Integration by substitution

Substitution for a single variable

Introduction (indefinite integrals)

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

Compute $\int (2x^3+1{)}^7(x^2)dx.$

Set ${\displaystyle u=2x^3+1.}$ This means $\frac{du}{dx}=6x^2,$ or as a differential form, $du=6x^{2}dx.$ Now consider $\int u^{7}du.$:

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where ${\displaystyle C}$ 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. $${\displaystyle \begin{array}{c}\frac{d}{dx}\left[\frac{1}{48}(2x^3+1{)}^8+C\right]=\frac{1}{6}(2x^3+1{)}^7(6x^2)=(2x^3+1{)}^7(x^2).\end{array}}$$ For definite integrals, the limits of integration must also be adjusted, but the procedure is mostly the same.

Statement for definite integrals

Let ${\displaystyle g:[a,b]\to I}$ be a differentiable function with a continuous derivative, where ${\displaystyle I\subset \mathbb{R}}$ is an interval. Suppose that ${\displaystyle f:I\to \mathbb{R}}$ is a continuous function. Then: $${\displaystyle {\int }_a^{b}f(g(x))\cdot g^{\prime }(x)dx={\int }_{g(a)}^{g(b)}f(u)du.}$$

In Leibniz notation, the substitution ${\displaystyle u=g(x)}$ yields: $${\displaystyle \frac{du}{dx}=g^{\prime }(x).}$$ Working heuristically with infinitesimals yields the equation $${\displaystyle du=g^{\prime }(x)dx,}$$ 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 ${\displaystyle f}$ and ${\displaystyle g}$ be two functions satisfying the above hypothesis that ${\displaystyle f}$ is continuous on ${\displaystyle I}$ and ${\displaystyle g^{\prime }}$ is integrable on the closed interval ${\displaystyle [a,b]}$. Then the function ${\displaystyle f(g(x))\cdot g^{\prime }(x)}$ is also integrable on ${\displaystyle [a,b]}$. Hence the integrals $${\displaystyle {\int }_a^{b}f(g(x))\cdot g^{\prime }(x)dx}$$ and $${\displaystyle {\int }_{g(a)}^{g(b)}f(u)du}$$ in fact exist, and it remains to show that they are equal.

Since ${\displaystyle f}$ is continuous, it has an antiderivative ${\displaystyle F}$. The composite function ${\displaystyle F\circ g}$ is then defined. Since ${\displaystyle g}$ is differentiable, combining the chain rule and the definition of an antiderivative gives: $${\displaystyle (F\circ g{)}^{\prime }(x)=F^{\prime }(g(x))\cdot g^{\prime }(x)=f(g(x))\cdot g^{\prime }(x).}$$

Applying the fundamental theorem of calculus twice gives: $${\displaystyle \begin{array}{cc}{\int }_a^{b}f(g(x))\cdot g^{\prime }(x)dx& ={\int }_a^b(F\circ g{)}^{\prime }(x)dx\\ & =(F\circ g)(b)-(F\circ g)(a)\\ & =F(g(b))-F(g(a))\\ & ={\int }_{g(a)}^{g(b)}{F}^{\prime }(u)du={\int }_{g(a)}^{g(b)}f(u)du,\end{array}}$$ which is the substitution rule.

Examples: Antiderivatives (indefinite integrals)

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

Example 1

Consider the integral: $${\displaystyle \int x\cos (x^2+1)dx.}$$ Make the substitution $u=x^2+1$ to obtain ${\displaystyle du=2xdx,}$ meaning $xdx=\frac{1}{2}du.$ Therefore: $${\displaystyle \begin{array}{cc}\int x\cos (x^2+1)dx& =\frac{1}{2}\int 2x\cos (x^2+1)dx\\ & =\frac{1}{2}\int \cos udu\\ & =\frac{1}{2}\sin u+C\\ & =\frac{1}{2}\sin (x^2+1)+C,\end{array}}$$ where ${\displaystyle C}$ is an arbitrary constant of integration.

Example 2: Antiderivatives of tangent and cotangent

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

Using the substitution ${\displaystyle u=\cos x}$ gives ${\displaystyle du=-\sin xdx}$ and $${\displaystyle \begin{array}{cc}\int \tan xdx& =\int \frac{\sin x}{\cos x}dx\\ & =\int -\frac{du}{u}\\ & =-\ln \left|u\right|+C\\ & =-\ln \left|\cos x\right|+C\\ & =\ln \left|\sec x\right|+C.\end{array}}$$

The cotangent function can be integrated similarly by expressing it as ${\displaystyle \cot x=\frac{\cos x}{\sin x}}$ and using the substitution ${\displaystyle u=\sin x,du=\cos xdx}$: $${\displaystyle \begin{array}{cc}\int \cot xdx& =\int \frac{\cos x}{\sin x}dx\\ & =\int \frac{du}{u}\\ & =\ln \left|u\right|+C\\ & =\ln \left|\sin x\right|+C.\end{array}}$$

Examples: Definite integrals

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

Example 1

Consider the integral: $${\displaystyle {\int }_0^2\frac{x}{\sqrt{x^2+1}}dx.}$$ Make the substitution $u=x^2+1$ to obtain ${\displaystyle du=2xdx,}$ meaning $xdx=\frac{1}{2}du.$ Therefore: $${\displaystyle \begin{array}{cc}{\int }_{x=0}^{x=2}\frac{x}{\sqrt{x^2+1}}dx& =\frac{1}{2}{\int }_{u=1}^{u=5}\frac{du}{\sqrt{u}}\\ & =\frac{1}{2}\left(2\sqrt{5}-2\sqrt{1}\right)\\ & =\sqrt{5}-1.\end{array}}$$ Since the lower limit ${\displaystyle x=0}$ was replaced with ${\displaystyle u=1,}$ and the upper limit ${\displaystyle x=2}$ with ${\displaystyle 2^2+1=5,}$ a transformation back into terms of ${\displaystyle x}$ was unnecessary.

Example 2: Trigonometric substitution

For the integral $${\displaystyle {\int }_0^1\sqrt{1-x^2}dx,}$$ a variation of the above procedure is needed. The substitution ${\displaystyle x=\sin u}$ implying ${\displaystyle dx=\cos udu}$ is useful because $\sqrt{1-{\sin }^{2}u}=\cos u.$ We thus have: $${\displaystyle \begin{array}{cc}{\int }_0^1\sqrt{1-x^2}dx& ={\int }_0^{\pi /2}\sqrt{1-{\sin }^{2}u}\cos udu\\ & ={\int }_0^{\pi /2}{\cos }^{2}udu\\ & ={\left[\frac{u}{2}+\frac{\sin (2u)}{4}\right]}_0^{\pi /2}\\ & =\frac{\pi }{4}+0\\ & =\frac{\pi }{4}.\end{array}}$$

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 ${\displaystyle \frac{\pi }{4}.}$