How to Integrate by Substitution .

Substitution for single variable

Relation to the fundamental theorem of calculus

Let

I \subseteq ℝ

be an interval and

ϕ : [a, b] \to I

be a continuously differentiable function. Suppose that

ƒ : I \to ℝ

is a continuous function. Then

$\int_{\phi(a)}^{\phi(b)} f(x)\,dx = \int_a^b f(\phi(t))\phi'(t)\, dt.$

Using Leibniz notation: the substitution

x = ϕ (t)

yields

dx / dt = ϕ' (t)

and thus, formally,

dx = ϕ' (t) dt

, which is the required substitution for

dx

. (One could view the method of integration by substitution as a major 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 used from left to right or from right to left in order to simplify a given integral. When used in the latter manner, it is sometimes known as u-substitution or w-substitution.
Integration by substitution can be derived from the fundamental theorem of calculus as follows. Let

ƒ

and

ϕ

be two functions satisfying the above hypothesis that

ƒ

is continuous on

I

and

ϕ'

is continuous on the closed interval

[a, b]

. Then the function

ƒ (ϕ (t)) ϕ' (t)

is also continuous on

[a, b]

. Hence the integrals

$\int_{\phi(a)}^{\phi(b)} f(x)\,dx$

and

$\int_a^b f(\phi(t))\phi'(t)\,dt$

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

ƒ

is continuous, it possesses an antiderivative

F

. The composite function

F \circ ϕ

is then defined. Since

F

and

ϕ

are differentiable, the chain rule gives

$(F \circ \phi)'(t) = F'(\phi(t))\phi'(t) = f(\phi(t))\phi'(t).$

Applying the fundamental theorem of calculus twice gives

$\begin{align} \int_a^b f(\phi(t))\phi'(t)\,dt & {} = (F \circ \phi)(b) - (F \circ \phi)(a) \\ & {} = F(\phi(b)) - F(\phi(a)) \\ & {} = \int_{\phi(a)}^{\phi(b)} f(x)\,dx, \end{align}$

which is the substitution rule.

Examples

Consider the integral

$\int_{0}^2 x \cos(x^2+1) \,dx$

If we make the substitution u = x² + 1, we obtain du = 2x dx and thence x dx = ½du
(1) Definite integral

$\begin{align} \int_{x=0}^{x=2} x \cos(x^2+1) \,dx & {} = \frac{1}{2} \int_{u=1}^{u=5}\cos(u)\,du \\ & {} = \frac{1}{2}(\sin(5)-\sin(1)). \end{align}$

It is important to note that since the lower limit x = 0 was replaced with u = 0² + 1 = 1, and the upper limit x = 2 replaced with u = 2² + 1 = 5, a transformation back into terms of x was unnecessary.
For the integral

$\int_0^1 \sqrt{1-x^2}\; dx$

the formula needs to be used from right to left: the substitution x = sin(u), dx = cos(u) du is useful, because $\sqrt{(1-\sin^2(u))} = \cos(u)$ :

$\int_0^1 \sqrt{1-x^2}\; dx = \int_0^\frac{\pi}{2} \sqrt{1-\sin^2(u)} \cos(u)\;du = \int_0^\frac{\pi}{2} \cos^2(u)\;du=\frac{\pi}{4}$

The resulting integral can be computed using integration by parts or a double angle formula 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 π/4.
(2) Antiderivatives
Substitution can be used to determine antiderivatives. One chooses a relation between x and u, determines the corresponding relation between dx and du by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between u and x is then undone.
Similar to our first example above, we can determine the following antiderivative with this method:

$\begin{align} & {} \quad \int x \cos(x^2+1) \,dx = \frac{1}{2} \int 2x \cos(x^2+1) \,dx \\ & {} = \frac{1}{2} \int\cos u\,du = \frac{1}{2}\sin u + C = \frac{1}{2}\sin(x^2+1) + C \end{align}$

where C is an arbitrary constant of integration.
Note that there were no integral boundaries to transform, but in the last step we had to revert the original substitution u = x² + 1.

Substitution for multiple variables

One may also use substitution when integrating functions of several variables. Here the substitution function (v₁,...,v_n) = φ(u₁, ..., u_n ) needs to be injective and continuously differentiable, and the differentials transform as

$dv_1 \cdots dv_n = |\det(\operatorname{D}\varphi)(u_1, \ldots, u_n)| \, du_1 \cdots du_n$

where det(Dφ)(u₁, ..., u_n ) denotes the determinant of the Jacobian matrix containing the partial derivatives of φ. 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 U be an open set in Rⁿ and φ : U → Rⁿ an injective differentiable function with continuous partial derivatives, the Jacobian of which is nonzero for every x in U. Then for any real-valued, compactly supported, continuous function f, with support contained in φ(U),

$\int_{\phi(U)} f(\mathbf{v})\, d \mathbf{v} = \int_U f(\phi(\mathbf{u})) \left|\det(\operatorname{D}\phi)(\mathbf{u})\right| \,d \mathbf{u}.$

The conditions on the theorem can be weakened in various ways. First, the requirement that φ be continuously differentiable can be replaced by the weaker assumption that φ be merely differentiable and have a continuous inverse (Rudin 1987, Theorem 7.26). This is guaranteed to hold if φ is continuously differentiable by the inverse function theorem. Alternatively, the requirement that Det(Dφ)≠0 can be eliminated by applying Sard's theorem (Spivak 1965).
For Lebesgue measurable functions, the theorem can be stated in the following form (Fremlin 2010, Theorem 263D):
Theorem. Let U be a measurable subset of Rⁿ and φ : U → Rⁿ an injective function, and suppose for every x in U there exists φ'(x) in R^n,n such that φ(y) = φ(x) + φ'(x) (y − x) + o(||y − x||) as y → x. Then φ(U) is measurable, and for any real-valued function f defined on φ(U),

$\int_{\phi(U)} f(v)\, dv \;=\; \int_U f(\phi(u)) \; \left|\det \phi'(u)\right| \,du$

in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.
Another very general version in measure theory is the following (Hewitt & Stromberg 1965, Theorem 20.3):
Theorem. Let X be a locally compact Hausdorff space equipped with a finite Radon measure μ, and let Y be a σ-compact Hausdorff space with a σ-finite Radon measure ρ. Let φ : X → Y be a continuous and absolutely continuous function (where the latter means that ρ(φ(E)) = 0 whenever μ(E) = 0). Then there exists a real-valued Borel measurable function w on X such that for every Lebesgue integrable function f : Y → R, the function (f $\circ$ φ)w is Lebesgue integrable on X, and

$\int_Y f(y)\,d\rho(y) = \int_X f\circ\phi(x)w(x)\,d\mu(x).$

Furthermore, it is possible to write

$w(x) = g\circ\phi(x)$

for some Borel measurable function g on Y.
In geometric measure theory, integration by substitution is used with Lipschitz functions. A bi-Lipschitz function is a Lipschitz function φ : U → Rⁿ which is one-to-one, and such that its inverse function φ⁻¹ : φ(U) → U 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 det D'φ is well-defined almost everywhere. The following result then holds:
Theorem. Let U be an open subset of Rⁿ and φ : U → Rⁿ be a bi-Lipschitz mapping. Let f : φ(U) → R be measurable. Then

$\int_U (f\circ \phi)|\det D\phi| = \int_{\phi(U)}f$

in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.

Application in probability

Substitution can be used to answer the following important question in probability: given a random variable $X$ with probability density $p_x$ and another random variable $Y$ related to $X$ by the equation $y=\phi(x)$ , what is the probability density for $Y$ ?
It is easiest to answer this question by first answering a slightly different question: what is the probability that $Y$ takes a value in some particular subset $S$ ? Denote this probability $P(Y \in S)$ . Of course, if $Y$ has probability density $p_y$ then the answer is

$P(Y \in S) = \int_S p_y(y)\,dy,$

but this isn't really useful because we don't know p_y; it's what we're trying to find in the first place. We can make progress by considering the problem in the variable $X$ . $Y$ takes a value in S whenever X takes a value in $\phi^{-1}(S)$ , so

$P(Y \in S) = \int_{\phi^{-1}(S)} p_x(x)\,dx.$

Changing from variable x to y gives

$P(Y \in S) = \int_{\phi^{-1}(S)} p_x(x)~dx = \int_S p_x(\phi^{-1}(y)) ~ \left|\frac{d\phi^{-1}}{dy}\right|~dy.$

Combining this with our first equation gives

$\int_S p_y(y)~dy = \int_S p_x(\phi^{-1}(y)) ~ \left|\frac{d\phi^{-1}}{dy}\right|~dy$

$p_y(y) = p_x(\phi^{-1}(y)) ~ \left|\frac{d\phi^{-1}}{dy}\right|.$

In the case where $X$ and $Y$ depend on several uncorrelated variables, i.e. $p_x=p_x(x_1\ldots x_n)$ , and $y=\phi(x)$ , $p_y$ can be found by substitution in several variables discussed above. The result is

$p_y(y) = p_x(\phi^{-1}(y)) ~ \left|\det \left[ D\phi ^{-1}(y) \right] \right|.$
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