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SASpecialist MathematicsSyllabus dot point

How does reversing the chain rule let us integrate composite functions?

Evaluate integrals using the method of substitution, including changing the limits for definite integrals.

Using substitution to reverse the chain rule, choosing u and replacing dx with du, handling definite integrals by changing the limits, and recognising when substitution applies.

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  1. What this dot point is asking
  2. The method
  3. A worked indefinite integral
  4. Handling a constant factor
  5. Definite integrals: change the limits

What this dot point is asking

You need to evaluate integrals by substitution, including correctly handling the limits of definite integrals.

The method

If u=g(x)u = g(x) then dudx=g′(x)\dfrac{du}{dx} = g'(x), so du=g′(x) dxdu = g'(x)\,dx. The substitution rule is

∫f(g(x)) g′(x) dx=∫f(u) du.\int f(g(x))\,g'(x)\,dx = \int f(u)\,du.

The aim is to pick uu so the integrand turns into f(u) duf(u)\,du with no xx left over. Look for a function whose derivative (up to a constant factor) is also in the integrand.

A worked indefinite integral

Handling a constant factor

Often dudu differs from the integrand by a constant. Solve for dxdx and carry the constant through.

For ∫xex2 dx\displaystyle\int x e^{x^2}\,dx, let u=x2u = x^2, so du=2x dxdu = 2x\,dx, hence x dx=12dux\,dx = \tfrac{1}{2}du:

∫xex2 dx=∫eu⋅12 du=12eu+C=12ex2+C.\int x e^{x^2}\,dx = \int e^{u}\cdot\tfrac{1}{2}\,du = \tfrac{1}{2}e^{u} + C = \tfrac{1}{2}e^{x^2} + C.

Definite integrals: change the limits

For a definite integral, the cleanest approach changes the limits to uu-values so you never convert back.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2025 SACE Stage 23 marksGiven that sin^3(x) = sin(x) - sin(x) cos^2(x), show that the integral from 0 to pi/3 of sin^3(x) dx = 5/24.
Show worked answer →

Rewrite the integrand using the given identity: integral from 0 to pi/3 of [sin(x) - sin(x) cos^2(x)] dx.

The first piece integrates directly; the second is a substitution. Let u = cos(x), so du = -sin(x) dx, i.e. sin(x) dx = -du. Change the limits: when x = 0, u = cos(0) = 1; when x = pi/3, u = cos(pi/3) = 1/2. [1 mark]

integral of sin(x) dx = -cos(x), and integral of sin(x) cos^2(x) dx = integral of u^2 (-du) = -u^3/3 = -cos^3(x)/3. So the antiderivative is -cos(x) + cos^3(x)/3. [1 mark]

Evaluate from 0 to pi/3:
At x = pi/3: -1/2 + (1/2)^3/3 = -1/2 + (1/8)/3 = -1/2 + 1/24 = -11/24.
At x = 0: -1 + 1/3 = -2/3 = -16/24.
Difference: -11/24 - (-16/24) = 5/24, as required. [1 mark]