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Integral Calculus

Integration by Substitution

Mathematics I 285 words Free to read

Reversing the Chain Rule

Just as differentiation has the chain rule for compositions, integration has substitution — the reverse of the chain rule, and the most-used integration technique. It transforms a hard integral into an easy one by a clever change of variable.

The idea: if the integrand contains an "inner function" u=g(x)u = g(x) together with (a multiple of) its derivative g(x)g'(x), substitute: f(g(x))g(x)dx=f(u)du,u=g(x), du=g(x)dx.\int f(g(x))\,g'(x)\, dx = \int f(u)\, du, \qquad u = g(x),\ du = g'(x)\, dx. Replacing g(x)g(x) by uu and g(x)dxg'(x)\,dx by dudu collapses the composite into a simple integral in uu, which you evaluate and then convert back. For example 2xcos(x2)dx\int 2x\cos(x^2)\, dx: let u=x2u = x^2, so du=2xdxdu = 2x\, dx, giving cosudu=sinu+C=sin(x2)+C\int \cos u\, du = \sin u + C = \sin(x^2) + C.

The recipe:

  1. Choose uu — usually the inner function whose derivative also appears (up to a constant).
  2. Compute du=g(x)dxdu = g'(x)\, dx and rewrite the whole integral in terms of uu.
  3. Integrate in uu, then substitute back u=g(x)u = g(x) (for an indefinite integral).

For a definite integral, either substitute back and use the original limits, or — cleaner — change the limits to uu-values (x=au=g(a)x = a \Rightarrow u = g(a)) and evaluate directly in uu, never converting back.

Common pitfall: substituting for uu but forgetting to convert dxdx into dudu (i.e. neglecting the g(x)dx=dug'(x)\,dx = du factor), and, in a definite integral, using the old xx-limits with the new uu-variable. Every xx (including the differential) must become uu: if you change the variable, change the limits too — or substitute back before applying them. Leaving a stray dxdx or mismatched limits is the defining error of substitution.

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Integral Calculus