Substitution and trigonometric integration: VCE Specialist Mathematics Units 3 & 4

1. Substitution reverses the chain rule

When a composite expression and its inner derivative occur together, integration by substitution replaces that composite expression with a simpler variable. If u = g(x), then du = g′(x) dx. This is shorthand for the chain-rule relationship that justifies the change of variable, not a licence to cancel arbitrary symbols. The goal is an integral written completely in u, with no unexplained x still present.

Look first for a repeated inner expression, a denominator whose derivative resembles the numerator, or a root whose inside can be simplified. For ∫x(x² + 3)⁴ dx, choose u = x² + 3. Because du = 2x dx, we replace x dx by du/2. The new integral is (1/2)∫u⁴ du, not ∫u⁴ du and not ∫xu⁴ du. A constant factor can be compensated; a missing variable factor requires further algebra.

A good substitution makes the remaining integral easier. Choosing u = x² for ∫eˣ dx is possible only with awkward branch choices and is unhelpful. Choosing u = x² + 3 for ∫x(x² + 3)⁴ dx removes exactly the structure that was causing difficulty. The method is a reasoned selection, not a rule that every bracket must become u.

2. Keep the differential and bounds together

For an indefinite integral, integrate in u and substitute the original expression back before giving the answer as a function of x. Include C once. For a definite integral, either change both bounds to u-values and stay in u until evaluation, or return to an antiderivative in x before applying the original bounds. Both approaches are valid; mixing them is not.

The order of transformed bounds is meaningful. If the substitution decreases as x increases, its differential or its reversed bounds accounts for the sign. Do not silently reorder bounds while retaining the same sign. When a substitution is used by solving for x in terms of u, choose a one-to-one interval and an appropriate branch; if necessary, split the original interval first.

The direct chain-rule identity can still hold for a non-monotonic g when the whole integrand is f(g(x))g′(x). Equal transformed bounds then correctly express cancellation. That situation differs from trying to express a general integral using a single-valued inverse x = g⁻¹(u). Knowing which justification you are using prevents unnecessary restrictions and hidden branch errors.

3. Simplify trigonometric structure first

Not every trigonometric product needs a named new variable immediately. Identities may reveal a constant term or a useful derivative pair. In radians, the double-angle identities give sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2. Therefore integrating either square produces a linear x-term as well as a sine term. Squaring an antiderivative would give neither relationship correctly.

For an odd power of sine multiplied by a power of cosine, reserve one sin x factor and replace the remaining sin²x factors by 1 − cos²x; then u = cos x often works. Similarly, reserve cos x when substituting u = sin x. For reciprocal functions, sec²x is the derivative of tan x, while sec x tan x is the derivative of sec x. Recognition of these pairs can be more useful than expanding everything.

For example, ∫sin³x cos²x dx becomes ∫sin x(1 − cos²x)cos²x dx. With u = cos x and du = −sin x dx, this is −∫(u² − u⁴) du = −u³/3 + u⁵/5 + C. Returning to x gives −cos³x/3 + cos⁵x/5 + C. The minus sign belongs to the differential and must survive every subsequent step.

4. Trigonometric substitutions remove selected roots

A root involving a² − x² suggests x = a sin θ with a > 0, because 1 − sin²θ = cos²θ. Choose −π/2 ≤ θ ≤ π/2 so that cos θ ≥ 0. Then √(a² − x²) = a cos θ and dx = a cos θ dθ. Without this branch statement the root is a|cos θ|, and dropping the absolute value could change the sign.

A sum a² + x² can suggest x = a tan θ on −π/2 < θ < π/2. On that interval sec θ is positive and dx = a sec²θ dθ. Use this choice only when the resulting trigonometric integral is simpler. Often a scaled arctangent can be recognised directly. The method is useful because an identity simplifies the full expression, not because a triangle must be drawn in every problem.

After an indefinite trigonometric substitution, return every symbol to x. For x = a sin θ, θ = arcsin(x/a), sin θ = x/a and cos θ = √(a² − x²)/a on the chosen branch. A right-triangle sketch can help with positive x, but the branch equations remain valid for negative x too. A triangle with unsigned lengths alone does not settle signs.

5. Diagnose a method before pressing enter

Write a three-part substitution record: the definition of u, the replaced differential, and the new bounds or planned back-substitution. Inspect the transformed line for mixed variables. A surviving x is not automatically wrong if you have an explicit expression for it in u, but that expression must be substituted before integration. This is especially important when roots force positive or negative branches.

With technology, compare a symbolic antiderivative’s derivative with the original integrand on its real domain, then evaluate definite integrals numerically as a separate check. Graphing can show whether the integrand is positive and whether the magnitude is plausible. It cannot justify an omitted differential or prove that two branches are interchangeable. Enter radians and bracket the entire denominator and square root.

In an unfamiliar question, test the proposed substitution by differentiating it before committing to lengthy work. If the derivative factor appears up to a constant, proceed. If an identity supplies the missing factor, show that identity. If neither happens, reconsider the choice. This deliberate pause is usually quicker than trying to repair an integral that has become more complicated.

6. Read a substitution as a change of measurement

The differential factor records how quickly the new variable changes compared with the old one. If u changes twice as fast as x locally, a small x-width corresponds to twice the u-width. That is why replacing the expression alone, without replacing dx, changes the accumulated quantity. Substitution preserves the integral by changing both the function description and the measurement of interval width.

A tangent substitution illustrates this balance neatly. For ∫1/(4 + x²) dx, set x = 2tan θ on −π/2 < θ < π/2. Then dx = 2sec²θ dθ and 4 + x² = 4sec²θ. The transformed quotient is simply (1/2)dθ, producing (1/2)arctan(x/2) + C. The secant factors cancel because both the denominator and the differential were transformed.

This agrees with direct inverse-circular recognition, so the substitution is useful as an explanation rather than the shortest calculation. In a more complicated root integral, the same identity may genuinely simplify an otherwise difficult expression. Compare the transformed integral with what you started with: a method should reveal structure, not merely replace one letter by another.

For a definite problem, a short endpoint table is often the clearest communication: write each original endpoint beside its transformed value. If the original interval reaches a place where the chosen inverse changes branch, split the interval before using that inverse. At the end, check whether the sign of the answer matches the sign pattern of the original integrand, not just the transformed notation.