# Substitution rule

 Topics in calculus Differentiation Integration

In calculus, the substitution rule is a tool for finding antiderivatives and integrals. Using the fundamental theorem of calculus often requires finding an antiderivative. For this and other reasons, the substitution rule is a relatively important tool for mathematicians. It is the counterpart to the chain rule of differentiation.

Suppose f(x) is an integrable function, and φ(t)  is a continuously differentiable function which is defined on the interval [a, b] and whose image is contained in the domain of f. Then

[itex]

\int_{\phi(a)}^{\phi(b)} f(x)\,dx = \int_{a}^{b} f(\phi(t)) \phi'(t)\,dt [itex]

The formula is best remembered using Leibniz' formalism: the substitution x = φ(t)  yields dx/dt = φ'(t)  and thus formally dx = φ'(t)dt , which is precisely the required substitution for dx. (In fact, one may view the substitution rule as a major justification of the Leibniz formalism for integrals and derivatives.)

The formula is used to transform an integral into another one which (hopefully) is easier to determine. 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.

## Examples

Consider the integral

[itex]

\int_{0}^2 t \cos(t^2+1) \,dt [itex] By using the substitution x = t2 + 1, we obtain dx = 2t dt and

[itex]

\int_{0}^2 t \cos(t^2+1) \,dt = \frac{1}{2} \int_{0}^2 \cos(t^2+1) 2t \,dt = \frac{1}{2} \int_{1}^{5}\cos(x)\,dx = \frac{1}{2}(\sin(5)-\sin(1)). [itex] Here we used the substitution rule "from right to left". Note how the lower limit t = 0 was transformed into x = 02 + 1 = 1 and the upper limit t = 2 into x = 22 + 1 = 5.

For the integral

[itex]

\int_0^1 \sqrt{1-x^2}\; dx [itex] the formula needs to be used from left to right: the substitution x = sin(t), dx = cos(t) dt is useful, because √(1-sin2(t)) = cos(t):

[itex]

\int_0^1 \sqrt{1-x^2}\; dx = \int_0^\frac{\pi}{2} \sqrt{1-\sin^2(t)} \cos(t)\;dt = \int_0^\frac{\pi}{2} \cos^2(t)\;dt [itex] The resulting integral can be computed using integration by parts.

## Antiderivatives

The substitution rule can be used to determine antiderivatives. One chooses a relation between x and t, determines the corresponding relation between dx and dt by differentiating, and performs the substitutions. An antiderivative for the substituted function can hopefully be determined; the original substitution between x and t is then undone.

Similar to our first example above, we can determine the following antiderivative with this method:

[itex]\int t \cos(t^2+1) \,dt = \frac{1}{2} \int \cos(t^2+1) 2t \,dt [itex]
[itex]\qquad = \frac{1}{2} \int\cos(x)\,dx = \frac{1}{2}\sin(x) + C = \frac{1}{2}\sin(t^2+1) + C.

[itex] Note that there were no integral boundaries to transform, but in the last step we had to revert the original substitution x = t2 + 1.

## Substitution rule for multiple variables

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

[itex]dv_1\cdots dv_n = |\det(\operatorname{D}\phi)(u_1, \ldots, u_n)| \, du_1\cdots du_n[itex]

where det(Dφ)(u1,...,un ) 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 given vectors equals the volume of the spanned parallelepiped.

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

Theorem. Let U, V  be open sets in Rn and φ : UV  a bijective differentiable function with continuous partial derivatives. Then for any integrable real-valued function f  on V

[itex] \int_V f(\mathbf{v}) d \mathbf{v} = \int_U f(\phi(\mathbf{u})) \left|\det(\operatorname{D}\phi)(\mathbf{u})\right| d \mathbf{u}.[itex]de:Integration durch Substitution
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