Topic 6.9

Choose a useful substitution

Builds on finding antiderivatives of basic functions. 8 minutes.

An inside function and its slope

The product \(x^2(x^3+2)^4\) has no antiderivative that a basic rule gives directly.

The expression \(x^3 + 2\) sits inside the fourth power, so call it \(g(x)\), the inside function.

The orange curve shows its slope \(g'(x) = 3x^2\), and the faint curve shows \(x^2\), the factor that stands outside the power.

Drag the point to the right. At \(x = 1\) the orange height is 3 and the faint height is 1, and at \(x = 2\) they are 12 and 4.

The orange height is 3 times the faint height at every \(x\), so \(x^2\) is the slope of \(g\) divided by 3.

Any inside function

The integrand is an outside function \(f\) applied to the inside function \(g(x)\), multiplied by a leftover factor \(L(x)\).

The orange curve shows the slope \(g'(x)\), and the faint curve shows \(L(x)\).

Drag the point. The orange height changes, and the faint dot stays at the same fraction of that height.

Press the button. The faint curve becomes twice as tall, so the fraction doubles, and it is again the same at every x.

A constant ratio is the test. If \(L(x)\) is a constant multiple of \(g'(x)\), then \(g\) is a good inside function.

Notation

  1. Name the inside function. Write \(u\) for its value.

    \[\cy{u = g(x)}\]
  2. A small change \(dx\) in \(x\) changes \(u\) by a small amount \(du\). The slope of \(g\) is \(g'(x)\), so \(du\) equals the slope times \(dx\).

    \[du = \cy{g'(x)}\,dx\]
  3. \(u = g(x)\) is a good choice when the rest of the integrand is a constant multiple \(k\) of the slope \(g'(x)\). Then \(g'(x)\,dx\) is \(du\), so the integral becomes \(k\int f(u)\,du\) with no \(x\) left.

    \[\text{integrand} = k\, f(g(x))\,\cy{g'(x)}\]
  4. The function \(g\) must have a derivative on the interval. The function \(f\) must be continuous where \(g\) takes its values.

When it fails

Replace the factor \(x^2\) with \(x\). Then drag the point and compare the two heights.

The orange height is still \(3x^2\), and the faint height is now \(x\). The ratio is 3 at \(x = 1\) and 6 at \(x = 2\), so \(x\) is not a constant multiple of \(g'(x)\). Then \(u = x^3 + 2\) does not remove every \(x\). Try a different inside function or a different method.

Example 1: a fraction

Choose an inside function for \(\displaystyle\int \frac{x}{x^2+4}\,dx\).

  1. The denominator \(x^2 + 4\) sits inside the reciprocal function \(1/u\). Try it as the inside function.

    \[g(x) = \cy{x^2 + 4}\]
  2. Its slope is \(2x\), which the orange line shows.

    \[g'(x) = \cy{2x}\]
  3. The leftover factor is \(x\), the faint line. At \(x = 2\) the orange height is 4 and the faint height is 2, and the ratio is 2 at every \(x\).

    \[\dfrac{2x}{x} = 2\]
  4. The ratio is constant, so \(u = x^2 + 4\) is a good choice.

Example 2: two candidates

Choose an inside function for \(\displaystyle\int \cos x\, e^{\sin x}\,dx\).

  1. The integrand has the factor \(\cos x\) and the exponent \(\sin x\). Test each one as the inside function.

  2. A tempting wrong move: Choose \(u = \cos x\), because it is the first factor.

    \[\begin{aligned} g(x) &= \cos x \\ g'(x) &= \cr{-\sin x} \end{aligned}\]

    The leftover factor \(e^{\sin x}\), the faint curve, is 1 at \(x = 0\). The slope \(-\sin x\), the red curve, is 0 there, so the two cannot have a constant ratio. Also, \(\cos x\) is not inside any other function.

  3. Choose the expression that sits inside another function. Here \(\sin x\) sits inside the exponential.

    \[g(x) = \cy{\sin x}\]
  4. Its slope is \(\cos x\), which is the other factor. The ratio is 1, so \(u = \sin x\) works.

    \[g'(x) = \cos x, \quad \dfrac{\cos x}{\cos x} = 1\]

Practice

Choose an inside function for \(\displaystyle\int \frac{(\ln x)^4}{x}\,dx\) and find the ratio of its slope to the leftover factor.

Find the expression that sits inside the fourth power. Then compare its slope with the other factor, the faint curve.

Solution
  1. For \(x > 0\), the expression \(\ln x\) sits inside the fourth power. Choose it.

    \[g(x) = \ln x\]
  2. Its slope is \(1/x\), which is also the leftover factor. The orange curve lies on the faint curve.

    \[g'(x) = \dfrac{1}{x}\]
  3. The ratio is 1, so \(u = \ln x\) works.

    \[\dfrac{1/x}{1/x} = 1\]

Questions

The orange curve is the slope \(g'(x)\) of the inside function \(g(x) = 2x^3 + 1\). The faint curve is a leftover factor \(L(x) = 3x^2\), and the faint dashed line is another one, \(M(x) = 2x\).

  1. What is the ratio \(g'(x)/L(x)\) of the orange height to the faint height?

    • 2 at every \(x\)
    • 3
    • \(\tfrac12\)
    • It changes with \(x\)
    Answer

    2 at every \(x\)

  2. For which leftover factor can you use \(u = g(x)\)?

    • Only \(L(x)\)
    • Only \(M(x)\)
    • Both
    • Neither
    Answer

    Only \(L(x)\)

  3. Which choice of \(u\) works for \(\displaystyle\int \frac{\sin x}{\cos^3 x}\,dx\)?

    • \(u = \cos x\)
    • \(u = \sin x\)
    • \(u = x\)
    Answer

    \(u = \cos x\)

Summary

You find an expression that sits inside another function, and you compare its slope with the rest of the integrand.

If the rest is a constant multiple of that slope, the substitution removes every x, because the slope times dx is du.

You test the ratio before you substitute, and you choose a different expression when the ratio is not constant.

This is one of four skills in topic 6.9 of Which & Why: AP Calculus.