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Integration by Substitution Type 2

Proposition

Let φ ⁣:[α,β]→R\varphi\colon[\alpha,\beta]\to\mathbb{R} be continuously differentiable, with φ(α)=a\varphi(\alpha)=a and φ(β)=b,\varphi(\beta)=b\text{,} and let ff be continuous on an interval containing the range of φ.\varphi\text{.} Then

∫abf(x) dx=x=φ(t)∫αβf(φ(t))φ′(t) dt.\int_a^b {\color{red}f(x)}\,{\color{blue}\mathrm{d}x} \overset{x=\varphi(t)}{=} \int_\alpha^\beta {\color{red}f(\varphi(t))}{\color{blue}\varphi'(t)\,\mathrm{d}t}.

Interpretation

Substitution of type 2 starts from the other side: instead of spotting an inner function, it writes the variable as x=φ(t)x=\varphi(t) and replaces dx\mathrm{d}x by φ′(t) dt\varphi'(t)\,\mathrm{d}t. It is useful when no derivative factor appears in the integrand, but an awkward expression such as 1−x2\sqrt{1-x^2} becomes simpler in terms of tt.

The chain rule differentiates an antiderivative composed with φ\varphi, producing f(φ(t))φ′(t)f(\varphi(t))\varphi'(t). This proves the Riemann integral identity above; the differential notation dx=φ′(t) dt\mathrm{d}x=\varphi'(t)\,\mathrm{d}t records that factor. The two types are therefore one identity read in opposite directions: type 1 goes from the side with g′(x)g'(x) to the simpler side, and type 2 deliberately creates the factor φ′(t)\varphi'(t).

In the Example, x=sin⁡tx=\sin t with tt from 00 to π2\frac{\pi}{2} gives 1−sin⁡2t=cos⁡t\sqrt{1-\sin^2t}=\cos t and dx=cos⁡t dt\mathrm{d}x=\cos t\,\mathrm{d}t, so ∫011−x2 dx=∫0π/2cos⁡2t dt=π4\int_0^1\sqrt{1-x^2}\,\mathrm{d}x=\int_0^{\pi/2}\cos^2t\,\mathrm{d}t=\frac{\pi}{4}, the area of a quarter of the unit disc. The limits only have to satisfy φ(α)=a\varphi(\alpha)=a and φ(β)=b\varphi(\beta)=b, and ff must be continuous on an interval containing all values of φ\varphi, not only those between aa and bb.

A common slip is to simplify a square root without checking the sign. In general 1−sin⁡2t=∣cos⁡t∣\sqrt{1-\sin^2t}=|\cos t|. The limits t=πt=\pi and t=π2t=\frac{\pi}{2} are also valid here, since sin⁡π=0\sin\pi=0 and sin⁡π2=1\sin\frac{\pi}{2}=1, but on [π2,π][\frac{\pi}{2},\pi] we have cos⁡t≤0\cos t\le0, and writing cos⁡t\cos t instead of ∣cos⁡t∣|\cos t| gives −π4-\frac{\pi}{4}.

Key Results

  • The function φ\varphi does not need to be invertible.
  • This formula is particularly useful for expressions such as a2−x2,\sqrt{a^2-x^2}\text{,} where a trigonometric parametrization can simplify the integrand.
  • This formula is often complementary to the t=g(x)t=g(x) approach, and usually only one of the two is effective for a given integral.

Examples

  • ∫011−x2 dx\displaystyle\int_0^1\sqrt{1-x^2}\,\mathrm{d}x can be evaluated by setting x=sin⁡t,x=\sin t\text{,} which changes the limits to 00 and π/2.\pi/2\text{.}
  • Trigonometric substitutions are common examples of parametrizations of this form.

Prerequisites

Sources

Aliases: inverse substitution Keywords: change of variables, substitution rule Symbols: alpha, beta, integral, phi, pi, real numbers, square root

Last updated: 20 October 2026 03:02