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FFR - Substitution and Differentials

This original textbook reading covers part of the AP topic. Follow the AP course outline for complete topic instruction and practice.

Translating derivative rules

Just as we had chain rule for derivatives, we have to have an equivalent way of calculating more complex integrals. This is where u-substitution comes in. For a complex integral where the argument is not just x, but another function of x, our best bet is to rewrite the integral in terms of another function.

However, we must also change the differential term, dx, to accommodate the change in variables. Suppose we wanted to change x to θ. We would also have to change the differential dx to dθ. For definite integrals, we have to also change the interval of integration (unless the beginning and ending values of the new function stay within the same interval, but this is rarely the case). Let’s learn this by example.

Example Problem 8.4.1

Convert the integral $\int_{0}^{1} \frac{x}{\sqrt{1 - x^{2}}} d x$ into one with a differential of $θ$ by substituting $x = \sin θ$ .

This question essentially spells out everything we need to do, but this will not be the case in the future.

We start by substituting x for sinθ. We can write this as $x = \sin θ$ .

Next, we substitute the differential dx. In order to solve for dx, we can differentiate both sides of the substitution we made.

$\frac{d x}{d θ} = \cos θ$

$d x = \cos θ d θ$

Here we run into a bit of a problem. There is no cosθdθ in our integral... or is there?

Recall we set $x = \sin θ$ earlier. We can solve for cosθ in terms of x using our famous trigonometric identity, $\sin^{2} θ + \cos^{2} θ = 1$ . Solving for cosθ, we get $\cos θ = \sqrt{1 - \sin^{2} θ}$

We can substitute this and x = sinθ into dx = cosθdθ, giving us:

$d x = \sqrt{1 - x^{2}} d θ$
Now divide both sides by the square root term

$\frac{d x}{\sqrt{1 - x^{2}}} = d θ$

We can see that now substituting dθ into the integral is possible and makes the integral much easier to solve.

However! We still have to change the interval of integration! When x ranges from 0 to 1, what is the range of θ?

$0 = \sin θ$ , $θ = 0$

$1 = \sin θ$ , $θ = \frac{π}{2}$

Before we substitute all of these into our main integral, recall that we have to make three substitutions.

Substitute x = sinθ (variable substitution)
Substitute dx = cosθdθ (differential substitution)
Substitute the limits of integration (from x₁ to x₂ into θ₁ to θ₂)

We must make these three substitutions whenever we want to rewrite an integral from one variable into another. Let’s start making these substitutions.

$\int_{\sin^{- 1} (0)}^{\sin^{- 1} (1)} x \cdot \frac{d x}{\sqrt{1 - x^{2}}}$

$= \int_{0}^{\frac{π}{2}} s i n θ d θ$
This is a much simpler integral to solve than the original. You can probably solve it in your head, but this concludes our problem.

Example Problem 8.4.2

Evaluate the indefinite integral $- \int \frac{1}{\sqrt{16 - x^{2}}} d x$

For this problem, we have to do something special. We have to factor out a 16 from the denominator so that we can substitute in and select a u. Doing this gives:

$- \int \frac{1}{\sqrt{16 (1 - \frac{1}{16} x^{2})}} d x$

$= - \frac{1}{4} \int \frac{1}{\sqrt{1 - {(\frac{1}{4} x)}^{2}}} d x$

Substituting $u = \frac{1}{4} x, d u = \frac{1}{4} d x, d x = 4 d u$ ,

$= - \frac{1}{4} \int \frac{4}{\sqrt{1 - {(u)}^{2}}} d u$

$= \cos^{- 1} (u) + C$

$= \cos^{- 1} (\frac{1}{4} x) + C$

Let’s briefly go over what we learned in this chapter:

Integrals help us measure the area under any curve.
Integrals are the same as an anti-derivative (First FTC)
The area under the curve between points a and b is equal to the anti-derivative at b minus the anti-derivative at a (Second FTC).
We can use the same techniques to evaluate integrals as we used to find derivatives, except backwards (ex: Inverse power rule, inverse trigonometric functions)
The chain rule pops up in evaluating some complex integrals, so we must keep an eye out for when an “outside” function is multiplied by its “inside” derivative.

Caution: It is recommended that you read through Chapter 9 before starting the problems for Chapter 8, because there is some overlap, and Chapter 9 will give you a stronger foundation for solving problems in Chapter 8; otherwise, you may feel like throwing a pillow against a wall or face-palming yourself constantly.

For the upcoming chapter problems you may need to refer to the inverse trigonometric integrals.

Show your work for Problem 1

Calculate the definite integral

\int_{0}^{\infty} e^{- 3 x} d x

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Show your work for Problem 2

Calculate the indefinite integral

\int \frac{x + 3}{x^{3}} d x

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Chapter 8 Problems

Calculate the definite integral $\int_{0}^{\infty} e^{- 3 x} d x$
Calculate the indefinite integral $\int \frac{x + 3}{x^{3}} d x$

Show your work for Problem 4

Calculate the definite integral

\int_{0}^{\infty} 3 e^{\frac{- x}{2}} d x

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Show your work for Problem 3

Evaluate the indefinite integral

\int \frac{x^{2} + 3 x + 3}{3 x^{4}} d x

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Evaluate the indefinite integral $\int \frac{x^{2} + 3 x + 3}{3 x^{4}} d x$
Calculate the definite integral $\int_{0}^{\infty} 3 e^{\frac{- x}{2}} d x$

Show your work for Problem 5

Evaluate the definite integral

\int_{- 0.5}^{0} - \frac{1}{\sqrt{1 - 4 x^{2}}} d x

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Show your work for Problem 6

Evaluate the indefinite integral

\int \frac{1}{\sqrt{9 x^{2} - 12 x + 5}} d x

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Evaluate the definite integral $\int_{- 0.5}^{0} - \frac{1}{\sqrt{1 - 4 x^{2}}} d x$
Evaluate the indefinite integral $\int \frac{1}{\sqrt{9 x^{2} - 12 x + 5}} d x$

Show your work for Problem 7

Evaluate the definite integral

\int_{0}^{6} \frac{1}{(4 x - 10) \sqrt{x^{2} - 5 x - 1}} d x

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Show your work for Problem 8

Evaluate the indefinite integral

\int \frac{2 x}{x^{4} - 8 x^{2} + 17} d x

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Evaluate the definite integral $\int_{0}^{6} \frac{1}{(4 x - 10) \sqrt{x^{2} - 5 x - 1}} d x$
Evaluate the indefinite integral $\int \frac{2 x}{x^{4} - 8 x^{2} + 17} d x$

Show your work for Problem 10

Evaluate the indefinite integral

\int \frac{1}{(2 x - 2) \sqrt{x}} d x

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Show your work for Problem 9

Evaluate the indefinite integral

\int \frac{2 x}{\sqrt{x^{4} + 3 x^{2} + 2}} d x

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Evaluate the indefinite integral $\int \frac{2 x}{\sqrt{x^{4} + 3 x^{2} + 2}} d x$
Evaluate the indefinite integral $\int \frac{1}{(2 x - 2) \sqrt{x}} d x$

Glossary and deeper reading · 4 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Chain rule
The chain rule differentiates a composition by multiplying the outer derivative at the inner output by the inner derivative. For differentiable functions, (f(g(x)))′=f′(g(x))g′(x). For example, the derivative of sin(x²) is cos(x²)·2x. Nested compositions require a factor for each successive layer.
Read more: 3.1 · Chain rule →
Derivative
A derivative gives the instantaneous rate of change of a function with respect to its input. At a point, it is the finite limit of the difference quotient when that limit exists. For example, the derivative of x² is 2x, so the tangent slope at x=3 is 6. A derivative function collects those rates across the inputs where the function is differentiable.
Read more: 2.2 · Derivative →
Differential
For a differentiable function y=f(x), the differential dy is defined as f′(x) dx. It represents the tangent-line estimate of the output change associated with an input change dx. For example, y=x² at x=3 and dx=0.01 gives dy=6·0.01=0.06. The actual change in y can differ from this linear estimate.
Read more: 4.6 · Differential →
Definite integral
A definite integral accumulates a function over a specified interval as a limit of Riemann sums when that limit exists. Its sign includes the function’s sign and the orientation of the limits. For example, the integral of the constant −2 from 0 to 3 is −6, although the geometric area is 6. In a rate model, the result has rate units multiplied by input units.
Read more: 6.3 · Definite integral →

Show your work for Problem 1

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Show your work for Problem 3

Show your work for Problem 5

Show your work for Problem 6

Show your work for Problem 7

Show your work for Problem 8

Show your work for Problem 10

Show your work for Problem 9

Chain rule

Derivative

Differential

Definite integral

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