Definite integrals by substitution
Master definite integrals by substitution for Year 12 Specialist Mathematics in Queensland (QCAA). This is the \(u\)-substitution method applied to bounded integrals, where the smart move is to change the limits of integration to \(u\)-values so the whole calculation finishes in \(u\).
You will learn to choose \(u=g(x)\), convert \(dx\) through \(du\), replace the limits with \(u=g(a)\) and \(u=g(b)\), and evaluate directly without back-substituting — the technique behind exact areas, logarithms and \(\pi\)-values that recurs across the integration topic and beyond.
Theory
Definite integrals by substitution extend the \(u\)-substitution method to bounded integrals in Year 12 Specialist Mathematics (QCAA, Queensland). The efficient approach is to change the limits of integration to \(u\)-values — if \(u=g(x)\), the limits \(x=a,\ x=b\) become \(u=g(a),\ u=g(b)\) — so the integral is finished entirely in \(u\) with no back-substitution. This page shows the full process with worked examples.
A definite integral \(\displaystyle\int_a^b f(x)\,dx\) has lower and upper limits of integration \(x=a\) and \(x=b\), and evaluates to a number. When the integrand has the form \(f(g(x))\,g'(x)\), the substitution \(u=g(x)\) works just as for an indefinite integral — but the limits must be handled as well.
The neat method is to change the limits to \(u\)-values. As \(x\) runs from \(a\) to \(b\), the new variable \(u=g(x)\) runs from \(g(a)\) to \(g(b)\). Rewriting everything — integrand, \(dx\) and both limits — gives an integral entirely in \(u\).
Because the limits are now \(u\)-values, you do not back-substitute. Once the antiderivative in \(u\) is found, evaluate it straight away at the new limits \(u=g(a)\) and \(u=g(b)\).
The bridge is still \(du=g'(x)\,dx\), and a missing constant factor is carried outside, for example \(x\,dx=\tfrac{1}{2}\,du\). A definite integral needs no \(+c\), and its value may be a fraction, a surd, a logarithm or a multiple of \(\pi\).
The change-of-variable rule for a definite integral: if \(u=g(x)\) with \(du=g'(x)\,dx\), then the limits change with the variable:
A constant factor is carried outside, for example when \(du=2x\,dx\):
Two standard results that substitution reaches after completing the square are the logarithm and inverse-tangent forms:
How to evaluate a definite integral by substitution
- Choose \(u\): pick the inner function \(g(x)\) whose derivative appears in the integrand, and find \(du=g'(x)\,dx\) (rearranging to a factor such as \(x\,dx=\tfrac{1}{2}\,du\) if needed).
- Change the limits: compute the new \(u\)-limits, \(u=g(a)\) at the lower end and \(u=g(b)\) at the upper end.
- Rewrite in \(u\): replace every \(x\), the \(dx\) and both limits so the integral is \(\displaystyle\int_{g(a)}^{g(b)} f(u)\,du\).
- Integrate and evaluate: find the antiderivative in \(u\) and evaluate it at the new limits — no back-substitution, no \(+c\).
Substitute \(u=x^2+1\) and relate \(x\,dx\) to \(du\):
| \(u\) | \(=\) | \(x^2+1\) |
| \(\dfrac{du}{dx}\) | \(=\) | \(2x\) |
| \(x\,dx\) | \(=\) | \(\dfrac{1}{2}\,du\) |
Change the limits to \(u\)-values:
| \(x=0\) | \(\Rightarrow\) | \(u=0^2+1=1\) |
| \(x=2\) | \(\Rightarrow\) | \(u=2^2+1=5\) |
Rewrite in \(u\) and evaluate — no back-substitution:
| \(\int_0^2 x\,(x^2+1)^3\,dx\) | \(=\) | \(\dfrac{1}{2}\int_1^5 u^3\,du\) |
| \(=\) | \(\dfrac{1}{2}\left[\dfrac{u^4}{4}\right]_1^5\) | |
| \(=\) | \(\dfrac{1}{8}\left(625-1\right)\) | |
| \(=\) | \(\dfrac{1}{8}\times 624\) | |
| \(=\) | \(78\) |
\(\displaystyle\int_0^2 x\,(x^2+1)^3\,dx=78\).
The numerator is exactly the derivative of the denominator, so \(u=x^2+x\):
| \(u\) | \(=\) | \(x^2+x\) |
| \(\dfrac{du}{dx}\) | \(=\) | \(2x+1\) |
| \(du\) | \(=\) | \((2x+1)\,dx\) |
Change the limits to \(u\)-values:
| \(x=1\) | \(\Rightarrow\) | \(u=1+1=2\) |
| \(x=2\) | \(\Rightarrow\) | \(u=4+2=6\) |
Rewrite as a \(\tfrac{1}{u}\) integral and evaluate:
| \(\int_1^2 \dfrac{2x+1}{x^2+x}\,dx\) | \(=\) | \(\int_2^6 \dfrac{1}{u}\,du\) |
| \(=\) | \(\left[\ln|u|\right]_2^6\) | |
| \(=\) | \(\ln 6-\ln 2\) | |
| \(=\) | \(\ln 3\) |
\(\displaystyle\int_1^2 \dfrac{2x+1}{x^2+x}\,dx=\ln 3\).
Write \(\tan x=\dfrac{\sin x}{\cos x}\); with \(u=\cos x\), \(\sin x\,dx=-du\):
| \(u\) | \(=\) | \(\cos x\) |
| \(\dfrac{du}{dx}\) | \(=\) | \(-\sin x\) |
| \(\sin x\,dx\) | \(=\) | \(-\,du\) |
Change the limits to \(u\)-values:
| \(x=0\) | \(\Rightarrow\) | \(u=\cos 0=1\) |
| \(x=\dfrac{\pi}{3}\) | \(\Rightarrow\) | \(u=\cos\dfrac{\pi}{3}=\dfrac{1}{2}\) |
Rewrite in \(u\); the minus sign flips the limits, then integrate:
| \(\int_0^{\pi/3} \tan x\,dx\) | \(=\) | \(-\int_1^{1/2} \dfrac{1}{u}\,du\) |
| \(=\) | \(\int_{1/2}^{1} \dfrac{1}{u}\,du\) | |
| \(=\) | \(\left[\ln|u|\right]_{1/2}^{1}\) | |
| \(=\) | \(\ln 1-\ln\dfrac{1}{2}\) | |
| \(=\) | \(\ln 2\) |
\(\displaystyle\int_0^{\pi/3} \tan x\,dx=\ln 2\).
Complete the square in the denominator:
| \(x^2-4x+8\) | \(=\) | \((x^2-4x+4)+4\) |
| \(=\) | \((x-2)^2+4\) |
Substitute \(u=x-2\) and change the limits:
| \(u\) | \(=\) | \(x-2\) |
| \(du\) | \(=\) | \(dx\) |
| \(x=2\) | \(\Rightarrow\) | \(u=0\) |
| \(x=4\) | \(\Rightarrow\) | \(u=2\) |
Rewrite as an arctan standard form and evaluate:
| \(\int_2^4 \dfrac{1}{(x-2)^2+4}\,dx\) | \(=\) | \(\int_0^2 \dfrac{1}{u^2+4}\,du\) |
| \(=\) | \(\dfrac{1}{2}\left[\arctan\dfrac{u}{2}\right]_0^2\) | |
| \(=\) | \(\dfrac{1}{2}\left(\arctan 1-\arctan 0\right)\) | |
| \(=\) | \(\dfrac{1}{2}\times\dfrac{\pi}{4}\) | |
| \(=\) | \(\dfrac{\pi}{8}\) |
\(\displaystyle\int_2^4 \dfrac{1}{x^2-4x+8}\,dx=\dfrac{\pi}{8}\).
Common pitfalls
Frequently asked questions
Do you change the limits when integrating by substitution?
For a definite integral, yes — changing the limits to \(u\)-values is the efficient method. If \(u=g(x)\) then the limits \(x=a,\ x=b\) become \(u=g(a),\ u=g(b)\), and you evaluate entirely in \(u\).
Why don't you back-substitute for a definite integral?
Because the limits are already written as \(u\)-values. Once the antiderivative in \(u\) is found, evaluate it at \(u=g(a)\) and \(u=g(b)\); returning to \(x\) is unnecessary.
What are the new limits if \(u=x^2+1\) and \(x\) runs from \(0\) to \(2\)?
Substitute each old limit: \(x=0\) gives \(u=0^2+1=1\), and \(x=2\) gives \(u=2^2+1=5\), so the new limits are \(u=1\) and \(u=5\).
Do you add \(+c\) to a definite integral?
No. The constant of integration cancels when you evaluate at the two limits, so a definite integral gives a single number with no \(+c\).
What happens to the sign when \(u=\cos x\)?
Differentiating gives \(du=-\sin x\,dx\), so \(\sin x\,dx=-du\). The minus sign is carried out front, or you can swap the two limits to absorb it.
When does a definite substitution give \(\pi\) or a logarithm?
Completing the square and substituting can turn an integrand into \(\dfrac{1}{u^2+p^2}\), giving an \(\arctan\) and often a multiple of \(\pi\); an \(\dfrac{f'(x)}{f(x)}\) integrand gives a natural logarithm.