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Business Mathematics and Statistics · Ch 7 — Integration

Integration by Substitution

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Integration by Substitution

Some integrands are not directly one of the standard forms, but become one after a change of variable. Integration by substitution (also called the change-of-variable method) is the technique for spotting and using this.

When substitution applies

Substitution works cleanly whenever the integrand can be seen as a function of some inner expression g(x)g(x), multiplied by (a constant multiple of) g′(x)g'(x) — exactly what the chain rule produces when differentiating a composite function, run in reverse:

∫f(g(x)) g′(x) dx=∫f(u) du,where u=g(x), du=g′(x) dx\int f\big(g(x)\big)\,g'(x)\,dx = \int f(u)\,du, \qquad \text{where } u=g(x),\ du=g'(x)\,dx

After integrating with respect to uu using the standard formulas, the last step is always to substitute u=g(x)u=g(x) back, so the final answer is written in terms of the original variable xx — an integral is never left in terms of uu.

Worked pattern: ∫3x2(x3+4)4 dx\displaystyle\int 3x^{2}(x^{3}+4)^{4}\,dx

Let u=x3+4u = x^{3}+4, so du=3x2 dxdu = 3x^{2}\,dx — and 3x2 dx3x^{2}\,dx is EXACTLY what multiplies (x3+4)4(x^{3}+4)^{4} in the integrand, so the substitution is a clean fit:

∫3x2(x3+4)4 dx=∫u4 du=u55+C=(x3+4)55+C\int 3x^{2}(x^{3}+4)^{4}\,dx = \int u^{4}\,du = \frac{u^{5}}{5}+C = \frac{(x^{3}+4)^{5}}{5}+C

Check by differentiating back: using the chain rule, ddx[(x3+4)55]=5(x3+4)4⋅3x25=3x2(x3+4)4\dfrac{d}{dx}\left[\dfrac{(x^{3}+4)^{5}}{5}\right] = \dfrac{5(x^{3}+4)^{4}\cdot 3x^{2}}{5} = 3x^{2}(x^{3}+4)^{4}, exactly the original integrand.

Spotting the right substitution …

Definition 1Integration by Substitution

∫f(g(x)) g′(x) dx=∫f(u) du\int f(g(x))\,g'(x)\,dx = \int f(u)\,du with u=g(x)u=g(x) — converts a composite integrand into a standard-formula integral in the new variable uu, then the result …