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 $v du$.

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.

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 $dv.$

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.

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.

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.

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.

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.

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.