26.6 Integration by Substitution
SL 5.10
- the four steps of integration by substitution (change of variable)
- how substitution handles integrals of the form \(\int k\,g'(x)f\big(g(x)\big)\,\mathrm{d}x\)
A substitution changes the variable so that the new integral is easy. On IB papers the substitution is given, unless the integral is of the form \(\int k\,g'(x)f\big(g(x)\big)\,\mathrm{d}x\).
The method
Integration by substitution
- Differentiate the substitution: find \(\dfrac{\mathrm{d}u}{\mathrm{d}x}\), and write \(\mathrm{d}x\) in terms of \(\mathrm{d}u\).
- Replace every part of the integrand, and \(\mathrm{d}x\), using the new variable \(u\). No \(x\) may remain.
- Integrate with respect to \(u\). (If this is not easier, the substitution was not a good choice.)
- Return to the original variable \(x\).
Not in the formula booklet.
Example 1 By using the substitution \(u = x^3 - 2\), find \(\displaystyle\int 3x^2\left(x^3 - 2\right)^5\mathrm{d}x\).
Step 1. \(\dfrac{\mathrm{d}u}{\mathrm{d}x} = 3x^2\), so \(\mathrm{d}x = \dfrac{\mathrm{d}u}{3x^2}\).
Step 2. \(\displaystyle\int 3x^2 u^5\,\frac{\mathrm{d}u}{3x^2} = \int u^5\,\mathrm{d}u\). The \(3x^2\) cancels, so no \(x\) is left.
Step 3. \(\displaystyle\int u^5\,\mathrm{d}u = \frac{u^6}{6} + C\).
Step 4. \(\displaystyle\int 3x^2\left(x^3 - 2\right)^5\mathrm{d}x = \frac{\left(x^3 - 2\right)^6}{6} + C\).
This is the reverse chain rule of Section 26.2, written out in full.
HL students: continue with 26.7 Further Integration by Substitution.
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| Replacing \((x^3 - 2)^5\) by \(u^5\) but keeping \(\mathrm{d}x\) | Replace \(\mathrm{d}x\) by \(\dfrac{\mathrm{d}u}{3x^2}\) | \(\mathrm{d}x\) and \(\mathrm{d}u\) are different. |
| Integrating while \(x\) and \(u\) are mixed | Write every \(x\) in terms of \(u\) first | You cannot integrate \(x\) with respect to \(u\). |
| Leaving the answer in \(u\) or \(\theta\) | Return to \(x\) | The question was in \(x\). |
Exercises
1 By using the substitution \(u = 2x^2 + 1\), find \(\displaystyle\int 4x\left(2x^2 + 1\right)^3\mathrm{d}x\).
\(\mathrm{d}u = 4x\,\mathrm{d}x\): \(\;\displaystyle\int u^3\,\mathrm{d}u = \frac{u^4}{4} + C = \frac{\left(2x^2 + 1\right)^4}{4} + C\).
2 By using the substitution \(u = x^2 + 9\), find \(\displaystyle\int \frac{x}{\sqrt{x^2 + 9}}\,\mathrm{d}x\).
\(\mathrm{d}u = 2x\,\mathrm{d}x\): \(\;\displaystyle\int \frac{1}{2}u^{-\frac{1}{2}}\,\mathrm{d}u = u^{\frac{1}{2}} + C = \sqrt{x^2 + 9} + C\).
3 By using the substitution \(u = \ln x\), find \(\displaystyle\int \frac{1}{x\ln x}\,\mathrm{d}x\), \(\;x > 1\).
\(\mathrm{d}u = \dfrac{1}{x}\,\mathrm{d}x\): \(\;\displaystyle\int \frac{1}{u}\,\mathrm{d}u = \ln\lvert u\rvert + C = \ln(\ln x) + C\).