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Notes for Introductory Calculus (Math 120)

Brenton LeMesurier

Section 5.5 Substitution

References.

So far we are rather limited in our ability to calculate antiderivatives and integrals because, unlike with derivatives, knowing indefinite integrals for two functions does not in general allow us to calculate the indefinite integral of their product, quotient, or composition. However, we can find a rule that will help up with some products and compositions, using the same strategy that lead us to our first few antiderivatives: take a fact about derivatives and “invert” it.
Surprisingly, it is the Chain Rule that is most useful, because the derivative of a composition is a certain product, and thus running it backwards gives an antiderivative for that product: almost a product rule for indefinite integrals.

Getting some integrals involving products, quotients and compositions.

To get an idea of how the Substitution Rule will work, let us first get a few examples of integrals of products by working backwards from some derivatives.
\begin{equation*} \frac{d}{dx}(\sin x)^3 = 3(\sin x)^{2} \frac{d}{dx}(\sin x) = 3 \sin^2 x \cos x \; \text{, so dividing by } 3, \end{equation*}
\begin{equation} \int \sin^2 x \cos x \, dx = \frac{1}{3} (\sin x)^3 + C\tag{5.5.1} \end{equation}
\begin{equation*} \frac{d}{dx}\ln(\cos x) = \frac{1}{\cos x} \frac{d}{dx}(\cos x) = -\frac{\sin x}{\cos x}\; \text{, so} \end{equation*}
\begin{equation} \int {\tan x} \, dx =\int \frac{\sin x}{\cos x} \, dx = -\ln(\cos x) + C\tag{5.5.2} \end{equation}

Reversing a Chain Rule Calculation.

Let us look at the first calculation above in reverse order. Factor \(\cos x\) has antiderivative \(\sin x\text{,}\) which also appears in the other factor, so
\begin{equation*} \sin^2 x \cos x = (\sin x)^2 \frac{d}{dx}(\sin x) \end{equation*}
Using the name \(u\) for this repeated term \(\sin x\text{,}\) this is \(\displaystyle u^2 \frac{du}{dx}.\)
Since \(u^2\) has antiderivative \(u^{3}/3\text{,}\)
\begin{equation*} u^2 \frac{du}{dx} = \frac{d}{du}\left(\frac{u^3}{3}\right)\frac{du}{dx}. \end{equation*}
This is what the Chain Rule gives for
\begin{equation*} \frac{d}{dx}\left(\frac{u^3}{3}\right), = \frac{d}{dx}\frac{(\sin x)^3}{3}. \end{equation*}
Thus \(\sin^2 x \cos x\) has antiderivative \(\displaystyle\frac{(\sin x)^3}{3}\text{.}\)

Reversing a Chain Rule Calculation.

In this calculation, the antiderivatives for \(\cos x\) and \(u^2\) have been combined to get this new antiderivative. In terms of indefinite integrals, we have used the two simple indefinite integrals
\begin{equation*} \int \cos x \, dx = \sin x + C \end{equation*}
and
\begin{equation*} \int u^2 du = \frac{u^3}{3} + C \end{equation*}
plus the Chain Rule \(\displaystyle \frac{d}{dx}f(u) = f'(u)\frac{du}{dx}\) to get
\begin{equation*} \int \sin^2 x \cos x \, dx = \frac{\sin^3 x}{3} + C. \end{equation*}

The Substitution Rule.

Suppose that we seek the (indefinite) integral of a function of the special “composition-product-derivative” form \(f(g(x))g'(x)\text{,}\) and we know an antiderivative \(F\) for \(f\text{.}\) Then the Chain Rule gives
\begin{equation*} \frac{d}{dx}[F(g(x))] = F'(g(x))g'(x) = f(g(x))g'(x) \end{equation*}
so \(F(g(x))\) is an antiderivative of this function, and
\begin{equation*} \int f(g(x))g'(x) \, dx = F(g(x)) + C. \end{equation*}
On the other hand, if we define \(u=g(x)\) then \(\displaystyle g'(x) = \frac{du}{dx}\text{,}\) \(F(g(x)) = F(u)\text{,}\) and \(\int f(u) \, du = F(u) + C\text{.}\) Combining these results
\begin{equation} \int f(g(x))g'(x) \, dx = \int f(u) \frac{du}{dx}\, dx = \int f(u) \, du = F(u) + C.\tag{5.5.3} \end{equation}
In practice, the emphasis is on choosing the new quantity \(u\) and changing to it as the variable, which involves getting a differential \(du\) in the integral formula in place of \(dx\text{.}\) To be precise,
That is, one can effectively make a substitution with the differential formula \(du = \displaystyle\frac{du}{dx}dx\) inside integral formulas, and this helps so long as the rest of the formula can also be expressed entirely in terms of the new variable \(u\text{.}\)
Note well: for this substitution method to be useful, one must completely convert to \(u\) from \(x\text{,}\) not have a mix of both variables in the transformed integral.

Choosing \(\mathbf u\).

The key in practice is finding a suitable choice of \(u\text{,}\) and there may be more than one worth trying.
One strategy is to use the quantity inside a composition as \(u\text{,}\) since \(u\) is the “inside” part of the Chain Rule.
This composition should then multiply some other factor containing just the derivative of \(u\text{.}\)
Another strategy is that to seek some quantity such that both its and its derivative appear in the integrand,
with the derivative simply multiplying the rest of the integrand.
Find \(\displaystyle\int x^3 \cos(x^4+2) \,dx\text{.}\)
Hint.
Start by thinking of options for \(u\text{.}\)
Find \(\displaystyle\int \cos x \sin x \, dx\text{.}\)
Hint.
Start by seeking several possible options for \(u\text{.}\)
Find \(\displaystyle\int \sqrt{2x+1} \, dx\text{.}\)
Here there is no product, just a composition, so what do we do about the need for factor \(du/dx\text{?}\)
Hint.
It still helps to try the inside of a composition as \(u\text{.}\)
Find \(\displaystyle\int \frac{x}{\sqrt{1-4x^2}} dx\text{.}\)
\(\displaystyle\int e^{3x} \, dx\)
\(\displaystyle\int \sqrt{1+x^2} \, x^5 \, dx\)
\(\displaystyle\int \tan x \, dx\)

Substitution in Definite Integrals.

Often the easiest way to deal with definite integrals is to first seek an indefinite integral, and then use the FTC. However with substitution, this involves the step of converting back from a function of new variable \(u\) to the original variable \(x\text{,}\) and it may be easier to avoid that by converting everything to the new variable, including the limits of integration.
I write the integral limits as “\(x=a\)”, “\(u=b\)” and so on to emphasize that \(u\) must completely displace \(x\text{,}\) in three places:
  • in the formula for the integrand, \(f(u)\) replaces \(f(g(x))\text{;}\)
  • in the differential, \(du\) replaces \(\displaystyle\frac{du}{dx}dx = g'(x)dx\text{;}\) and
  • in the limits of integration, \(c=g(a)\) and \(d=g(b)\) replace \(a\) and \(b\text{.}\)
\(\displaystyle\int_0^4 \sqrt{2x+1} \, dx\)
\(\displaystyle\int_1^2 \frac{1}{(3-5x)^2} \, dx, \; = \int_1^2 \frac{dx}{(3-5x)^2}\)
Evaluate \(\displaystyle\int_1^e \frac{\ln x}{x} dx\text{.}\)
Try it both ways: using the above formula , and by first finding the indefinite integral as a function of \(x\) and then using the FTC.

Short-cuts From Symmetry.

For even and odd functions integrated over a symmetric interval \([-a,a]\text{,}\) the intervals simplify:
  • (a) If \(f(x)\) is odd \(\big[ f(-x) = -f(x) \big]\) then \(\displaystyle\int_{-a}^a f(x) \, dx = 0.\)
  • (b) If \(f(x)\) is even \(\big[ f(-x) = f(x) \big]\) then \(\displaystyle\int_{-a}^a f(x) \, dx = 2\int_{0}^a f(x) \, dx.\)
Evaluate \(\displaystyle\int_{-2}^2 x^6+1 \; dx\text{.}\)
Evaluate \(\displaystyle\int_{-1}^1 \frac{\tan x}{1+\sec^2 x} dx\text{.}\)

Exercises Exercises

Study Calculus Volume 1, Section 5.5 2 ; in particular Theorem 7, the Problem Solving Strategy that follows it, Examples 30–33 (and maybe 34 and 35), Checkpoints 25–28, (and maybe 29 and 30), and one or several exercises from each of the following ranges: 256–260, 261–270, 271–287 and 292–297; Some suggested selections are Exercises 257, 261, 265, 271, 275, 281, 293, 297.
As noted above, for definite integrals one can either do it as described there (Theorem 8, Examples 34 and 35, Checkpoints 29 and 30) or (a) first get the indefinite integral \(\int f(x) dx = F(x) + C\) using substitution and then (b) use FTC: \(\int_a^b f(x) dx = F(b) -F(a)\text{.}\)
openstax.org/books/calculus-volume-1/pages/5-5-substitution
openstax.org/books/calculus-volume-1/pages/5-5-substitution
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