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[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

Integration by Parts.

One of the most powerful methods of integration comes from the formula for finding the derivative of the product of two functions. If [math]u[/math] and [math]v[/math] are both differentiable functions of [math]x[/math], we recall that [math]\frac{d}{dx} uv = u\frac{dv}{dx}+ v\frac{du}{dx}[/math]. As a consequence, of course, it is true that [math]u \frac{dv}{dx} = \frac{d}{dx}uv - v \frac{du}{dx}[/math]. From this last equation it follows that

[[math]] \int u \frac{dv}{dx} dx = \int \Bigl(\frac{d}{dx} uv - v \frac{du}{dx} \Bigr) dx. [[/math]]

Since the integral of a difference is equal to the difference of the integrals, the last equation is equivalent to

[[math]] \int u \frac{dv}{dx} dx = \int \frac{d}{dx} (uv) dx - \int v \frac{du}{dx} dx. [[/math]]

But [math]\int \frac{d}{dx}(uv) dx = uv + c[/math]. Hence, leaving the constant of integration as a by-product of the last integral, we obtain

Theorem

[[math]] \int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx. [[/math]]

This is the formula for the technique of integration by parts. To use it, the function to be integrated must be factored into a product of two functions, one to be labeled [math]u[/math] and the other [math]\frac{dv}{dx}[/math]. If we can recover [math]v[/math] from [math]\frac{dv}{dx}[/math], then we hope that [math]\int v \frac{du}{dx} dx[/math] is easier to find than [math]\int u \frac{dv}{dx} dx[/math]. As we shall see, the trick is to find the right factorization. Sometimes it is obvious, and sometimes it is not.

Example Integrate

[[math]] \mbox{(a)}\; \int \ln x dx, \;\;\; \mbox{(b)}\; \int x \sin x dx. [[/math]]

For (a), consider the factorization [math]\ln x \cdot 1[/math], and let [math]u = \ln x[/math] and [math]\frac{dv}{dx} = 1[/math]. Then [math]\frac{du}{dx} = \frac{1}{x}[/math] and [math]v = x[/math] (we normally do not concern ourselves with a constant of integration at this point---we look for some [math]v[/math], not the most general [math]v[/math]). Integrating by parts, we have

[[math]] \begin{array}{ccccl} \int \ln x \cdot 1 \cdot dx &=& (\ln x)(x) &-& \int x \cdot \frac{1}{x} \cdot dx,\\ \updownarrow \;\;\; \updownarrow & &\updownarrow \;\;\;\;\; \updownarrow & & \;\;\;\updownarrow \;\;\;\updownarrow\\ u \;\;\; \frac{dv}{dx} & & u\;\;\;\;\; v & & \;\;\; v \;\;\; \frac{du}{dx} \end{array} [[/math]]

or

[[math]] \begin{eqnarray*} \int \ln x dx &=& x \ln x - \int dx\\ &=& x \ln x - x + c. \end{eqnarray*} [[/math]]

If this is the correct answer, its derivative will be the integrand. Checking, we get

[[math]] \begin{eqnarray*} \frac{d}{dx} (x \ln x - x + c) &=& 1 \cdot \ln x+ x \cdot \frac{1}{x} - 1 + 0 \\ &=& \ln x + 1 - 1 = \ln x. \end{eqnarray*} [[/math]]

For (b) we have choices. We can try letting [math]u = x[/math] and [math]\frac{dv}{dx} = \sin x[/math], or we can try [math]u = \sin x[/math] and [math]\frac{dv}{dx} = x[/math], or even [math]u = x \sin x[/math] and [math]\frac{dv}{dx} = 1[/math]. Trial and error shows that the first suggestion works and the others do not. If [math]u = x[/math] and [math]\frac{dv}{dx} = \sin x[/math], then we have [math]\frac{du}{dx} = 1[/math] and [math]v = -\cos x[/math]. The integration-by-parts formula implies that

[[math]] \int x \sin x dx = (x)(-\cos x) - \int (-\cos x)(1) dx, [[/math]]

or

[[math]] \begin{eqnarray*} \int x \sin x dx &=& - x \cos x + \int \cos x dx \\ &=& -x \cos x + \sin x + c. \end{eqnarray*} [[/math]]

Although it is not true that every function can be written as a product of functions which will lead to a simplification of integrals---this technique is of little use if [math]\int v \frac{du}{dx} dx[/math] is not easier to integrate than [math]\int u \frac{dv}{dx} dx[/math]---it is true that many can and that they can be integrated as were the two integrals in Example 1. The problem is to find the best function to call [math]u[/math]. Sometimes the technique of integration by parts must be used more than once in a problem.

Example

Integrate [math]\int (3x^2 - 4x + 7) e^{2x} dx[/math]. If we let [math]u = 3x^2 - 4x + 7[/math] and [math]\frac{dv}{dx} = e^{2x}[/math], then [math]\frac{du}{dx} = 6x - 4[/math] and [math]v = \frac{1}{2}e^{2x}[/math].

[[math]] \begin{eqnarray*} \int (3x^2 - 4x + 7)e^{2x} dx &=& (3x^2 - 4x + 7)(\frac{1}{2}e^{2x}) - \int (\frac{1}{2} e^{2x})(6x - 4) dx\\ &=& \frac{1}{2}(3x^2 - 4x + 7)e^{2x} - \int (3x - 2)e^{2x}dx. \end{eqnarray*} [[/math]]

This last integral again involves a product of a polynomial and [math]e^{2x}[/math], so we apply the technique again. Let [math]u_1 = 3x - 2[/math] and [math]\frac{dv_1}{dx} = e^{2x}[/math]. Then [math]\frac{du_1}{dx} = 3[/math] and [math]v_1 = \frac{1}{2}e^{2x}[/math]. Thus

[[math]] \begin{eqnarray*} \int (3x - 2)e^{2x} dx &=& (3x - 2)(\frac{1}{2}e^{2x}) - \int (\frac{1}{2}e^{2x})(3) dx \\ &=& \frac{1}{2} (3x - 2)e^{2x} - \frac{3}{2} \int e^{2x}dx \\ &=& \frac{1}{2} (3x - 2)e^{2x} - \frac{3}{4}e^{2x} + c_1. \end{eqnarray*} [[/math]]

Substituting, we have

[[math]] \begin{eqnarray*} \int (3x^2 - 4x + 7)e^{2x}dx &=& \frac{1}{2}(3x^2 - 4x + 7)e^{2x} - [\frac{1}{2}(3x - 2)e^{2x} - \frac{3}{4} e^{2x} + c_1] \\ &=& [(\frac{3}{2}x^2 - 2x + \frac{7}{2} - \frac{3}{2}x + 1 + \frac{3}{4})e^{2x} ] + c\\ &=& (\frac{3}{2} x^2 - \frac{7}{2} x + \frac{21}{4})e^{2x} + c. \end{eqnarray*} [[/math]]

Generally, faced with the product of a polynomial and a trigonometric or exponential function, it is best to let the polynomial be u and the transcendental function be [math]\frac{dv}{dx}[/math]. In this way, the degree of the polynomial is reduced by one each time the product is integrated by parts. However, faced with a product of transcendental functions, the choice may not be quite so obvious. Example

Integrate [math]\int e^{2x} \cos 3xdx[/math]. Let us select [math]e^{2x}[/math] as [math]u[/math] and [math]\cos 3x[/math] as [math]\frac{dv}{dx}[/math]. Then [math]\frac{du}{dx} = 2e^{2x}[/math] and [math]v = \frac{1}{3} \sin 3x[/math]. Thus

[[math]] \begin{eqnarray*} \int e^{2x} \cos 3x dx = (e^{2x})(\frac{1}{3} \sin 3x) - \int (\frac{1}{3} \sin 3x)(2e^{2x}) dx \\ = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \int e^{2x} \sin 3x dx. \end{eqnarray*} [[/math]]

At this point a second integration by parts is necessary. The reader should check that the choice of [math]\sin 3x[/math] for [math]u[/math] will lead back to an identity. We choose [math]e^{2x}[/math] for [math]u_1[/math] and [math]\sin 3x[/math] for [math]\frac{dv_1}{dx}[/math]. Then [math]\frac{du_1}{dx} = 2e^{2x}[/math] and [math]v_1 = - \frac{1}{3} \cos 3x[/math]. Hence

[[math]] \begin{eqnarray*} \int e^{2x} \sin 3x dx &=& (e^{2x})(-\frac{1}{3} \cos 3x) - \int (- \frac{1}{3} \cos 3x)(2e^{2x}) dx \\ &=& - \frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} \int e^{2x} \cos 3x dx. \end{eqnarray*} [[/math]]

Substituting, we have

[[math]] \int e^{2x} \cos 3x dx = \frac{1}{3} e^{2x} \sin 3x - \frac{2}{3} \Bigl( -\frac{1}{3} e^{2x} \cos 3x + \frac{2}{3} \int e^{2x} \cos3x dx \Bigr) [[/math]]

or

[[math]] \int e^{2x} \cos 3x dx = \frac{1}{3} e^{2x} \sin 3x + \frac{2}{9} e^{2x} \cos 3x - \frac{4}{9} \int e^{2x} \cos 3x dx. [[/math]]

This does not look much simpler, since we have found an integral in terms of itself. But, if we add [math]\frac{4}{9} \int e^{2x} \cos 3x dx[/math] to each side of the equation (supplying the constant of integration at the same time), we have

[[math]] \frac{13}{9} \int e^{2x} \cos 3x dx = \frac{e^{2x}}{9} (3 \sin 3x + 2 \cos 3x) + c_1. [[/math]]

Finally, therefore,

[[math]] \int e^{2x} \cos 3x dx = \frac{e^{2x}}{13} (3 \sin 3x + 2 \cos 3x) + c. [[/math]]

If we had chosen [math]\cos 3x[/math] as [math]u[/math] in the first integration by parts and [math]\sin 3x[/math] as [math]u_1[/math] in the second integration by parts, the integral could be found in the same way that we just found it.

The differential of a function was introduced in Section 6 of Chapter 2 and was shown to satisfy the equation [math]dF(x) = F'(x) dx[/math]. As a result, the symbol [math]dx[/math] which occurs in an indefinite integral [math]\int f(x) dx[/math] may be legitimately regarded as a differential since, if [math]F'(x) = f(x)[/math], then [math]dF(x) = f(x)dx[/math]. Moreover, if [math]u[/math] is a differentiable function of [math]x[/math], then

[[math]] dF(u) = F'(u) du = f (u) du, [[/math]]

so we write

[[math]] F(u) + c = \int f (u) du. [[/math]]

[see (6.7) on page 216]. Using differentials, we obtain a very compact form for the formula for integration by parts. Since [math]du = \frac{du}{dx}dx[/math] and [math]dv = \frac{dv}{dx} dx[/math], substitution in (1.1) yields \medskip Theorem (7.1.1')

[[math]] \int udv = uv - \int v du. [[/math]]

\medskip We have less to write when we use this form of the formula, but the result is the same.

Example

Integrate [math]\int x \ln x dx[/math]. lf we use (1.1'), we set [math]u= \ln x[/math] and [math]dv = x dx[/math]. Then [math]du = \frac{1}{x}dx[/math] and [math]v = \frac{1}{2} x^2[/math]. Hence

[[math]] \begin{eqnarray*} \int x \ln x dx &=& (\ln x)(\frac{1}{2}x^2) - \int (\frac{1}{2}x^2 ) \Bigl( \frac{1}{x} dx \Bigr) \\ &=& \frac{1}{2} x^{2} \ln x - \frac{1}{2} \int x dx\\ &=& \frac{1}{2} x^{2} \ln x - \frac{1}{4} x^2 + c. \end{eqnarray*} [[/math]]

In the remainder of this chapter we shall take full advantage of the streamlined notation offered by the differential and shall use it freely when making substitutions in indefinite integrals. Example

Find [math]\int x \ln(x + 1) dx[/math]. This example illustrates the fact that a judicious choice of a constant of integration ean sometimes simplify the computation. Set [math]u = \ln(x + 1)[/math] and [math]dv = xdx[/math]. Then [math]du = \frac{1}{x + 1} dx[/math] and we may take [math]v = \frac{x^2}{2}[/math]. If we do this, we have

[[math]] \begin{eqnarray*} \int x \ln(x + 1) dx &=& \ln(x + 1 ) \frac{x^2}{2} - \int \frac{ x^2}{2} \frac{1}{x + 1} dx\\ &=& \frac{1}{2} x^{2} \ln(x + 1) - \frac{1}{2} \int \frac{x^2}{x + 1} dx. \end{eqnarray*} [[/math]]

Dividing, we find that [math]\frac{x^2}{x + 1} = x - 1 + \frac{1}{x + 1}[/math], and hence

[[math]] \begin{eqnarray*} \int x \ln(x + 1) dx &=& \frac{1}{2} x^{2} \ln(x + 1) - \frac{1}{2} \int \Bigl(x - 1 + \frac{1}{x + 1} \Bigr) dx\\ &=& \frac{1}{2} x ^{2} \ln(x + 1) - \frac{1}{2} (\frac{1}{2} x^2 - x + \ln |x + 1|) + c\\ &=& \frac{1}{2}(x^2 - 1) \ln(x + 1) - \frac{1}{4} x^2 + \frac{1}{2} x + c. \end{eqnarray*} [[/math]]

[Since [math]\ln(x + 1)[/math] makes sense only if [math]x + 1 \gt 0[/math], we have replaced [math]\ln |x + 1|[/math] by [math]\ln(x + 1).][/math] The same result is reached more quickly if we take [math]v = \frac{x^2}{2} + c = \frac{x^2 + k}{2}[/math]. Then integration by parts gives

[[math]] \int x \ln(x+ 1)dx = \ln(x+ 1)\frac{x^2 + k}{2} - \int \frac{x^2 + k}{2} \frac{1}{x+ 1} dx. [[/math]]

Since we have a free choice for [math]k[/math], we shall choose [math]k = -1[/math] and then [math]\frac{x^2 + k}{2} \frac{1}{ x + 1} = \frac{x^2 - 1}{2} \frac{1}{x + 1} = \frac{x - 1}{2}[/math]. The problem becomes

[[math]] \begin{eqnarray*} \int x \ln(x + 1)dx &=& \ln(x + 1) \frac{x^2 - 1}{2} - \int \frac{ x - 1}{2} dx\\ &=& \frac{1}{2} (x^2 - 1) \ln(x + 1) - \frac{1}{4} x^2 + \frac{1}{2} x + c. \end{eqnarray*} [[/math]]

The method of integration by parts can frequently be used to obtain a recursion formula for one integral in terms of a simpler one. As an example, we derive the useful identity

Theorem

For every integer [math]n \geq 2[/math],

[[math]] \int \cos^{n} x dx = \frac{\cos^{n - 1}x \sin x}{n} + \frac{n-1}{n} \int \cos ^{n-2} x dx. [[/math]]

Show Proof

We write [math]\cos^{n} x[/math] as the product [math]\cos^{n - 1} x \cos x[/math], and let [math]u = \cos^{n - 1} x[/math] and [math]dv = \cos x dx[/math]. Then [math]v = \sin x[/math] and [math]du = (n - 1) \cos^{n-2} x (-\sin x dx) = - (n - 1) \cos^{n-2} x \sin x dx[/math]. Hence

[[math]] \begin{eqnarray*} \int \cos^{n} x dx &=& \cos^{n-1} x \sin x - \int \sin x[ -(n - 1) \cos^{n-2}x \sin x dx]\\ &=& \cos^{n-1} x \sin x + (n - 1) \int \cos^{n-2}x \sin^{2}x dx. \end{eqnarray*} [[/math]]

Replacing [math]\sin^{2}x[/math] by [math]1 - \cos^{2}x[/math], the equation becomes

[[math]] \begin{eqnarray*} \int \cos^{n}x dx &=& \cos^{n-1} x \sin x + (n -1)\int \cos^{n-2} x (1 - \cos^{2}x) dx \\ &=& \cos^{n - 1} x \sin x + (n -1) \int \cos^{n - 2} x dx - (n -1) \int \cos^{n} x dx . \end{eqnarray*} [[/math]]

Adding [math](n-1) \int \cos^{n} x dx[/math] to both sides of the equation, we obtain

[[math]] n \int \cos^{n} x dx = \cos^{n - 1} x \sin x + (n - 1) \int \cos^{n -2} x dx, [[/math]]

whence follows at once upon division by [math]n[/math].

■

Thus by repeated applications of the recursion formula (1.2), the integral [math]\int \cos^{n} x dx[/math] can be reduced eventually to a polynomial in [math]\sin x[/math] and [math]\cos x[/math]. If [math]n[/math] is even, the final integral is

[[math]] \int \cos^{0} x dx = \int dx = x + c, [[/math]]

and, if [math]n[/math] is odd, it is

[[math]] \int \cos x dx = \sin x + c. [[/math]]

Example

Use the recursion formula (1.2) to find [math]\int cos^{5}2x dx[/math]. We first write [math]\int \cos^{5}2x dx = \frac{1}{2} \int \cos^{5} 2x d(2x)[/math] and then

[[math]] \int \cos^{5}2x d(2x) = \frac{\cos^{4} 2x \sin 2x}{5} + \frac{4}{5} \int \cos^{3}2x d(2x). [[/math]]

A second application of the formula yields

[[math]] \int \cos^{3} 2x d(2x) = \frac{\cos^{2} 2x \sin 2x}{3} + \frac{2}{3} \int \cos 2x d(2x), [[/math]]

and, of course,

[[math]] \int \cos 2x d(2x) = \sin 2x + c_1. [[/math]]

Combining, we have

[[math]] \begin{eqnarray*} \int \cos^{5} 2x dx &=& \frac{1}{2} \Bigl[ \frac{\cos^{4} 2x\sin 2x}{5} + \frac{4}{5} \Big\{ \frac{ (\cos^{2} 2x \sin 2x}{3} + \frac{2}{3} ( \sin 2x + c_1) \Big\} \Bigr] \\ &=& \frac{1}{10} \cos^{4} 2x \sin 2x + \frac{2}{15} \cos^{2} 2x \sin 2x + \frac{4}{15} \sin 2x + c. \end{eqnarray*} [[/math]]

\end{exercise}

General references

Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.