Definite integrals by substitution
Master definite integrals by substitution in Year 12 VCE Specialist Mathematics. 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 \(u\). It sits in the Calculus area of study of the VCE Mathematics Study Design (VCAA), within the Integral calculus topic of Unit 4.
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. 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.