may be the key step in solving a problem that doesn't even involve definite integrals. True, the important part of the Fundamental Theorem says that we can always use definite integration to a variable limit as a way to find an antiderivative. But when a more explicit form for an answer is readily found, it can often save a lot of work in subsequent computations.

In this section we will see that applying the Chain Rule and the Product Rule to integrals - first indefinite, then definite - greatly expands our repertoire of "easy" integrals.

8.4.1 Algebraic Substitutions

We begin by asking you to calculate an indefinite integral, i.e., to guess an antiderivative.

Activity 1

Calculate \(\displaystyle\int_\,^\,x \sqrt{1+x^2}\;dx\).

Comment on Activity 1

Whatever procedure you found for Activity 1, it is an example of an important technique, one that we now formalize for use with other problems as well.

Let's start again. Let \(u(x)=1+x^2\). If we calculate the differential on each side, we have \(du = 2x\,dx\). Thus,

Now think of \(u\) as the independent variable, and try to evaluate \(\displaystyle\int_\,\frac{1}{2} \sqrt{u}\;du\). That's easy:

Think of \(u\) as the dependent variable again - substitute \(u=1+x^2\) to obtain

The Chain Rule justifies our alternating between thinking of \(u\) first as a dependent variable, then as an independent variable, and then as a dependent variable again. If all this seems familiar, you are right. We did the same thing in Chapter 6 when we were finding solutions of differential equations by separating the variables.

Notice that we have finally justified our notational scheme for integration: The differential \(dx\) in the notation

helps us carry out substitution arguments. The process of antidifferentiation can be thought of either as finding antiderivatives or as finding antidifferentials - it matters little in the final result. However, it matters a lot in terms of designing and using powerful notation to do much of the work for us. As we noted in Chapter 4, the particular powerful notation \(dx\) was one of Leibniz's many contributions to the calculus. The notation we use for integrals - both definite and indefinite - is also due to Leibniz.

Checkpoint 1

Next we study application of substitution to definite integrals.

Activity 2

1. Verify that ∫ 1 3 x 1 + x 2 d x and ∫ 2 10 1 2 u d u have the same value.

2. How are the limits on the second integral related to those on the first?

3. Fill in the upper and lower limits on the second definite integral to make the equality a true statement:

   \(\displaystyle\int_a^{\,b} x \sqrt{1+x^2}\;dx = \int_{\_}^{\,\_} \frac{1}{2} \sqrt{u}\;du\).

4. Use the substitution technique to evaluate

   \(\displaystyle\int_0^{\,\pi/2}\sin^2(\theta)\,\cos(\theta)\,d \theta\).

Comment on Activity 2