The Substitution and Composition Theorems

Substitution Theorem

Definite integrals

Let φ : [a, b] → I be a differentiable function with a continuous derivative, where I ⊆ R is an interval. Suppose that f : I → R is a continuous function. Then

$\int _{a}^{b}f(\varphi (x))\varphi '(x)\,dx=\int _{\varphi (a)}^{\varphi (b)}f(u)\,du.$

In Leibniz notation, the substitution u = φ(x) yields

${\frac {du}{dx}}=\varphi '(x).$

Working heuristically with infinitesimals yields the equation

$du=\varphi '(x)\,dx,$

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 f and φ be two functions satisfying the above hypothesis that f is continuous on I and φ′ is integrable on the closed interval [a,b]. Then the function f(φ(x))φ′(x) is also integrable on [a,b]. Hence the integrals

$\int _{a}^{b}f(\varphi (x))\varphi '(x)\,dx$

and

$\int _{\varphi (a)}^{\varphi (b)}f(u)\,du$

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

Since f is continuous, it has an antiderivative F. The composite function F ∘ φ is then defined. Since φ is differentiable, combining the chain rule and the definition of an antiderivative gives

$(F\circ \varphi )'(x)=F'(\varphi (x))\varphi '(x)=f(\varphi (x))\varphi '(x).$

Applying the fundamental theorem of calculus twice gives

${\begin{aligned}\int _{a}^{b}f(\varphi (x))\varphi '(x)\,dx&=\int _{a}^{b}(F\circ \varphi )'(x)\,dx\\&=(F\circ \varphi )(b)-(F\circ \varphi )(a)\\&=F(\varphi (b))-F(\varphi (a))\\&=\int _{\varphi (a)}^{\varphi (b)}f(u)\,du,\end{aligned}}$

which is the substitution rule.

Examples

Example 1

Consider the integral

$\int _{0}^{2}x\cos(x^{2}+1)dx.$

Make the substitution $u=x^{2}+1$ to obtain $du=2xdx$, meaning $\textstyle xdx={\frac {1}{2}}du$. Therefore,

${\begin{aligned}\int _{x=0}^{x=2}x\cos(x^{2}+1)dx&={\frac {1}{2}}\int _{u=1}^{u=5}\cos(u)\,du\\[6pt]&={\frac {1}{2}}(\sin(5)-\sin(1)).\end{aligned}}$

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

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

Example 2

For the integral

$\int _{0}^{1}{\sqrt {1-x^{2}}}\,dx,$

a variation of the above procedure is needed. The substitution $x=\sin u$ implying $dx=\cos u\,du$ is useful because ${\sqrt {1-\sin ^{2}u}}=\cos(u)$. We thus have

${\begin{aligned}\int _{0}^{1}{\sqrt {1-x^{2}}}\,dx&=\int _{0}^{\pi /2}{\sqrt {1-\sin ^{2}u}}\cos(u)\,du\\[6pt]&=\int _{0}^{\pi /2}\cos ^{2}u\,du\\[6pt]&=\left[{\frac {u}{2}}+{\frac {\sin(2u)}{4}}\right]{\Biggl |}_{0}^{\pi /2}\\[6pt]&={\frac {\pi }{4}}+0={\frac {\pi }{4}}.\end{aligned}}$

The resulting integral can be computed using integration by parts or a double angle formula, $2\cos ^{2}u=1+\cos(2u)$, 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 $\pi /4$.

Indefinite Integrals

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 $x$ and $u$ is then undone.

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

${\begin{aligned}\int x\cos(x^{2}+1)\,dx&={\frac {1}{2}}\int 2x\cos(x^{2}+1)\,dx\\[6pt]&={\frac {1}{2}}\int \cos u\,du\\[6pt]&={\frac {1}{2}}\sin u+C={\frac {1}{2}}\sin(x^{2}+1)+C,\end{aligned}}$

where $C$ is an arbitrary constant of integration.

There were no integral boundaries to transform, but in the last step reverting the original substitution $u=x^{2}+1$ 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.

The tangent function can be integrated using substitution by expressing it in terms of the sine and cosine:

$\int \tan x\,dx=\int {\frac {\sin x}{\cos x}}\,dx$

Using the substitution $u=\cos x$ gives $du=-\sin x\,dx$ and

${\begin{aligned}\int \tan x\,dx&=\int {\frac {\sin x}{\cos x}}\,dx\\&=\int -{\frac {du}{u}}\\&=-\ln |u|+C\\&=-\ln |\cos x|+C=\ln |\sec x|+C.\end{aligned}}$

Composition Theorem

Let $f$ be Riemann integrable on $[a,b]$, with $f([a,b])\subseteq [c,d]$, and let $\phi :[c,d]\to \mathbb {R}$ be continuous. Then the composition $\varphi \circ f$ is Riemann integrable on $[a,b]$ as well. That is, if both $\int _{a}^{b}f$ and $\int _{c}^{d}\varphi$ exist, and $f([a,b])\subseteq [c,d]$, then $\int _{a}^{b}\varphi \circ f$ exists.

Product Theorem

If $f$ and $g$ are both Riemann integrable on $[a,b]$, then the product $fg$ is Riemann integrable on $[a,b]$ as well; that is, if both $\int _{a}^{b}f$ and $\int _{a}^{b}g$ exist, then $\int _{a}^{b}fg$ exists.

Licensing

Content obtained and/or adapted from:

• Integration by substitution, Wikipedia under a CC BY-SA license