Английская Википедия:Integration by substitution

In calculus, integration by substitution, also known as u-substitution, reverse chain rule or change of variables,^[1] is a method for evaluating integrals and antiderivatives. It is the counterpart to the chain rule for differentiation, and can loosely be thought of as using the chain rule "backwards."

Substitution for a single variable

Introduction (indefinite integrals)

Before stating the result rigorously, consider a simple case using indefinite integrals.

Compute <math display="inline">\int(2x^3+1)^7(x^2)\,dx.</math>^[2]

Set <math>u=2x^3+1.</math> This means <math display="inline">\frac{du}{dx}=6x^2,</math> or as a differential form, <math display="inline">du=6x^2\,dx.</math> Now:

<math display="block">\begin{aligned}

   \int(2x^3 +1)^7(x^2)\,dx &= \frac{1}{6}\int\underbrace{(2x^3+1)^{7}}_{u^{7}}\underbrace{(6x^2)\,dx}_{du} \\
                            &= \frac{1}{6}\int u^{7}\,du \\
                            &= \frac{1}{6}\left(\frac{1}{8}u^{8}\right)+C \\
                            &= \frac{1}{48}(2x^3+1)^{8}+C,

\end{aligned}</math>

where <math>C</math> is an arbitrary constant of integration.

This procedure is frequently used, but not all integrals are of a form that permits its use. In any event, the result should be verified by differentiating and comparing to the original integrand.

<math display="block">\frac{d}{dx}\left[\frac{1}{48}(2x^3+1)^{8}+C\right]=\frac{1}{6}(2x^3+1)^{7}(6x^2) = (2x^3+1)^7(x^2).</math>

For definite integrals, the limits of integration must also be adjusted, but the procedure is mostly the same.

Statement for definite integrals

Let <math>g:[a,b]\rightarrow I</math> be a differentiable function with a continuous derivative, where <math>I \subset \mathbb{R}</math> is an interval. Suppose that <math>f:I\rightarrow\mathbb{R}</math> is a continuous function. Then:^[3] <math display="block">\int_a^b f(g(x))\cdot g'(x)\, dx = \int_{g(a)}^{g(b)} f(u)\ du. </math>

In Leibniz notation, the substitution <math>u=g(x)</math> yields: <math display="block">\frac{du}{dx} = g'(x).</math> Working heuristically with infinitesimals yields the equation <math display="block">du = g'(x)\,dx,</math> which suggests the substitution formula above. (This equation may be put on a rigorous foundation by interpreting it as a statement about differential forms.) One may view the method of integration by substitution as a partial justification of Leibniz's notation for integrals and derivatives.

The formula is used to transform one integral into another integral that is easier to compute. Thus, the formula can be read from left to right or from right to left in order to simplify a given integral. When used in the former manner, it is sometimes known as u-substitution or w-substitution in which a new variable is defined to be a function of the original variable found inside the composite function multiplied by the derivative of the inner function. The latter manner is commonly used in trigonometric substitution, replacing the original variable with a trigonometric function of a new variable and the original differential with the differential of the trigonometric function.

Proof

Integration by substitution can be derived from the fundamental theorem of calculus as follows. Let <math>f</math> and <math>g</math> be two functions satisfying the above hypothesis that <math>f</math> is continuous on <math>I</math> and <math>g'</math> is integrable on the closed interval <math>[a,b]</math>. Then the function <math>f(g(x))\cdot g'(x)</math> is also integrable on <math>[a,b]</math>. Hence the integrals

and

in fact exist, and it remains to show that they are equal.

Since <math>f</math> is continuous, it has an antiderivative <math>F</math>. The composite function <math>F \circ g</math> is then defined. Since <math>g</math> is differentiable, combining the chain rule and the definition of an antiderivative gives:

Applying the fundamental theorem of calculus twice gives:

<math display="block"> \begin{align} \int_a^b f(g(x)) \cdot g'(x)\ dx &= \int_a^b (F \circ g)'(x)\ dx \\ &= (F \circ g)(b) - (F \circ g)(a) \\ &= F(g(b)) - F(g(a)) \\ &= \int_{g(a)}^{g(b)} f(u)\ du, \end{align} </math>

which is the substitution rule.

Examples: Definite integrals

Example 1

Consider the integral: <math display="block">\int_0^2 x \cos(x^2+1)\ dx.</math>

Make the substitution <math display="inline">u = x^{2} + 1</math> to obtain <math>du = 2x\ dx,</math> meaning <math display="inline">x\ dx = \frac{1}{2}\ du.</math> Therefore:

<math display="block">\begin{align} \int_{x=0}^{x=2} x \cos(x^2+1) \ dx &= \frac{1}{2} \int_{u=1}^{u=5}\cos(u)\ du \\[6pt] &= \frac{1}{2}(\sin(5)-\sin(1)). \end{align}</math>

Since the lower limit <math>x = 0</math> was replaced with <math>u = 1,</math> and the upper limit <math>x = 2</math> with <math>2^{2} + 1 = 5,</math> a transformation back into terms of <math>x</math> was unnecessary.

Alternatively, one may fully evaluate the indefinite integral (see below) first then apply the boundary conditions. This becomes especially handy when multiple substitutions are used.

Example 2

For the integral <math display="block">\int_0^1 \sqrt{1-x^2}\,dx,</math> a variation of the above procedure is needed. The substitution <math>x = \sin u</math> implying <math>dx = \cos u \,du</math> is useful because <math display="inline">\sqrt{1-\sin^2u} = \cos(u).</math> We thus have:

<math display="block">\begin{align} \int_0^1 \sqrt{1-x^2}\ dx &= \int_0^{\pi/2} \sqrt{1-\sin^2u} \cos(u)\ du \\[6pt] &= \int_0^{\pi/2} \cos^2u\ du \\[6pt] &= \left[\frac{u}{2} + \frac{\sin(2u)}{4}\right]_0^{\pi/2} \\[6pt] &= \frac{\pi}{4} + 0 \\ &= \frac{\pi}{4}. \end{align}</math>

The resulting integral can be computed using integration by parts or a double angle formula, <math display="inline">2\cos^{2} u = 1 + \cos (2u),</math> followed by one more substitution. One can also note that the function being integrated is the upper right quarter of a circle with a radius of one, and hence integrating the upper right quarter from zero to one is the geometric equivalent to the area of one quarter of the unit circle, or <math>\frac\pi 4.</math>

Examples: Antiderivatives

Substitution can be used to determine antiderivatives. One chooses a relation between <math>x</math> and <math>u,</math> determines the corresponding relation between <math>dx</math> and <math>du</math> by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between <math>x</math> and <math>u</math> is then undone.

Similar to example 1 above, the following antiderivative can be obtained with this method:

<math display="block">\begin{align} \int x \cos(x^2+1) \,dx &= \frac{1}{2} \int 2x \cos(x^2+1) \,dx \\[6pt] &= \frac{1}{2} \int\cos u\,du \\[6pt] &= \frac{1}{2}\sin u + C \\ &= \frac{1}{2}\sin(x^2+1) + C, \end{align}</math>

where <math>C</math> is an arbitrary constant of integration.

There were no integral boundaries to transform, but in the last step reverting the original substitution <math>u = x^{2} + 1</math> was necessary. When evaluating definite integrals by substitution, one may calculate the antiderivative fully first, then apply the boundary conditions. In that case, there is no need to transform the boundary terms.

Trigonometric functions

The tangent function can be integrated using substitution by expressing it in terms of the sine and cosine: <math>\tan x = \frac{\sin x}{\cos x}</math>.

Using the substitution <math>u = \cos x</math> gives <math>du = -\sin x\,dx</math> and <math display="block">\begin{align}

  \int \tan x \,dx &= \int \frac{\sin x}{\cos x} \,dx \\
   &= \int -\frac{du}{u} \\
   &= -\ln |u| + C \\
   &= -\ln |\cos x| + C \\
   &= \ln |\cos x|^{-1} + C \\
   &= \ln |\sec x| + C.
\end{align}</math>

The cotangent function can be integrated similarly by expressing it as <math>\cot x = \frac{\cos x}{\sin x}</math> and using the substitution <math>u = \sin{x}, du = \cos{x}\,dx</math>: <math display="block">\begin{align}

  \int \cot x \,dx &= \int \frac{\cos x}{\sin x} \,dx \\
   &= \int \frac{du}{u} \\
   &= \ln |u| + C \\
   &= \ln |\sin x| + C.
\end{align}</math>

Substitution for multiple variables

One may also use substitution when integrating functions of several variables.

Here, the substitution function Шаблон:Math needs to be injective and continuously differentiable, and the differentials transform as:

<math display="block">dv_1 \cdots dv_n = \left|\det(D\varphi)(u_1, \ldots, u_n)\right| \, du_1 \cdots du_n,</math>

where Шаблон:Math denotes the determinant of the Jacobian matrix of partial derivatives of Шаблон:Math at the point Шаблон:Math. This formula expresses the fact that the absolute value of the determinant of a matrix equals the volume of the parallelotope spanned by its columns or rows.

More precisely, the change of variables formula is stated in the next theorem:

Theorem. Let Шаблон:Math be an open set in Шаблон:Math and Шаблон:Math an injective differentiable function with continuous partial derivatives, the Jacobian of which is nonzero for every Шаблон:Mvar in Шаблон:Mvar. Then for any real-valued, compactly supported, continuous function Шаблон:Mvar, with support contained in Шаблон:Math:

<math display="block">\int_{\varphi(U)} f(\mathbf{v})\, d\mathbf{v} = \int_U f(\varphi(\mathbf{u})) \left|\det(D\varphi)(\mathbf{u})\right| \,d\mathbf{u}.</math>

The conditions on the theorem can be weakened in various ways. First, the requirement that Шаблон:Mvar be continuously differentiable can be replaced by the weaker assumption that Шаблон:Mvar be merely differentiable and have a continuous inverse.^[4] This is guaranteed to hold if Шаблон:Mvar is continuously differentiable by the inverse function theorem. Alternatively, the requirement that Шаблон:Math can be eliminated by applying Sard's theorem.^[5]

For Lebesgue measurable functions, the theorem can be stated in the following form:^[6]

Theorem. Let Шаблон:Mvar be a measurable subset of Шаблон:Math and Шаблон:Math an injective function, and suppose for every Шаблон:Mvar in Шаблон:Mvar there exists Шаблон:Math in Шаблон:Math such that Шаблон:Math as Шаблон:Math (here Шаблон:Mvar is little-o notation). Then Шаблон:Math is measurable, and for any real-valued function Шаблон:Mvar defined on Шаблон:Math: <math display="block">\int_{\varphi(U)} f(v)\, dv = \int_U f(\varphi(u)) \left|\det \varphi'(u)\right| \,du</math> in the sense that if either integral exists (including the possibility of being properly infinite), then so does the other one, and they have the same value.

Another very general version in measure theory is the following:^[7]

Theorem. Let Шаблон:Mvar be a locally compact Hausdorff space equipped with a finite Radon measure Шаблон:Mvar, and let Шаблон:Mvar be a σ-compact Hausdorff space with a σ-finite Radon measure Шаблон:Mvar. Let Шаблон:Math be an absolutely continuous function (where the latter means that Шаблон:Math whenever Шаблон:Math). Then there exists a real-valued Borel measurable function Шаблон:Mvar on Шаблон:Mvar such that for every Lebesgue integrable function Шаблон:Math, the function Шаблон:Math is Lebesgue integrable on Шаблон:Mvar, and <math display="block">\int_Y f(y)\,d\rho(y) = \int_X (f\circ \varphi)(x)\,w(x)\,d\mu(x).</math> Furthermore, it is possible to write <math display="block">w(x) = (g\circ \varphi)(x)</math> for some Borel measurable function Шаблон:Mvar on Шаблон:Mvar.

In geometric measure theory, integration by substitution is used with Lipschitz functions. A bi-Lipschitz function is a Lipschitz function Шаблон:Math which is injective and whose inverse function Шаблон:Math is also Lipschitz. By Rademacher's theorem, a bi-Lipschitz mapping is differentiable almost everywhere. In particular, the Jacobian determinant of a bi-Lipschitz mapping Шаблон:Math is well-defined almost everywhere. The following result then holds:

Theorem. Let Шаблон:Mvar be an open subset of Шаблон:Math and Шаблон:Math be a bi-Lipschitz mapping. Let Шаблон:Math be measurable. Then <math display="block">\int_U (f\circ \varphi)(x) |\det D\varphi(x)|\,dx = \int_{\varphi(U)} f(x)\,dx</math> in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.

The above theorem was first proposed by Euler when he developed the notion of double integrals in 1769. Although generalized to triple integrals by Lagrange in 1773, and used by Legendre, Laplace, and Gauss, and first generalized to Шаблон:Mvar variables by Mikhail Ostrogradsky in 1836, it resisted a fully rigorous formal proof for a surprisingly long time, and was first satisfactorily resolved 125 years later, by Élie Cartan in a series of papers beginning in the mid-1890s.^[8]^[9]

Application in probability

Substitution can be used to answer the following important question in probability: given a random variable Шаблон:Mvar with probability density Шаблон:Math and another random variable Шаблон:Mvar such that Шаблон:Mvar for injective (one-to-one) Шаблон:Mvar what is the probability density for Шаблон:Mvar?

It is easiest to answer this question by first answering a slightly different question: what is the probability that Шаблон:Mvar takes a value in some particular subset Шаблон:Mvar? Denote this probability Шаблон:Math Of course, if Шаблон:Mvar has probability density Шаблон:Math, then the answer is:

but this is not really useful because we do not know Шаблон:Math it is what we are trying to find. We can make progress by considering the problem in the variable Шаблон:Mvar. Шаблон:Mvar takes a value in Шаблон:Mvar whenever Шаблон:Mvar takes a value in <math display="inline">\phi^{-1}(S),</math> so:

Changing from variable Шаблон:Mvar to Шаблон:Mvar gives:

<math display="block">P(Y \in S) = \int_{\phi^{-1}(S)} p_X(x)\,dx = \int_S p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|\,dy.</math>

Combining this with our first equation gives:

<math display="block">\int_S p_Y(y)\,dy = \int_S p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|\,dy,</math>

so:

<math display="block">p_Y(y) = p_X(\phi^{-1}(y)) \left|\frac{d\phi^{-1}}{dy}\right|.</math>

In the case where Шаблон:Mvar and Шаблон:Mvar depend on several uncorrelated variables (i.e., <math display="inline">p_X=p_X(x_1, \ldots, x_n)</math> and <math>y=\phi(x)</math>), <math>p_Y</math>can be found by substitution in several variables discussed above. The result is:

<math display="block">p_Y(y) = p_X(\phi^{-1}(y)) \left|\det D\phi ^{-1}(y) \right|.</math>

Notes

Шаблон:Reflist

References

External links

Шаблон:Wikibooks Шаблон:Wikiversity

Шаблон:Calculus topics Шаблон:Integrals

es:Métodos de integración#Método de integración por sustitución

[1] Шаблон:Harvnb

[2] Шаблон:Harvnb

[3] Шаблон:Harvnb

[4] Шаблон:Harvnb

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[6] Шаблон:Harvnb

[7] Шаблон:Harvnb

[8] Шаблон:Harvnb

[9] Шаблон:Harvnb

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