Integration by substitution
Master integration by substitution for Year 12 Specialist Mathematics in Queensland (QCAA). This is the technique that reverses the chain rule: when an integrand looks like a function of an inner function times that inner function's derivative, a well-chosen substitution turns it into a standard integral.
You will learn to recognise the \(f(g(x))\,g'(x)\) pattern, choose \(u=g(x)\), convert \(dx\) through \(du\), and back-substitute to finish — the essential integration technique behind areas, volumes and differential equations later in the course.
Theory
Integration by substitution reverses the chain rule and is a core technique of Year 12 Specialist Mathematics (QCAA, Queensland). When an integrand has the form \(f(g(x))\,g'(x)\), setting \(u=g(x)\) turns it into a standard integral in \(u\). This page shows how to choose \(u\), convert \(dx\) through \(du\), integrate and then back-substitute.
Integration by substitution (the \(u\)-substitution method) is the reverse of the chain rule. It works when the integrand can be written as an outer function of an inner function, multiplied by the derivative of that inner function: \(f(g(x))\,g'(x)\).
The inner function \(g(x)\) is the expression you would differentiate last by the chain rule — often what sits inside a power, a root, an exponential, a logarithm or a trig function. You rename it \(u=g(x)\).
Differentiating gives \(du=g'(x)\,dx\). This is the bridge that converts every \(x\) and the \(dx\) into \(u\) and \(du\). If the derivative \(g'(x)\) is present only up to a constant multiple, carry that constant outside the integral.
After rewriting, the integral \(\int f(u)\,du\) is a standard one. Integrate it, then back-substitute \(u=g(x)\) so the answer is in terms of \(x\), and add the constant \(+c\) (these are indefinite integrals).
The substitution rule: if \(u=g(x)\) then \(du=g'(x)\,dx\), and
A constant factor can be moved outside; for example when \(du=2x\,dx\):
A special case is the logarithm form, where the numerator is the derivative of the denominator:
How to integrate by substitution
- Choose \(u\): pick the inner function \(g(x)\) whose derivative \(g'(x)\) also appears in the integrand (at least up to a constant).
- Differentiate: find \(du=g'(x)\,dx\), and if needed rearrange to make the factor in the integrand, e.g. \(x\,dx=\tfrac{1}{2}\,du\).
- Rewrite: replace every \(x\) and the \(dx\) so the integral is entirely in \(u\), \(\int f(u)\,du\).
- Integrate and back-substitute: integrate in \(u\), replace \(u=g(x)\), and add \(+c\).
Let \(u\) be the inner cubic; its derivative \(3x^2\) is exactly \(du\):
| \(u\) | \(=\) | \(x^3+4\) |
| \(du\) | \(=\) | \(3x^2\,dx\) |
Rewrite the whole integral in \(u\), then integrate:
| \(\int 3x^2(x^3+4)^5\,dx\) | \(=\) | \(\int u^5\,du\) |
| \(=\) | \(\dfrac{u^6}{6}+c\) |
Back-substitute \(u=x^3+4\):
| \(=\) | \(\dfrac{(x^3+4)^6}{6}+c\) |
\(\displaystyle\int 3x^2(x^3+4)^5\,dx=\dfrac{(x^3+4)^6}{6}+c\).
Here \(du=2x\,dx\), so the factor \(x\,dx\) is only half of \(du\):
| \(u\) | \(=\) | \(x^2+5\) |
| \(du\) | \(=\) | \(2x\,dx\) |
| \(x\,dx\) | \(=\) | \(\dfrac{1}{2}\,du\) |
Rewrite in \(u\), carrying the \(\tfrac{1}{2}\) outside, and integrate:
| \(\int x(x^2+5)^4\,dx\) | \(=\) | \(\dfrac{1}{2}\int u^4\,du\) |
| \(=\) | \(\dfrac{1}{2}\cdot\dfrac{u^5}{5}+c\) |
Back-substitute \(u=x^2+5\):
| \(=\) | \(\dfrac{(x^2+5)^5}{10}+c\) |
\(\displaystyle\int x(x^2+5)^4\,dx=\dfrac{(x^2+5)^5}{10}+c\).
Let \(u=\cos x\); because \(du=-\sin x\,dx\), the factor \(\sin x\,dx=-du\):
| \(u\) | \(=\) | \(\cos x\) |
| \(du\) | \(=\) | \(-\sin x\,dx\) |
| \(\sin x\,dx\) | \(=\) | \(-\,du\) |
Rewrite in \(u\); the minus sign comes out front, then integrate:
| \(\int \sin x\,e^{\cos x}\,dx\) | \(=\) | \(-\int e^{u}\,du\) |
| \(=\) | \(-e^{u}+c\) |
Back-substitute \(u=\cos x\):
| \(=\) | \(-e^{\cos x}+c\) |
\(\displaystyle\int \sin x\,e^{\cos x}\,dx=-e^{\cos x}+c\).
The numerator is the derivative of the denominator, so this is the \(\tfrac{f'}{f}\) form:
| \(u\) | \(=\) | \(x^3+1\) |
| \(du\) | \(=\) | \(3x^2\,dx\) |
Rewrite as \(\int\tfrac{1}{u}\,du\) and integrate:
| \(\int \dfrac{3x^2}{x^3+1}\,dx\) | \(=\) | \(\int \dfrac{1}{u}\,du\) |
| \(=\) | \(\ln\lvert u\rvert+c\) |
Back-substitute \(u=x^3+1\):
| \(=\) | \(\ln\lvert x^3+1\rvert+c\) |
\(\displaystyle\int \dfrac{3x^2}{x^3+1}\,dx=\ln\lvert x^3+1\rvert+c\).
Common pitfalls
Frequently asked questions
When should you use integration by substitution?
Use it when the integrand is a composite function times the derivative of its inner part, \(f(g(x))\,g'(x)\) — for example something inside a power, root, exponential, logarithm or trig function, with that inner expression's derivative present as a factor.
How do you choose \(u\) in a substitution?
Choose \(u=g(x)\), the inner function, so that its derivative \(g'(x)\) appears in the integrand (at least up to a constant). A good check is that after substituting, no \(x\) is left.
What is \(du\) and why do you need it?
Differentiating \(u=g(x)\) gives \(du=g'(x)\,dx\). This converts the \(dx\) into \(du\); without it the integral cannot be fully rewritten in \(u\).
What if the derivative is off by a constant?
If the integrand has, say, \(x\,dx\) but \(du=2x\,dx\), then \(x\,dx=\tfrac{1}{2}\,du\); carry the constant \(\tfrac{1}{2}\) outside the integral.
Do you have to back-substitute at the end?
Yes, for an indefinite integral. After integrating in \(u\), replace \(u=g(x)\) so the answer is in \(x\), and include \(+c\).
When does substitution give a natural logarithm?
When the integrand is \(\dfrac{f'(x)}{f(x)}\) — the derivative of the denominator over the denominator — the result is \(\ln\lvert f(x)\rvert+c\).