Integration by Substitution (AA SL)
Once the thing you're integrating is a power, exponential, sine or cosine of a linear expression like \(2x+3\) rather than of \(x\) alone, the plain power rule isn't enough - you need to account for that inner linear function too. This page covers integration by substitution with worked examples and the mistakes that lose marks. It's part of the broader Integration topic.
17 questions on this sub-topic.
Integration by substitution
Covered under IB syllabus reference SL5.10: the indefinite integral of \(x^n\) (\(n\in\mathbb{Q}\)), \(\sin x\), \(\cos x\), \(e^x\) and \(\dfrac{1}{x}\), and the composites of any of these with the linear function \(ax+b\), integrated by substitution. These composite results aren't separately listed in the formula booklet - you build them from the basic integrals using a substitution.
Substitution, \(u = g(x)\)
If the integrand contains \(g'(x)\) multiplied by a function of \(g(x)\), let \(u=g(x)\), so \(du=g'(x)\,dx\).
For example, in \(\displaystyle\int 2x(x^2+1)^3\,dx\) let \(u=x^2+1\), so \(du=2x\,dx\) and the integral becomes \(\displaystyle\int u^3\,du\).
Substitution, \(u = ax+b\)
Let \(u=ax+b\), so \(\dfrac{du}{dx}=a\) and \(dx = \dfrac{1}{a}\,du\).
Rewrite the integral entirely in terms of \(u\), integrate, then substitute \(ax+b\) back in for \(u\). The result always carries a factor \(\tfrac1a\), for example \(\displaystyle\int e^{ax+b}\,dx=\dfrac1a e^{ax+b}+C\).
Need the full syllabus wording and formula-booklet reference table? See Integration.
Worked examples
Find \(\displaystyle\int (2x+3)^4\,dx\).
Worked solution
Let \(u=2x+3\), so \(\dfrac{du}{dx}=2\), i.e. \(dx=\dfrac12\,du.\)
\(\int(2x+3)^4\,dx=\dfrac12\int u^4\,du\) M1
\(=\dfrac12\cdot\dfrac{u^5}{5}+C=\dfrac{u^5}{10}+C\)
\(=\dfrac{(2x+3)^5}{10}+C.\) A1
(a)(i) Find \(\displaystyle\int e^{3x}\,dx\).
(a)(ii) Find \(\displaystyle\int 2e^{-x}\,dx\).
Worked solution
(a)(i) Let \(u=3x\), so \(du=3\,dx\), i.e. \(dx=\dfrac13\,du.\)
\(\int e^{3x}\,dx=\dfrac13\int e^{u}\,du\) M1
\(=\dfrac13e^{u}+C=\dfrac13e^{3x}+C.\) A1
(a)(ii) Let \(u=-x\), so \(du=-dx\), i.e. \(dx=-du.\)
\(\int2e^{-x}\,dx=-2\int e^{u}\,du\) A1
\(=-2e^{u}+C=-2e^{-x}+C.\) A1
Using the substitution \(u = x^2 - 1\), find \(\displaystyle\int_1^2 2x\sqrt{x^2-1}\,dx.\)
Worked solution
Let \(u = x^2 - 1,\) so \(du = 2x\,dx.\) M1
Change limits: \(x=1\Rightarrow u=0;\) \(x=2\Rightarrow u=3.\) A1
\(\displaystyle\int_0^3 \sqrt{u}\,du = \int_0^3 u^{1/2}\,du = \left[\tfrac23 u^{3/2}\right]_0^3 = \tfrac23 \cdot 3^{3/2} = \tfrac23 \cdot 3\sqrt{3}\) M1
\(= 2\sqrt{3}.\) A1
\(2\sqrt{3}.\) A1
Common mistakes
- Forgetting the factor from \(dx\). With \(u=2x+3\), \(dx=\tfrac12\,du\), so \(\int(2x+3)^4\,dx\) is not \(\dfrac{(2x+3)^5}{5}\) - the \(\tfrac12\) from the substitution must appear in the answer.
- Losing the sign with a negative coefficient. \(\int e^{-x}\,dx = -e^{-x}+C\), not \(e^{-x}+C\) - with \(u=-x\), \(dx=-du\) flips the sign, and it's easy to drop under time pressure.
- Substituting \(u=ax+b\) but forgetting to convert \(dx\). If you let \(u=ax+b\), then \(dx=\tfrac{1}{a}\,du\) has to appear in the new integral too - dropping that factor of \(\tfrac{1}{a}\) is the most common substitution error.
Ready to practise properly?
18 substitution questions, marked instantly like the real exam.
Quick answers
How do you integrate a composite of \(ax+b\) by substitution?
Let \(u=ax+b\), so \(du=a\,dx\) and \(dx=\tfrac1a\,du\). Rewrite the integral in terms of \(u\), integrate, then substitute \(ax+b\) back in for \(u\). The result always carries a factor \(\tfrac1a\). This works for powers, exponentials, sines and cosines of a linear expression.
When should I choose a different substitution?
If the integrand contains \(g'(x)\) multiplied by a function of \(g(x)\), let \(u=g(x)\) so that \(du=g'(x)\,dx\). For example, for \(2x(x^2+1)^3\) let \(u=x^2+1\). For a definite integral, change the limits as well. See the parent Integration page for GDC guidance.