Integration by substitution (WACE Year 12 Mathematics Specialist Unit 4)

What this dot point is asking

SCSA wants confident use of substitution for both indefinite and definite integrals, including recognising when the integrand contains a function and (a multiple of) its derivative.

The method

Choose $u = g(x)$ so that the integrand simplifies. Differentiate to get $du = g'(x)\,dx$, and replace every $x$-expression, including $dx$, by its $u$-equivalent. The integral must end up entirely in $u$ with no stray $x$ remaining. Integrate, then for an indefinite integral substitute $x$ back.

Choosing the substitution

Look for an inner function whose derivative is also present (up to a constant factor). Common choices: the expression under a root, the exponent of $e$, the argument of a trig function, or the denominator. If the derivative is present only up to a constant, adjust by that constant.

Why substitution reverses the chain rule

The substitution rule is the chain rule read in reverse. If $F$ is an antiderivative of $f$, then by the chain rule $\dfrac{d}{dx}F(g(x)) = f(g(x))\,g'(x)$. Integrating both sides recovers $\displaystyle\int f(g(x))\,g'(x)\,dx = F(g(x)) + C$, which is exactly what the substitution $u = g(x)$, $du = g'(x)\,dx$ computes when it turns the integral into $\displaystyle\int f(u)\,du = F(u) + C$. Understanding the rule this way makes it clear why the integrand must contain the derivative $g'(x)$ of the inner function: that factor is the leftover from differentiating the composite, and without it (or a constant multiple of it) the substitution will not close.

Recognition versus full working

With practice, many substitutions can be done by recognition, writing the answer directly. An integral of the form $\displaystyle\int g'(x)\,[g(x)]^n\,dx$ is immediately $\dfrac{[g(x)]^{n+1}}{n+1} + C$, and $\displaystyle\int \dfrac{g'(x)}{g(x)}\,dx$ is immediately $\ln|g(x)| + C$. SCSA accepts recognition, but in the calculator-free section it is safer to show the substitution explicitly for any non-trivial case, because a single dropped constant factor loses the answer. A good habit is to differentiate your proposed antiderivative mentally as a check: it should return the original integrand exactly.

Definite integrals: change the limits

This is cleaner and avoids errors from reverting. If you do keep $x$-limits, you must back-substitute before applying them.

Substitutions that simplify rather than match

Not every substitution exploits a derivative already present; some are chosen simply to simplify an awkward expression. A linear substitution $u = ax + b$ handles integrands like $\sqrt{2x + 1}$ or $(3x - 4)^7$, where $du = a\,dx$ supplies a constant factor. A trigonometric substitution such as $x = a\sin\theta$ converts $\sqrt{a^2 - x^2}$ into $a\cos\theta$ via the Pythagorean identity, turning a root into a clean trig integrand; this is how integrals leading to $\arcsin$ are handled. The guiding principle is to pick the substitution that removes the structural obstacle, whether that is a composition, a root, or a sum of squares.

A reliable checking habit

Because substitution has several places to slip, build in a check. After finding an antiderivative, differentiate it mentally and confirm it returns the original integrand, including any constant factor. For a definite integral, sanity-check the sign and rough size of the answer against the graph: an integral of a positive function over an interval must be positive, and its magnitude should be comparable to the area under the curve. These quick checks catch the most common errors, a dropped constant or an unconverted limit, before they cost marks.