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 the reverse chain rule (integrating by inspection or substitution) with worked examples and the mistakes that lose marks. It's part of the broader Integration topic.
17 questions on this sub-topic.
The reverse chain rule
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 inspection or substitution. These composite results aren't separately listed in the formula booklet - you build them from the basic derivatives and reverse the chain rule yourself.
Reverse chain rule
For \((ax+b)^n\): integrate as if it's a power of \(x\), then divide by \(a\).
\(\displaystyle\int (ax+b)^n\,dx = \dfrac{(ax+b)^{n+1}}{a(n+1)}+C\). The same "integrate, then divide by \(a\)" idea applies to \(e^{ax+b}\), \(\sin(ax+b)\) and \(\cos(ax+b)\).
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 \(u\). Slower than inspection, but safer when the coefficient \(a\) is easy to mishandle.
Need the full syllabus wording and formula-booklet reference table? See Integration.
Worked examples
Find \(\displaystyle\int (2x+3)^4\,dx\).
Worked solution
Reverse chain rule. For \((ax+b)^n\), integrate as if a power, then divide by \(a(n+1)\).
\(\int(2x+3)^4\,dx=\dfrac{(2x+3)^5}{5\cdot 2}+C.\) M1
\(\dfrac{(2x+3)^5}{5\cdot 2}\)
\(=\dfrac{(2x+3)^5}{10}\)
\(=\dfrac{(2x+3)^5}{10}+C.\) A1
(a)(i) Find \(\displaystyle\int e^{3x}\,dx\).
(a)(ii) Find \(\displaystyle\int 2e^{-x}\,dx\).
Worked solution
(a)(i) Reverse chain for exponentials (divide by the coefficient of \(x\)): \(\int e^{3x}\,dx=\tfrac13 e^{3x}+C.\) M1
\(\int e^{3x}\,dx=\tfrac13 e^{3x}+C.\) A1
(a)(ii) \(\int 2e^{-x}\,dx=-2e^{-x}+C.\) A1
\(\int 2e^{-x}\,dx=-2e^{-x}+C.\) 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 to divide by the coefficient \(a\). \(\int(2x+3)^4\,dx\) is not \((2x+3)^5\) - you must divide by \(a(n+1)\), not just \((n+1)\), because the inner function isn't just \(x\).
- Losing the sign with a negative coefficient. \(\int e^{-x}\,dx = -e^{-x}+C\), not \(e^{-x}+C\) - dividing by \(a=-1\) 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
What is the reverse chain rule for integration?
For a composite of \(ax+b\), integrate as if it were the outer function alone, then divide by \(a\). This works for powers, exponentials, sines and cosines of a linear expression.
When do I need integration by substitution instead of by inspection?
Simple composites of \(ax+b\) can be integrated by inspection using the reverse chain rule directly. Writing out \(u=ax+b\) explicitly helps when the coefficient is easy to mishandle. See the parent Integration page for GDC guidance.