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Applied Mathematics · Ch 4 — Integration and Its Application

Integration by Substitution

4.2

Integration by Substitution

Not every integral can be evaluated directly from the standard formulas — many integrands are composite functions in disguise, and integration by substitution is the technique for simplifying them by changing the variable of integration.

The idea: if we substitute g(x)=tg(x) = t, then differentiating gives g′(x) dx=dtg'(x)\,dx = dt. This lets us rewrite an integral of the form ∫f(g(x)) g′(x) dx\int f(g(x))\,g'(x)\,dx entirely in terms of tt:

∫f(g(x)) g′(x) dx=∫f(t) dt,where t=g(x)\displaystyle\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt, \quad \text{where } t = g(x)

Once the integral is evaluated in terms of tt, we substitute t=g(x)t = g(x) back to express the answer in terms of xx. For instance, in ∫2xx2+1 dx\int \dfrac{2x}{x^2+1}\,dx, putting t=x2+1t = x^2+1 gives dt=2x dxdt = 2x\,dx, turning the integral into the elementary ∫1t dt\int \dfrac{1}{t}\,dt.

Recognising which substitution to try is mostly a matter of pattern recognition. A few substitutions that work well in common situations:

  • For f(x)\sqrt{f(x)}, put f(x)=tf(x) = t (or t2t^2).
  • For log⁡x\log x, put log⁡x=t\log x = t, or equivalently x=etx = e^t.
  • For a composite f(g(x))f(g(x)), put g(x)=tg(x) = t.
  • For [f(x)]m/n[f(x)]^{m/n}, put f(x)=tnf(x) = t^n. …