When the integrand contains a composite function whose “inner derivative” appears (up to constants), it is convenient to change variable. The method is organised into five steps.

In brief — The 5 steps

  1. Choose the substitution t=g(x)t = g(x).
  2. Compute the differential dt=g(x)dxdt = g'(x)\,dx.
  3. Obtain dxdx: dx=dtg(x)dx = \dfrac{dt}{g'(x)} and substitute.
  4. Compute the antiderivative in tt.
  5. Inverse substitution t=g(x)t = g(x).

Example — 0πxsin(x2)dx\int_0^\pi x\sin(x^2)\,dx

Set t=x2t = x^2, dt=2xdxdt = 2x\,dx, dx=dt/(2x)dx = dt/(2x): 0πxsin(t)dt2x=12sin(t)dt=12[cos(t)]=12[cos(x2)]0π=12[1cos(π2)].\int_0^\pi x\sin(t)\cdot\frac{dt}{2x} = \frac{1}{2}\int\sin(t)\,dt = \frac{1}{2}\bigl[-\cos(t)\bigr] = \frac{1}{2}\bigl[-\cos(x^2)\bigr]_0^\pi = \frac{1}{2}\bigl[1-\cos(\pi^2)\bigr].

Example — 01x1+x2dx\int_0^1\dfrac{x}{1+x^2}\,dx

Set t=1+x2t=1+x^2, dt=2xdxdt=2x\,dx: 12dtt=12lnt=12[ln(1+x2)]01=12ln2.\frac{1}{2}\int\frac{dt}{t} = \frac{1}{2}\ln|t| = \frac{1}{2}\bigl[\ln(1+x^2)\bigr]_0^1 = \frac{1}{2}\ln 2.

The sign that the substitution works is that, after step 3, the variable xx disappears completely and there remains an integral in tt alone.

Topics: Integrale
Concepts: Differenziale · Integrale per sostituzione
Methods: Integrale sostituzione
Skills: Calcolare · Integrare