OBJECTIVES

When you finish this section, you should be able to:

1. Integrate by parts (p. 472)
2. Derive a formula using integration by parts (p. 476)

IN WORDS

The integral of the product \(u\) \(dv\) equals \(u\) times \(v\) minus the integral of the product \(v du\).

Integrating by Parts

Find \(\int x e^x~dx\).

Solution Choose \(u\) and \(dv\) so that \[ \int u dv = \int x e^x~dx \]

Suppose we choose \[ u = x\qquad \text{and} \quad dv = e^x~dx \]

Then \[ du = dx\qquad \text{and} \qquad v = \int dv = \int e^x~dx=e^x \]

Notice that we did not add a constant. Only a particular antiderivative of \(dv\) is required at this stage; we will add the constant of integration at the end. Using the integration by parts formula, we have \[ \int \underset{\color{#0066A7}{\hbox{\(u\)}}}{\underbrace{x}} \underset{\color{#0066A7}{dv}}{\underbrace{e^{x} dx}} = \underset{\color{#0066A7}{uv}}{\underbrace{x e^{x}}}-\int \underset{\color{#0066A7}{v}}{\underbrace{e^{x}}}\underset{\color{#0066A7}{\hbox{\(du\)}}} {\underbrace{\hbox{\(dx\)}}}=x e^{x}-e^{x}+C=e^{x}(x-1)+C \]

IN WORDS

We choose \(dv\) so that it can be easily integrated and choose \(u\) so that \(\int v du\) is simpler than \(\int u dv\).

NOW WORK

Problem 3.

Integrating by Parts

Find \(\int x\) sin \(x~dx\).

Solution We use the integration by parts formula with \[ u=x \qquad\hbox{and}\qquad dv=\sin x\,dx\qquad {\color{#0066A7}{\int}} {\color{#0066A7}{udv=}} {\color{#0066A7}{\int x\sin x\,dx}} \]

Then \[ du = dx \quad and \quad v = \int \sin x~dx = -cos x \]

and \[ \begin{eqnarray*} \int x\sin x\,dx \underset{\underset{\color{#0066A7}{\hbox{\(\int udv= uv - \int vdu\)}}}{\color{#0066A7}{\uparrow}}} {=} -x\cos x+\int \cos x\,dx=-x\cos x+\sin x+C \\[-9pt] \end{eqnarray*} \]

NOW WORK

Problem 5.

Integration by Parts: Guidelines for Choosing \(u\) and \(dv\)

• \(dx\) is always part of \(dv.\)
• \(dv\) should be easy to integrate.
• \(u\) and \(dv\) are chosen so that \(\int v du\) is no more difficult to integrate than the original integral \(\int u dv\).
• If the new integral is more complicated, try different choices for \(u\) and \(dv.\)

Integrating by Parts to Find \(\int \ln x\, dx\)

Derive the formula \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int \ln x\,dx=x\ln x-x+C } \]

Solution We use the integration by parts formula with \[ \begin{equation*} \hbox{ }u=\ln x\qquad \hbox{and}\qquad dv=dx \end{equation*} \]

Then \[ du=\frac{1}{x}\,dx\qquad {\rm and}\qquad v=\int dx=x \ \]

Now \[ \int \ln x\,dx= {{x}}\cdot {{\ln x}}-\int {{x}}\cdot {{\frac{1}{x}\,dx}}\,=x\,\ln x-\int dx=x\,\ln x-x+C \]

NOTE

The integral \(\int \ln x\,dx\) can be found in the list of integrals on the inserts at the front and back of the book and is a basic integral.

Integrating by Parts to Find \(\int \tan^{-1} x\, dx\)

Derive the formula \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int \tan ^{-1}x\,dx=x\,\tan ^{-1}x-\dfrac{1}{2}\ln \,(1+x^{2})+C } \]

Solution We use the integration by parts formula with \[ u=\tan ^{-1}x\qquad \hbox{and }\qquad dv=dx \]

Then \[ du=\frac{1}{1+x^{2}}\,dx\qquad \hbox{and}\qquad v=\int dx=x \] and \[ \int \tan ^{-1}x\,dx=x\cdot \tan ^{-1}x-\int \frac{x}{1+x^{2}}dx \]

To find the integral \(\int \dfrac{x}{1+x^{2}}dx\), we use the substitution \( t=1+x^{2}\). Then \(dt=2x\,dx\), or equivalently, \(x~dx=\dfrac{dt}{2}.\) \[ \begin{equation*} \int \frac{x}{1+x^{2}}dx=\frac{1}{2}\int \frac{dt}{t}=\frac{1}{2}\ln \vert \,t\vert =\frac{1}{2}\ln (1+x^{2}) \end{equation*} \]

As a result, \(\int \tan ^{-1} x \,dx=x \tan ^{-1}x-\dfrac{1}{2}\ln (1+x^{2})+C\).

NEED TO REVIEW?

The method of substitution is discussed in Section 5.6, pp. 387-393.

NOW WORK

Problem 9.

Integrating by Parts

Find \(\int x^{2}\,e^{x}\,dx\).

Solution We use the integration by parts formula with \[ u=x^{2}\qquad \hbox{and}\qquad dv=e^{x}\,dx \]

Then \[ du=2x\,dx\qquad \hbox{and}\qquad v=\int e^{x}\,dx=e^{x} \]

and \[ \int x^{2} e^{x}\,dx=x^{2}e^{x}-2\int xe^{x}\,dx \]

The integral on the right is simpler than the original integral. To find it, we use integration by parts a second time. (Refer to Example 1.)

\[ \begin{equation*} \int xe^{x}\,dx=xe^{x}-e^{x} \end{equation*} \]

Then \[ \int x^{2}e^{x}\,dx=x^{2}e^{x}-2( xe^{x}-e^{x}) +C=e^{x}(x^{2}-2x+2)+C \]

NOW WORK

Problem 13.

Finding the Area Under the Graph of \(f(x)=x\ln x\)

Find the area under the graph of \(f( x) =x\ln x\) from \(1\) to \(2.\)

Figure 1 \(f(x)=x\ln x\)

Solution See Figure 1 for the graph of \(f( x) =x\ln x\). The area \(A\) under the graph of \(f\) from \(1\) to \(2\) is \( A=\int_{1}^{2}x\ln x\,dx.\) We use the integration by parts formula with \[ \begin{equation*} u=\ln x\qquad \hbox{and}\qquad dv=x\,dx \end{equation*} \]

Then \[ \begin{equation*} du=\frac{1}{x} dx\qquad \hbox{and}\qquad v=\int x dx=\frac{x^{2}}{2} \end{equation*} \]

and \[ \begin{eqnarray*} A=\int_{1}^{2}x\ln x\,dx&=&\left[ \frac{x^{2}}{2}\ln x\right] _{1}^{2}-\int_{1}^{2}\frac{x^{2}}{2}\left( \frac{1}{x} dx\right) =2\ln 2-\dfrac{1}{2}\int_{1}^{2}x\,dx\\[5pt] &=&2\ln 2- \dfrac{1}{2}\left[ \frac{x^{2}}{2}\right] _{1}^{2}=2\ln 2-\dfrac{3}{4} \end{eqnarray*} \]

NOW WORK

Problem 37.

Deriving a Formula

Derive the formula \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int e^{ax}\cos (bx) \,dx=\dfrac{e^{ax}[b\sin (bx) +a\cos (bx)] }{a^{2}+b^{2}}+C\qquad b ≠ 0} \tag{1} \]

Solution We use the integration by parts formula with \[ u=e^{ax}\qquad \hbox{and} \qquad dv=\cos (bx) \,dx \]

Then \[ \begin{equation*} du=ae^{ax}\,dx\qquad \hbox{and}\qquad v=\int \cos (bx) \,dx=\frac{1}{b}\sin (bx) \end{equation*} \]

and \[ \begin{equation*} \int e^{ax}\cos (bx) \,dx=e^{ax}\,\dfrac{\sin (bx) }{b}-\frac{a}{b}\int e^{ax}\sin (bx) \,dx \tag{2} \end{equation*} \]

The new integral on the right, \(\int e^{ax}\sin (bx) \,dx,\) is different from the original integral, but it is essentially of the same form. We use integration by parts again with this integral by choosing \[ \begin{equation*} u=e^{ax}\qquad \hbox{and}\qquad dv=\sin (bx) \,dx \end{equation*} \]

Then \[ \begin{equation*} du=ae^{ax}\,dx\qquad \hbox{and}\qquad v=\int \sin (bx) \,dx=- \frac{1}{b}\cos (bx) \end{equation*} \]

and \[ \begin{equation*} \int e^{ax}\sin (bx) \,dx=-\frac{1}{b}\,e^{ax}\cos (bx) +\frac{a}{b}\int e^{ax}\cos (bx) \,dx\tag{3} \end{equation*} \]

Substituting the result from (3) into (2), we obtain \[ \begin{eqnarray*} \int e^{ax}\cos (bx) \,dx& =& \frac{1}{b}e^{ax}\sin (bx) -\frac{a}{b}\left[ -\frac{1}{b}e^{ax}\cos (bx) + \frac{a}{b}\int e^{ax}\cos (bx) \,dx\right] \\[8pt] \int e^{ax}\cos (bx) \,dx& =&\frac{1}{b}\,e^{ax}\sin (bx) +\frac{a}{b^{2}}\,e^{ax}\cos (bx) -\frac{a^{2}}{b^{2}}\int e^{ax}\cos (bx) \,dx \end{eqnarray*} \]

Now we solve for \(\int e^{ax}\cos (bx) \,dx\) and simplify. \[ \begin{eqnarray*} \int e^{ax}\cos (bx) \,dx+\frac{a^{2}}{b^{2}}\int e^{ax}\cos (bx) \,dx& =&\frac{1}{b}\,e^{ax}\sin (bx) +\frac{a}{ b^{2}}\,e^{ax}\cos (bx) \nonumber \\[5pt] \left( 1+\frac{a^{2}}{b^{2}}\right) \int e^{ax}\cos (bx) \,dx &=&\frac{1}{b^{2}}\,e^{ax} [b\sin (bx) +a\cos (bx) ] \\ \int e^{ax}\cos (bx) \,dx &=&\frac{e^{ax}[b\sin (bx)+a\cos (bx)]}{a^{2}+b^{2}}+C \end{eqnarray*} \]

NOW WORK

Problem 41.

Deriving a Formula

Derive the formula \[\bbox[5px, border:1px solid black, #F9F7ED]{ \int \sec ^{n} x\,dx=\dfrac{\sec ^{n-2}x\,\tan \,x}{n-1} +\dfrac{n-2}{n-1}\int \sec ^{n-2} x\,dx\qquad n \geq 3}\tag{4} \]

Solution We begin by writing \(\sec ^{n}x=\sec ^{n-2}x\sec ^{2}x\), and choose \[ u=\sec ^{n-2}x\qquad \hbox{and}\qquad dv=\sec ^{2}x\,dx \]

This choice makes \(\int dv\) easy to integrate. Then \[ \begin{eqnarray*} du&=&[ (n-2)\sec ^{n-3}x\cdot \sec x\,\tan x] \,dx =[ (n-2)\sec ^{n-2}x\,\tan x] \,dx \\[6pt] v&=&\int \sec^{2}x\,dx=\tan x \end{eqnarray*} \]

Using integration by parts, we get \[ \begin{equation*} \int \sec ^{n}x\,dx=\sec ^{n-2}x\,\tan x-(n-2)\int \sec ^{n-2}x\,\tan ^{2}x\,dx \end{equation*} \]

To express the integrand on the right in terms of \(\sec x,\) we use the trigonometric identity, \(\tan ^{2}x+1=\sec ^{2}x,\) and replace \(\tan ^{2}x\) by \(\sec ^{2}x-1\), obtaining \[ \begin{eqnarray*} \int \sec ^{n}x\,dx& =&\sec ^{n-2}x\tan x-(n-2)\int \sec ^{n-2}x(\sec^{2}x-1)\,dx \\[4pt] \int \sec ^{n}x\,dx& =& \sec ^{n-2}x\tan x-(n-2)\int \sec ^{n}x\,dx+(n-2)\int \sec ^{n-2}x\,dx \end{eqnarray*} \]

Moving the middle term on the right to the left, we obtain \[ (n-1)\int \sec ^{n}x\,dx=\sec ^{n-2}x\,\tan x+(n-2)\int \sec ^{n-2}x\,dx \]

Finally, divide both sides by \(n-1\): \[ \int \sec ^{n}x\,dx=\frac{\sec ^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int \sec ^{n-2}x\,dx \]

NOW WORK

Problem 59.