\hrulefill}\PARA{038
\hfill Page \thepage } %} \input{tcilatex} \begin{document} \begin{abstract} A Laboratory report created with Scientific Notebook \end{abstract} \section{\protect\vspace{1pt}Ma 115 Lecture 11/15/99} \subsection{Integration by Parts --- Choosing the ``parts''} \vspace{1pt} In applying the integration by parts formula, \begin{center} $\vspace{1pt}\dint u\,dv=u\,v-\dint v\,du\;\;$where\ $du=\dfrac{du}{dx}\,dx$ and $dv=\dfrac{dv}{dx}\,dx$\\[0pt] \end{center} the difficult part is deciding what to pick as $u$ and as $dv$. \ Here are some guidelines. \vspace{1pt} \QTR{blue}{General Rules:} \begin{itemize} \item In general, pick $u$ as a function which is easy to differentiate and pick $dv$ as a function which is easy to integrate. \item Pick $u$ and $dv$ so that $v\,du$ is simpler than $u\,\,dv.$ \end{itemize} \QTR{blue}{Special Rules:} \begin{itemize} \item Polynomials are easy to differentiate or integrate, but the derivative is simpler. \ So it is a good idea to take a polynomial for $u$. \item The functions $e^{x}$, $\sin x$ and $\cos x$ are easy to differentiate or integrate and don't get more complicated. \ So it is a good idea to take these functions for $dv$. \item The functions $\ln x$, $\arcsin x$ and $\arctan x$ are easy to differentiate but hard to integrate. \ So it is a good idea to take these functions for $u$. \end{itemize} \begin{center} \vspace{1pt}\FRAME{itbpF}{3.928in}{0.0623in}{0in}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 3.928in;height 0.0623in;depth 0in;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.2619";cropright "1";cropbottom "0.2607";filename 'maroon3.wmf';file-properties "XNPEU";}} \end{center} \begin{example} Compute $\int x\sqrt{x-1}\,dx$ \end{example} \begin{itemize} \item \emph{Solution.} \ (... by Substitution.) This is an example easily computed using the Method of Substitution. Recall, we take $u=x-1,\;du=dx.$ Then \begin{eqnarray*} \int x\sqrt{x-1}\,dx &=&\int \left( u+1\right) \sqrt{u}\,du=\int \left( u^{3/2}+u^{1/2}\right) \,du \\ &=&\dfrac{2}{5}u^{5/2}+\dfrac{2}{3}u^{3/2}+C \\ &=&\dfrac{2}{5}\left( x-1\right) ^{5/2}+\dfrac{2}{3}\left( x-1\right) ^{3/2}+C \end{eqnarray*} (... by \ Integration by parts.) \ This same integral can be computed using integration by parts. \ Following the first \emph{special rule} above, we set $u$ as the polynomial and $dv$ as what remains. Thus \[ \begin{array}{ll} u=x & dv=\sqrt{x-1}dx \\ du=dx & v=\dfrac{2}{3}\left( x-1\right) ^{3/2}% \end{array}% \] Then, \begin{eqnarray*} \int x\sqrt{x-1}\,dx &=&x\left( \dfrac{2}{3}\left( x-1\right) ^{3/2}\right) -\int \dfrac{2}{3}\left( x-1\right) ^{3/2}\,dx \\ &=&\dfrac{2}{3}x\left( x-1\right) ^{3/2}-\dfrac{2}{3}\left( \dfrac{2}{5}% \left( x-1\right) ^{5/2}\right) +C \\ &=&\dfrac{2}{3}x\left( x-1\right) ^{3/2}-\dfrac{4}{15}\left( x-1\right) ^{5/2}+C \end{eqnarray*} Check by differentiating (using the product rule). \[ \dfrac{d}{dx}\left( \dfrac{2}{3}x\left( x-1\right) ^{3/2}-\dfrac{4}{15}% \left( x-1\right) ^{5/2}+C\right) =x\sqrt{\left( x-1\right) } \] \medskip\ Which of the two ways to compute this integral is better? \ Both are OK. \ \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.4407";cropright "1";cropbottom "0.8847";filename 'maroon3.wmf';file-properties "XNPEU";}} \end{itemize} By the way, using Scientific Notebook to compute the integral, we get \[ \int x\sqrt{x-1}\,dx=\frac{2}{3}\left( \sqrt{\left( x-1\right) }\right) ^{3}+% \frac{2}{5}\left( \sqrt{\left( x-1\right) }\right) ^{5} \] Note the constant $C$ is missing. \ This we know about. \ Scientific Notebook does not supply the constant of integration. \ \QTR{blue}{Note also }the expression of the answer is given in powers of the radical $\sqrt{x-1}% \; $ --- \ as opposed to fractional powers of $x-1.$ \ Both are the same ....... and of course both are correct. \begin{center} $\FRAME{itbpF}{4.1943in}{0.0623in}{0in}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0in;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.7183";cropright "1";cropbottom "0.6071";filename 'maroon3.wmf';file-properties "XNPEU";}}$ \vspace{0in} \end{center} \vspace{0in} \subsection{Example --- Choosing the ``parts''} \vspace{1pt}Recall the rules........ \QTR{blue}{General Rules} \begin{itemize} \item In general, pick $u$ as a function which is easy to differentiate and pick $dv$ as a function which is easy to integrate. \item Pick $u$ and $dv$ so that $v\,du$ is simpler than $u\,\,dv.$ \end{itemize} \QTR{blue}{Special Rules} \begin{itemize} \item Polynomials are easy to differentiate or integrate, but the derivative is simpler. \ So it is a good idea to take a polynomial for $u$. \item The functions $e^{x}$, $\sin x$ and $\cos x$ are easy to differentiate or integrate and don't get more complicated. \ So it is a good idea to take these functions for $dv$. \item The functions $\ln x$, $\arcsin x$ and $\arctan x$ are easy to differentiate but hard to integrate. \ So it is a good idea to take these functions for $u$. \end{itemize} \begin{center} \vspace{1pt}\FRAME{itbpF}{3.928in}{0.0623in}{0in}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 3.928in;height 0.0623in;depth 0in;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.2619";cropright "1";cropbottom "0.2607";filename 'maroon3.wmf';file-properties "XNPEU";}} \end{center} \begin{example} Compute\qquad $\dint x^{4}\ln x\,dx.$ \end{example} \emph{Solution.} \ Normally we would take the polynomial $x^{4}$ as $u$. \ However, since it is hard to integrate $\ln x$, we take \[ \begin{array}{ll} u=\ln x & dv=x^{4}\,dx \\ du=\dfrac{1}{x}\,dx\qquad & v=\dfrac{x^{5}}{5}% \end{array}% \] Then, integrating by parts \fbox{\underline{% \CustomNote[Formula]{Margin Hint}{% Integration by parts formula \par \begin{eqnarray*} \dint u\,dv=u\,v-\dint v\,du\;\; \\ \text{where}\ du=\dfrac{du}{dx}\,dx\text{ and }dv=\dfrac{dv}{dx}\,dx\newline \end{eqnarray*} }}} gives \begin{center} \begin{eqnarray*} \dint x^{4}\ln x\,dx &=&\dfrac{x^{5}}{5}\ln x-\dint \dfrac{x^{5}}{5}\dfrac{1% }{x}\,dx \\ &=&\dfrac{x^{5}}{5}\ln x-\dint \dfrac{x^{4}}{5}\,dx \\ &=&\dfrac{x^{5}}{5}\ln x-\dfrac{x^{5}}{25}+C\medskip \end{eqnarray*} \end{center} Check by differentiating (using the product rule): If $\ f\left( x\right) =\dfrac{x^{5}}{5}\ln x-\dfrac{x^{5}}{25}+C$ \ then $% \; $% \[ f\,^{\prime }\left( x\right) =\left( x^{4}\ln x+\dfrac{x^{5}}{5}\dfrac{1}{x}% \right) -\dfrac{x^{4}}{5}=x^{4}\ln x \] \ \thinspace which is the integrand we started with. \begin{center} \FRAME{itbpF}{3.928in}{0.0623in}{0in}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 3.928in;height 0.0623in;depth 0in;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.2619";cropright "1";cropbottom "0.2607";filename 'maroon3.wmf';file-properties "XNPEU";}} \end{center} \begin{remark} In this example, the general rule about selecting for $dv$ a function easy to integrate supersedes the rule about taking for $u$ the variable raised to a power. \end{remark} \begin{center} \FRAME{itbpF}{3.928in}{0.0623in}{0in}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 3.928in;height 0.0623in;depth 0in;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.2619";cropright "1";cropbottom "0.2607";filename 'maroon3.wmf';file-properties "XNPEU";}}\qquad \vspace{0in} \end{center} \subsection{Integrals of \QTR{redbig}{ln} $x$, \QTR{redbig}{arcsin} $x$ and \QTR{redbig}{arctan} $x$} \vspace{1pt} Integration by parts can be used to find the integrals of many functions. \begin{example} Compute $\;\dint \arcsin x\,dx.$ \end{example} \emph{Solution:} \ We do not appear to have the product of two functions here. \ However, one of the functions can be $1$. \ That is, we write $% \arcsin x=1\cdot \arcsin x$. \ Take \[ \begin{array}{ll} u=\arcsin x & dv=1\,dx=dx \\ du=\dfrac{1}{\sqrt{1-x^{2}}}\,dx\qquad & v=x% \end{array}% \] Thus \begin{center} $\dint \arcsin x\,dx=x\arcsin x-\dint \dfrac{x}{\sqrt{1-x^{2}}}\,dx$ \end{center} To compute the second integral, use the substitution $u=1-x^{2}$. \ Then $% du=-2x\,dx$ and $x\,dx=-\dfrac{1}{2}\,du$. \ Thus, \begin{eqnarray*} \dint \dfrac{x}{\sqrt{1-x^{2}}}\,dx &=&-\dfrac{1}{2}\dint \dfrac{1}{\sqrt{u}}% \,du \\ &=&-\sqrt{u}+C \\ &=&-\sqrt{1-x^{2}}+C \end{eqnarray*} Therefore, \begin{center} \begin{eqnarray*} \dint \arcsin x\,dx\allowbreak &=&x\arcsin x-\dint \dfrac{x}{\sqrt{1-x^{2}}}% \,dx \\ &=&x\arcsin x+\sqrt{1-x^{2}}+C \end{eqnarray*} $\medskip $ \end{center} Check by differentiating. \ Let $f\left( x\right) =x\arcsin x+\sqrt{1-x^{2}}% +C$, then \[ f\,^{\prime }\left( x\right) =\left( \arcsin x+x\dfrac{1}{\sqrt{1-x^{2}}}% \right) +\dfrac{1}{2}\dfrac{-2x}{\sqrt{1-x^{2}}}=\arcsin x \] which is the original integrand. \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.6628";cropright "1";cropbottom "0.6626";filename 'maroon3.wmf';file-properties "XNPEU";}} Apply integration by parts to determine the following integrals. \begin{exercise} Compute\qquad $\dint \ln x\,dx.$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.1505in}{0.1833in}{0in% }}{}{}{}]{Margin Hint}{% Take$\quad u=\ln x$ and $dv=dx$\emph{.}}...% \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\dint \ln x\,dx=x\ln x-x+C$}...% \CustomNote[\hyperref{\TCIIcon{BITMAPSETSolution}{0.1609in}{0.1609in}{0in}}{% }{}{}]{Margin Hint}{% Take $ \begin{array}{ll} u=\ln x & dv=dx \\ du=\dfrac{1}{x}\,dx\qquad & v=x% \end{array} $% \par Then \par $\dint \ln x\,dx=x\ln x-\dint x\dfrac{1}{x}\,dx=x\ln x-x+C$} \end{exercise} \begin{exercise} Compute\qquad $\dint \arctan x\,dx.$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETProbSolvHint}{0.1505in}{0.1833in}{0in% }}{}{}{}]{Margin Hint}{% Take$\quad u=\arctan x$ and $dv=dx\emph{.}$}...% \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\dint \arctan x\,dx=x\arctan x-\dfrac{1}{2}\ln \left( 1+x^{2}\right) +C$}...% \CustomNote[\hyperref{\TCIIcon{BITMAPSETSolution}{0.1609in}{0.1609in}{0in}}{% }{}{}]{Margin Hint}{% Take $ \begin{array}{ll} u=\arctan x & dv=dx \\ du=\dfrac{1}{1+x^{2}}\,dx\qquad & v=x% \end{array} $% \par Then $\dint \arctan x\,dx=x\dfrac{1}{1+x^{2}}-\dint \dfrac{x}{1+x^{2}}\,dx$% \par Now make the substitution $u=1+x^{2}$. \ Then $du=2x\,dx$. \par So \begin{eqnarray*} \dint \arctan x\,dx=x\dfrac{1}{1+x^{2}}-\dfrac{1}{2}\dint \dfrac{1}{u}\,du\ \\ =x\dfrac{1}{1+x^{2}}-\dfrac{1}{2}\ln \left| u\right| +C\ \\ =x\dfrac{1}{1+x^{2}}-\dfrac{1}{2}\ln \left( 1+x^{2}\right) +C \end{eqnarray*} Note, the absolute values signs are dropped in the last step. \ (That is because the function $1+x^{2}$ is always positive.)} \end{exercise} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.7183";cropright "1";cropbottom "0.6071";filename 'maroon3.wmf';file-properties "XNPEU";}}\qquad As a result of these computations, we now know: \begin{center} $ \begin{array}{l} \dint \ln x\,dx=x\ln x-x+C \\ \dint \arcsin x\,dx=x\arcsin x+\sqrt{1-x^{2}}+C \\ \dint \arctan x\,dx=x\arctan x-\dfrac{1}{2}\ln \left( 1+x^{2}\right) +C% \end{array} $ $\FRAME{itbpF}{4.1943in}{0.0761in}{0in}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0761in;depth 0in;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "1.164";cropright "1";cropbottom "0.3862";filename 'maroon3.wmf';file-properties "XNPEU";}}$ \end{center} \qquad \subsection{ Integration by Parts for Definite Integrals} The integration by parts formula also has a version for definite integrals: \vspace{1pt} \begin{theorem} (\QTR{blue}{Integration by Parts for definite integrals}) \ Suppose $u(x)$ and $v\left( x\right) $ are functions for which the following integrals are defined. \ Then \end{theorem} $\vspace{1pt}\qquad \qquad $% \begin{eqnarray*} \dint_{a}^{b}u\,dv &=&\left[ u\,v\rule{0in}{0.15in}\right] _{a}^{b}-\dint_{a}^{b}v\,du\text{ \ } \\ \text{where \ }du &=&\dfrac{du}{dx}\,dx\text{ \ and \ }dv=\dfrac{dv}{dx}\,dx% \newline \end{eqnarray*} $\vspace{1pt}$ \vspace{1pt} \begin{example} Compute\qquad $\dint_{0}^{1}3x^{2}\arctan x\,dx.$ \end{example} \emph{Solution:} \ We apply integration by parts with \[ \begin{array}{ll} u=\arctan x & dv=3x^{2}\,dx \\ du=\dfrac{1}{1+x^{2}}\,dx\qquad & v=x^{3}% \end{array}% \] So \begin{center} \begin{eqnarray*} \dint_{0}^{1}3x^{2}\arctan x\,dx &=&\left[ x^{3}\arctan x\right] _{0}^{1}-\dint_{0}^{1}\dfrac{x^{3}}{1+x^{2}}\,dx\medskip \\ &=&\left. \left( x^{3}\arctan x\right) \right| _{x=1}-\left. \left( x^{3}\arctan x\right) \right| _{x=0}-\dint_{0}^{1}\dfrac{x^{3}}{1+x^{2}}\,dx \end{eqnarray*} \end{center} For the integrated part, recall $\arctan 1=\dfrac{\pi }{4}$ and $\arctan 0=0$% . \ For the remaining integral, we substitute $u=1+x^{2}$. \ Then\ $% du=2x\,dx $ and so $x\,dx=\dfrac{1}{2}\,du$ and $x^{2}=u-1$. \ Thus \begin{center} \begin{eqnarray*} \dint_{0}^{1}3x^{2}\arctan x\,dx &=&\left[ 1^{3}\arctan 1\right] -\dint_{x=0}^{1}\dfrac{u-1}{u}\,\dfrac{1}{2}\,du\ \\ &=&\dfrac{\pi }{4}-\dfrac{1}{2}\dint_{x=0}^{1}\left( 1-\dfrac{1}{u}\right) \,du\ \\ &=&\dfrac{\pi }{4}-\dfrac{1}{2}\left[ u-\ln u\rule{0in}{0.15in}\right] _{x=0}^{1}\ \\ &=&\dfrac{\pi }{4}-\dfrac{1}{2}\left[ 1+x^{2}-\ln \left( 1+x^{2}\right) \rule% {0in}{0.15in}\right] _{x=0}^{1}\ \\ &=&\dfrac{\pi }{4}-\dfrac{1}{2}\left[ 2-\ln \left( 2\right) \right] +\dfrac{1% }{2}\left[ 1\right] \ \\ &=&\dfrac{\pi }{4}-\dfrac{1}{2}+\dfrac{1}{2}\ln \left( 2\right) \end{eqnarray*} \end{center} \FRAME{dtbpF}{4.1943in}{0.0761in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0761in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "1.014";cropright "1";cropbottom "0.5362";filename 'maroon3.wmf';file-properties "XNPEU";}}\qquad Of course, you can always compute a definite integral by first finding the indefinite integral and then evaluating at the limits. \begin{center} \vspace{0in} \end{center} \vspace{0in} \subsection{ Example --- Integration by Parts for Definite Integrals} Recall.... \begin{theorem} (\QTR{blue}{Integration by Parts for definite integrals}) \ Suppose $u(x)$ and $v\left( x\right) $ are functions for which the following integrals are defined. \ Then \end{theorem} $\vspace{1pt}\qquad \qquad $% \begin{eqnarray*} \dint_{a}^{b}u\,dv &=&\left[ u\,v\rule{0in}{0.15in}\right] _{a}^{b}-\dint_{a}^{b}v\,du\text{ \ } \\ \text{where \ }du &=&\dfrac{du}{dx}\,dx\text{ \ and \ }dv=\dfrac{dv}{dx}\,dx% \newline \end{eqnarray*} $\vspace{1pt}$ \vspace{1pt}\FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.9405";cropright "1";cropbottom "0.3849";filename 'maroon3.wmf';file-properties "XNPEU";}} \begin{example} Compute $\,\dint_{-1}^{3}xe^{-4x}\,dx.$ \end{example} \emph{Solution:} \ We apply integration by parts with \[ \begin{array}{ll} u=\ x & dv=e^{-4x}\,dx \\ du=\,dx\qquad & v=-\dfrac{1}{4}e^{-4x}% \end{array}% \] So \begin{center} \begin{eqnarray*} \dint_{-1}^{3}xe^{-4x}\,dx &=&\left[ x\left( -\dfrac{1}{4}e^{-4x}\right) % \right] _{-1}^{3}+\dfrac{1}{4}\dint_{-1}^{3}e^{-4x}\,dx \\ &=&\left. \left( x\left( -\dfrac{1}{4}e^{-4x}\right) \right) \right| _{x=3}-\left. \left( x\left( -\dfrac{1}{4}e^{-4x}\right) \right) \right| _{x=-1}+\dfrac{1}{4}\dint_{-1}^{3}e^{-4x}\,dx \\ &=&\left( -\dfrac{3}{4}e^{-12}\right) -\left( \dfrac{1}{4}e^{4}\right) +% \dfrac{1}{4}\dint_{-1}^{3}e^{-4x}\,dx \end{eqnarray*} \end{center} For the remaining integral, we substitute $u=-4x$. \ Then\ $du=-4\,dx$ and so $\,dx=-\dfrac{1}{4}\,du$. \ Thus$\ $% \[ \dint e^{-4x}\,\,dx=-\dfrac{1}{4}\int e^{u}\,\,du=-\dfrac{1}{4}e^{u}=-\dfrac{% 1}{4}e^{-4x} \] leaving off the constant of integration. Finally, \begin{eqnarray*} \dint_{-1}^{3}xe^{-4x}\,dx &=&\left( -\dfrac{3}{4}e^{-12}\right) -\left( \dfrac{1}{4}e^{4}\right) -\left. \dfrac{1}{16}e^{-4x}\right| _{-1}^{3} \\ &=&\left( -\dfrac{3}{4}e^{-12}\right) -\left( \dfrac{1}{4}e^{4}\right) -% \dfrac{1}{16}\left( e^{-12}-e^{4}\right) \\ &=&-\frac{13}{16}e^{-12}-\frac{3}{16}e^{4} \end{eqnarray*} the last step coming from simplification. \fbox{% \CustomNote[\underline{Why can we leave off the constant of integration above?}]{Margin Hint}{% Because it will cancel when the evaluations are performed. \ Recall \par \[ \left. F\left( x\right) \right| _{a}^{b}=F\left( b\right) -F\left( a\right) \]% Now put in the constant. \par \begin{eqnarray*} \left. \left( F\left( x\right) +C\right) \,\right| _{a}^{b} &=&\left( F\left( b\right) +C\right) -\left( F\left( a\right) +C\right) \\ &=&F\left( b\right) -F\left( a\right) \end{eqnarray*}% }} \FRAME{dtbpF}{4.1943in}{0.0761in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0761in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.7753";cropright "1";cropbottom "0.7749";filename 'maroon3.wmf';file-properties "XNPEU";}} \qquad \subsection{ Reduction Formulas} Oftentimes, integrals of the form \[ \int x^{n}\,f\,\left( x\right) \,dx \] can be computed a multiple use of integration by parts, reducing the power by one at each successive integration. \ We have already computed an \hyperref{example}{}{}{sintpartsindef1.tex}. \ In this section, a few more examples will be considered. \ Forms for which reduction works especially well are \begin{eqnarray*} &&\int x^{n}e^{ax}\,dx \\ &&\int x^{n}\cos ax\,dx\; \\ &&\int x^{n}\,\sin ax\,dx \end{eqnarray*} The best way to learn these is by experience. \ But the general principle is to select the parts as \[ \begin{array}{ll} u=x^{n} & dv=f\,\left( x\right) dx \\ du=nx^{n-1}\,dx & v=F\left( x\right)% \end{array}% \] \vspace{1pt}where $F\left( x\right) $ is an antiderivative of $f\,\left( x\right) $. \begin{example} Compute $\int x^{2}e^{x}\,dx.$ \end{example} Solution. \ Following the ``template'' above, \[ \begin{array}{ll} u=x^{2} & dv=e^{x}dx \\ du=2x\,dx & v=e^{x}% \end{array}% \] Thus \[ \int x^{2}e^{x}\,dx=x^{2}e^{x}-2\int xe^{x}\,dx \] Now apply the parts \[ \begin{array}{ll} u=x & dv=e^{x}dx \\ du=\,dx & v=e^{x}% \end{array}% \] Then \begin{eqnarray*} \int x^{2}e^{x}\,dx &=&x^{2}e^{x}-2\left( xe^{x}-\int e^{x}\,dx\right) \\ &=&x^{2}e^{x}-2xe^{x}+2e^{x}+C \end{eqnarray*} \begin{center} \FRAME{dtbpF}{4.2384in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0692in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.7968";cropright "1";cropbottom "0.5965";filename 'maroon3.wmf';file-properties "XNPEU";}} \vspace{0in} \end{center} \subsection{ Reduction Formulas} Oftentimes, integrals of the form \[ \int x^{n}\,f\,\left( x\right) \,dx \] can be computed a multiple use of integration by parts, reducing the power by one at each successive integration. \ The general principle is to select the parts as \[ \begin{array}{ll} u=x^{n} & dv=f\,\left( x\right) dx \\ du=nx^{n-1}\,dx & v=F\left( x\right)% \end{array}% \] \vspace{1pt}where $F\left( x\right) $ is an antiderivative of $f\,\left( x\right) $.\FRAME{dtbpF}{4.2384in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0692in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.9967";cropright "1";cropbottom "0.3966";filename 'maroon3.wmf';file-properties "XNPEU";}} \begin{example} Compute $\int x^{3}\sin 2x\,dx.$ \end{example} Solution. \ Following the ``template'', take \[ \begin{array}{ll} u=x^{3} & dv=\sin 2xdx \\ du=3x^{2}\,dx & v=-\dfrac{1}{2}\cos 2x% \end{array}% \] Thus \[ \int x^{3}\sin 2x\,dx=-\dfrac{1}{2}x^{3}\cos 2x+\dfrac{3}{2}\int x^{2}\cos 2x\,dx \] $\allowbreak $Now apply the parts \[ \begin{array}{ll} u=x^{2} & dv=\cos 2xdx \\ du=\,2xdx & v=\dfrac{1}{2}\sin 2x% \end{array}% \] Then \[ \int x^{2}\sin 2x\,dx=-\dfrac{1}{2}x^{3}\cos 2x+\dfrac{3}{2}\left( \dfrac{1}{% 2}x^{2}\sin 2x-\int x\sin 2x\,dx\right) \] Finally, use the parts \[ \begin{array}{ll} u=x^{{}} & dv=\sin 2xdx \\ du=\,dx & v=-\dfrac{1}{2}\sin 2x% \end{array}% \] Then \begin{eqnarray*} \int x^{3}\sin 2x\,dx &=&-\dfrac{1}{2}x^{3}\cos 2x+\dfrac{3}{4}x^{2}\sin 2x+% \dfrac{3}{2}\left( -\dfrac{1}{2}x\sin 2x+\dfrac{1}{2}\int \sin 2x\,dx\right) \\ &=&-\dfrac{1}{2}x^{3}\cos 2x+\dfrac{3}{4}x^{2}\sin 2x-\dfrac{3}{4}x\sin 2x-% \dfrac{3}{8}\cos 2x+C \end{eqnarray*} \FRAME{dtbpF}{4.2384in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0692in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.9967";cropright "1";cropbottom "0.3966";filename 'maroon3.wmf';file-properties "XNPEU";}} \vspace{0in} \subsection{Recurrence of the integral} \vspace{1pt} Often times, no matter what parts are selected, the integral never seems to get simpler. \ In special cases, usually involving $\sin bx$ or $\cos bx$, a repeated integration by parts yields the recurrence of the original integral. \ \ Such is the case in the following example. \ \emph{Resolution of such problems always requires two successive integrations by parts.} \vspace{1pt} \begin{example} Compute $\int e^{x}\sin x\,dx$ \end{example} \emph{Solution. }Try the parts \[ \begin{array}{lll} u=e^{x} & & dv=\sin x\,dx \\ du=e^{x}\,dx & & v=-\cos x% \end{array}% \] Applying integration by parts, we obtain \begin{eqnarray*} \int e^{x}\sin x\,dx &=&e^{x}\left( -\cos x\right) +\int e^{x}\cos x\,dx \\ &=&-e^{x}\cos x+\int e^{x}\cos x\,dx \end{eqnarray*} The second integral looks so much like the first that we may suspect that the wrong parts were selected. \ In this case, this is not so. \ Proceed to perform integration by parts again. \ Take \[ \begin{array}{lll} u=e^{x} & & dv=\cos x\,dx \\ du=e^{x}\,dx & & v=\sin x% \end{array}% \] Then the second integral becomes \[ \int e^{x}\cos x\,dx=e^{x}\sin x-\int e^{x}\sin x\,dx \] Putting it all together yields \begin{eqnarray*} \int e^{x}\sin x\,dx &=&-e^{x}\cos x+\int e^{x}\cos x\,dx \\ &=&-e^{x}\cos x+\left( e^{x}\sin x-\int e^{x}\sin x\,dx\right) \end{eqnarray*} Now solve for $\int e^{x}\sin x\,dx.$ \ \ This gives \begin{eqnarray*} 2\int e^{x}\sin x\,dx &=&-e^{x}\cos x+e^{x}\sin x\;\;\;\text{or} \\ \int e^{x}\sin x\,dx &=&\dfrac{1}{2}\left( -e^{x}\cos x+e^{x}\sin x\right) +C \end{eqnarray*} where in the last step we have inserted the required constant of integration $C.$ \vspace{1pt} \FRAME{dtbpF}{4.1943in}{0.0761in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0761in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "1.0529";cropright "1";cropbottom "0.4973";filename 'maroon3.wmf';file-properties "XNPEU";}}\qquad The general forms where this always works are \[ \int e^{ax}\sin bx\,dx\;\;\text{and\ \ }\int e^{ax}\cos bx\,dx \] For the first one, the correct selection of parts is \[ \begin{array}{lll} u=e^{ax} & & dv=\sin bx\,dx \\ du=ae^{x}\,dx & & v=-\dfrac{1}{b}\cos x% \end{array}% \] It is important to make the same type of parts selection for the second integration. \begin{exercise} What is the correct selection of parts for the second integral, $\int e^{ax}\cos bx\,dx\,?$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% The correct selection of parts is \par \[ \begin{array}{lll} u=e^{ax} & & dv=\cos bx\,dx \\ du=ae^{x}\,dx & & v=\dfrac{1}{b}\sin x% \end{array}% \]% } \end{exercise} \FRAME{dtbpF}{4.1943in}{0.0623in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";display "PICT";valid_file "F";width 4.1943in;height 0.0623in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.9418";cropright "1";cropbottom "0.6084";filename 'maroon3.wmf';file-properties "XNPEU";}} \begin{center} \vspace{0in} \end{center} \subsection{Example --- Recurrence of the integral} \vspace{1pt} There are other examples where the recurrence of the original integral occurs. \vspace{1pt} \begin{example} Compute $\int \sec ^{3}\,x\,dx$ \end{example} \emph{Solution. }Try the parts \[ \begin{array}{lll} u=\sec x & & dv=\sec ^{2}x\,dx \\ du=\sec x\tan x\,dx & & v=\tan x% \end{array}% \]% Applying integration by parts, we obtain \begin{eqnarray*} \int \sec ^{3}x\,dx &=&\sec x\tan x-\int \sec x\tan ^{2}x\,dx \\ &=&\sec x\tan x-\int \sec x\tan ^{2}x\,dx \end{eqnarray*}% Now we know than $\tan ^{2}x+1=\sec ^{2}x.$ \ Hence \begin{eqnarray*} \int \sec ^{3}x\,dx &=&\sec x\tan x-\int \sec x\left( \sec ^{2}x-1\right) \,dx \\ &=&\sec x\tan x-\int \sec ^{3}x\,dx+\int \sec x\,dx \end{eqnarray*}% Thus \begin{eqnarray*} 2\int \sec ^{3}x\,dx &=&\sec x\tan x+\int \sec x\,dx \\ &=&\sec x\tan x+\ln \,\left| \sec x+\tan x\right| +C\;\;\;\text{or} \\ \int \sec ^{3}x\,dx &=&\dfrac{1}{2}\sec x\tan x+\frac{1}{2}\ln \,\left| \left( \sec x+\tan x\right) \right| +C \end{eqnarray*}% where in the last step we made the usual replacement step of $C$ for $\dfrac{% 1}{2}C.$ \ The last integral, $\int \sec x\,dx,$ was computed as follows: \begin{eqnarray*} \int \sec x\,dx &=&\int \sec x\,\dfrac{\sec x+\tan x}{\sec x+\tan x}dx \\ &=&\int \,\dfrac{\sec ^{2}x+\sec x\tan x}{\sec x+\tan x}dx \end{eqnarray*}% Now let $u=\sec x+\tan x,$ and $du=\left( \sec x\tan x+\sec ^{2}x\right) \,dx $. \ Hence \begin{eqnarray*} \int \sec x\,dx &=&\int \,\dfrac{\sec ^{2}x+\sec x\tan x}{\sec x+\tan x}dx \\ &=&\int \dfrac{du}{u}=\ln \,\left| u\right| \\ &=&\ln \,\left| \sec x+\tan x\right| \end{eqnarray*}% we have omitted the constant $C,$ which is inserted later.\medskip \begin{remark} This example is by no means simple. \ \QTR{blue}{Study it carefully.} \end{remark} \FRAME{dtbpF}{4.1943in}{0.0761in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.1943in;height 0.0761in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.7753";cropright "1";cropbottom "0.7749";filename 'maroon3.wmf';file-properties "XNPEU";}} \subsection{Exercises} \qquad \begin{enumerate} \item Instructions \item $\int xe^{2x}\,dx$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int xe^{2x}\,dx=\allowbreak \frac{1}{2}xe^{2x}-\frac{1}{4% }e^{2x}+C$% \par \emph{Solution.} \ For the parts take \[ \begin{array}{ll} u=x & dv=e^{2x}dx \\ du=dx & v=\dfrac{1}{2}e^{2x}% \end{array}% \] Then \begin{eqnarray*} \int xe^{2x}\,dx=\dfrac{1}{2}xe^{2x}-\dfrac{1}{2}\int e^{2x}\,dx \\ =\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+C \end{eqnarray*}% } \item $\int xe^{-3x}\,dx$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int xe^{-3x}\,dx=\allowbreak -\frac{1}{3}xe^{-3x}-\frac{1% }{9}e^{-3x}+C$% \par \emph{Solution.} \ For the parts take \[ \begin{array}{ll} u=x & dv=e-3 \\ du=dx & v=-\dfrac{1}{3}e^{-3x}% \end{array}% \] Then \begin{eqnarray*} \int xe^{2x}\,dx=-\dfrac{1}{3}xe^{-3x}+\dfrac{1}{3}\int e^{-3x}\,dx \\ =-\dfrac{1}{3}xe^{-3x}-\dfrac{1}{9}e^{-3x}+C \end{eqnarray*}% } \item $\int x\ln |x+1|\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x\ln \left| x+1\right| \ dx\;=\allowbreak \frac{1}{4}% \left( x+1\right) \left( 2x\ln \left| x+1\right| -x-2\ln \left| x+1\right| +3\right) \allowbreak +C$} \item $\int (x-2)e^{x+1}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (x-2)e^{x+1}\ dx=\allowbreak -4e^{x+1}+\left( x+1\right) e^{x+1}+C$} \item $\int 3x\sqrt{2x+1}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 3x\sqrt{2x+1}\ dx=\allowbreak x\left( 2x+1\right) ^{3/2}-\dfrac{1}{5}\left( 2x+1\right) ^{5/2}+C$% \par \emph{Solution.} \ Take the parts \[ \begin{array}{ll} u=3x & dv=\sqrt{2x+1}\,dx \\ du=3dx & v=\dfrac{1}{3}\left( 2x+1\right) ^{3/2}% \end{array}% \] \lbrack To find $v$ we used substitution with the substitution taken for $% 2x+1.$\rbrack\ \ Then \begin{eqnarray*} \int 3x\sqrt{2x+1}\ dx=3x\left( \dfrac{1}{3}\left( 2x+1\right) ^{3/2}\right) -\int \left( 2x+1\right) ^{3/2}\,dx \\ =x\left( 2x+1\right) ^{3/2}-\dfrac{1}{5}\left( 2x+1\right) ^{5/2}+C \end{eqnarray*}% \par {}} \item $\int t\sqrt{t+2}\ dt\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int t\sqrt{t+2}\ dt\;=\allowbreak \frac{2}{5}\left( \sqrt{\left( t+2\right) }\right) ^{5}-\frac{4}{3}\left( \sqrt{\left( t+2\right) }\right) ^{3}$% \par {} \par \emph{Solution.} \ Take the parts \[ \begin{array}{ll} u=t & dv=\sqrt{t+2}\,dt \\ du=dt & v=\dfrac{2}{3}\left( t+2\right) ^{3/2}% \end{array}% \] \lbrack To find $v$ we used substitution with the substitution taken for $% t+2 $\rbrack\ \ Then \begin{eqnarray*} \int t\sqrt{t+2}\ dt\;=t\left( \dfrac{2}{3}\left( t+2\right) ^{3/2}\right) -\int \dfrac{2}{3}\left( t+2\right) ^{3/2}\,dt \\ =t\left( \dfrac{2}{3}\left( t+2\right) ^{3/2}\right) -\dfrac{4}{15}\left( t+2\right) ^{5/2}+C \end{eqnarray*} where the last integral was computed by using the substitution method with $% u=t+1.$% \par \vspace{1pt} \par {}} \item $\int \dfrac{2x}{e^{2x}}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \dfrac{2x}{e^{2x}}\ dx=\int 2xe^{-x}\,dx=\allowbreak -2xe^{-x}-2e^{-x}+C$% \par \emph{Solution.} \ Take the parts \[ \begin{array}{ll} u=2x & dv=e^{-x}\,dx \\ du=2dx & v=-e^{-x}% \end{array}% \] Then \begin{eqnarray*} \int 2xe^{-x}\,dx=-2xe^{-x}+2\int e^{-x}\,dx \\ =-2xe^{-x}-2e^{-x}+C \end{eqnarray*}% \par {}} \item $\int (1+x)e^{-3x}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (1+x)e^{-3x}\ dx=\allowbreak -\frac{4}{9}e^{-3x}-% \frac{1}{3}xe^{-3x}+C$% \par Use the parts $u=x+1$ and $dv=e^{-3x}\,dx$.} \item $\int \left( x^{2}+1\right) \sin 2x\,dx$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \left( x^{2}+1\right) \sin 2x\,dx=\allowbreak -\frac{% 1}{4}\cos 2x-\frac{1}{2}x^{2}\cos 2x+\frac{1}{2}x\sin 2x+C$% \par Use the parts $u=x^{2}+1$ and $dv=\sin 2x\,dx$} \item $\int \left( x^{2}-4\right) \cos 3x\,dx$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \left( x^{2}-4\right) \cos 3x\,dx=\allowbreak \frac{1% }{3}x^{2}\sin 3x-\frac{38}{27}\sin 3x+\frac{2}{9}x\cos 3x+C$% \par Use the parts $u=x^{2}-4$ and $dv=\cos 3x\,dx$.} \item $\int x(x^{3}+1)\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x(x^{3}+1)\ dx=\allowbreak \frac{1}{5}x^{5}+\frac{1}{% 2}x^{2}+C$% \par Use the parts $u=x$ and $dv=\left( x^{3}+1\right) \,dx$. \ \par \vspace{1pt} \par Hoqwever, this problem can also be done using primitive rules as follows. \par \begin{eqnarray*} \int x(x^{3}+1)\ dx=\int \left( x^{4}+x\right) \,dx \\ =\int x^{4}dx+\int x\,dx \\ =\dfrac{1}{4}x^{5}+\dfrac{1}{2}x^{2}+C \end{eqnarray*}% } \item $\int x^{2}\sqrt{x+3}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{2}\sqrt{x+3}\ dx=\allowbreak \frac{2}{7}\left( \sqrt{\left( x+3\right) }\right) ^{7}-\frac{12}{5}\left( \sqrt{\left( x+3\right) }\right) ^{5}+6\left( \sqrt{\left( x+3\right) }\right) ^{3}+C$} \item $\int x\sec ^{2}x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x\sec ^{2}x\,dx=\allowbreak x\tan x+\ln \left| \cos x\right| +C$% \par Solution. \ Use the parts $u=x$ and $dv=\sec ^{2}x.$ \ Then $du=dx$ and $% v=\tan x.$ \ This gives \begin{eqnarray*} \int x\sec ^{2}x\,dx=\allowbreak x\tan x-\int \tan x\,dx \\ =\allowbreak x\tan x-\int \dfrac{\sin x}{\cos x}\,dx \end{eqnarray*}% \par Now use the substitution $u=\cos x$, with $du=-\sin x\,dx$ to get \begin{eqnarray*} \int x\sec ^{2}x\,dx=\allowbreak x\tan x-\int \dfrac{1}{u}\,dx \\ =\allowbreak x\tan x-\ln \left| u\right| +C \\ =\allowbreak x\tan x-\ln \left| \cos x\right| +C \end{eqnarray*}% } \item $\int x\csc ^{2}2x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x\csc ^{2}2x\,dx=-\frac{1}{2}x\cot 2x+\frac{1}{4}\ln \left| \sin 2x\right| +C$% \par Solution. \ Use the parts $u=x$ and $dv=\csc ^{2}2x.$ \ Then $du=dx$ and $v=-% \dfrac{1}{2}\cot 2x.$ \ This gives \begin{eqnarray*} \int x\csc ^{2}2x\,dx=\allowbreak -\dfrac{1}{2}x\cot 2x+\dfrac{1}{2}\int \cot 2x\,dx \\ =\allowbreak x\tan x+\dfrac{1}{2}\int \dfrac{\cos 2x}{\sin 2x}\,dx \end{eqnarray*}% \par Now use the substitution $u=\sin 2x$, with $du=2\cos 2x\,dx$ to get \begin{eqnarray*} \int x\csc ^{2}2x\,dx=\allowbreak -\dfrac{1}{2}x\cot 2x+\dfrac{1}{4}\int \dfrac{1}{u}\,dx \\ =-\dfrac{1}{2}x\cot 2x+\dfrac{1}{4}\ln \left| u\right| +C \\ =-\dfrac{1}{2}x\cot 2x+\dfrac{1}{4}\ln \left| \sin 2x\right| +C \end{eqnarray*}% } \item $\int \left( x+1\right) \sin \left( x+1\right) dx$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \left( x+1\right) \sin \left( x+1\right) dx=\allowbreak \sin \left( x+1\right) -\left( x+1\right) \cos \left( x+1\right) $\ $+C$% \par Use the parts $u=x+1$ and $dv=\sin \left( x+1\right) \,dx$% \par \vspace{1pt} \par \QTR{red}{Do not} expand the function in the problem for example by writing \[ \sin \left( x+1\right) =\sin x\cos 1+\cos x\sin 1 \] This unnecesssarily complicates the solution.} \item $\int \left( x+2\right) \cos \left( x+1\right) dx$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \left( x+2\right) \cos \left( x+1\right) dx=\allowbreak \sin \left( x+1\right) +\cos \left( x+1\right) +\left( x+1\right) \sin \left( x+1\right) +C$% \par First, expand as \[ \int \left( x+2\right) \cos \left( x+1\right) dx=\int \left( x+1\right) \cos \left( x+1\right) dx+\int \cos \left( x+1\right) dx \]% \par for the first integral on the right use the parts $u=x+1$ and $dv=\cos \left( x+1\right) \,dx.$ \ The second integral is elementary. \par \vspace{1pt} \par Do not expand the function in the problem for example by writing \[ \cos \left( x+1\right) =\cos x\cos 1-\sin x\sin 1 \] This unnecesssarily complicates the solution.} \item $\int x(e^{x}+2)(e^{x}-2)\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x(e^{x}+2)(e^{x}-2)\ dx=\allowbreak \frac{1}{2}% xe^{2x}-\frac{1}{4}e^{2x}-2x^{2}+C$% \par It is best to multiply the integrand and collect simple terms. \ Thus, rewrite as \begin{eqnarray*} \int x(e^{x}+2)(e^{x}-2)\ dx=\int x\left( e^{2x}-4\right) dx \\ =\int xe^{2x}\,dx-4\int x\,dx \end{eqnarray*} For the first integral on the right use the parts $u=x$ and $dv=e^{2x}\,dx$.} \item $\int p^{2}e^{-p}\ dp\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int p^{2}e^{-p}\ dp=\allowbreak -p^{2}e^{-p}-2pe^{-p}-2e^{-p}+C$} \item $\int p^{3}e^{p^{2}}\ dp\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int p^{3}e^{p^{2}}\ dp=\allowbreak \frac{1}{2}% p^{2}e^{p^{2}}-\frac{1}{2}e^{p^{2}}+C$% \par Selecting the parts for this one is more subtle. \ \ Take for the parts $% u=p^{2}$ and $dv=pe^{p^{2}}\,dp$. \ Then $du=2pdp$ and $v=\dfrac{1}{2}% e^{p^{2}}.$}\newline \item $\int s^{2}\ln s\ ds\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int s^{2}\ln s\ ds=\allowbreak \frac{1}{3}s^{3}\ln s-% \frac{1}{9}s^{3}+C$} \item $\int (x^{2}+x-1)e^{3x}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (x^{2}+x-1)e^{3x}\ dx=\allowbreak \frac{1}{3}% x^{2}e^{3x}+\frac{1}{9}xe^{3x}-\frac{10}{27}e^{3x}+C$% \par Take for the parts $u=x^{2}+x-1$ and $dv=e^{3x}\,dx.$ \ You will need two integrations by parts to complete the computation.} \item $\int (x^{2}-x+1)e^{ax}\ dx\;$\ \ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (x^{2}-x+1)e^{ax}\ dx=\allowbreak \dfrac{1}{a}\left( \dfrac{a^{2}x^{2}e^{ax}-2axe^{ax}+2e^{ax}}{a^{2}}-\dfrac{axe^{ax}-e^{ax}}{a}% +e^{ax}\right) +C$% \par \vspace{1pt} \par Take for the parts $u=x^{2}-x+1$ and $dv=e^{ax}\,dx.$ \ You will need two integrations by parts to complete the computation. \par {}} \item $\int (x-1)^{2}e^{-x}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (x-1)^{2}e^{-x}\ dx=\allowbreak -e^{-x}-x^{2}e^{-x}+C $} \item $\int \dfrac{\ln |x|}{x^{2}}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \dfrac{\ln \left| x\right| }{x^{2}}\ dx=\allowbreak -% \dfrac{\ln \left| x\right| +1}{x}+C$% \par Take $u=\ln x$ and $dv=\dfrac{1}{x}\,dx$. \ Then $du=\dfrac{1}{x^{2}}\,dx$ and $v=-x^{-1}$. \par Hence \begin{eqnarray*} \int \dfrac{\ln \left| x\right| }{x^{2}}\ dx=-x^{-1}\ln \left| x\right| +\int \dfrac{1}{x^{2}}\,dx \\ =-x^{-1}\ln \left| x\right| -\frac{1}{x}+C \end{eqnarray*}% } \item $\int 3x\ln (x^{2})\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int 3x\ln (x^{2})\ dx\;=3x^{2}\ln x-\frac{3}{2}x^{2}+C$% \par Remember to first apply the logarithm rule for powers to convert this integral to \[ \int 6x\ln x\,dx \] where we assume tacitly that $x>0$ (Why?) \ The solution follow directly by taking the parts $u=6x$ and $dv=\ln x\,dx.$% \par \vspace{1pt} \par {}} \item $\int x^{3}(x^{2}+1)^{1/2}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% \begin{eqnarray*} \int x^{3}(x^{2}+1)^{1/2}\ dx &=&x^{2}\left( \dfrac{1}{3}\left( x^{2}+1\right) ^{3/2}\right) -\dfrac{2}{3}\int x\left( x^{2}+1\right) ^{3/2}dx \\ &=&\dfrac{1}{3}x^{2}\left( x^{2}+1\right) ^{3/2}-\dfrac{2}{15}\left( x^{2}+1\right) ^{5/2}+C \end{eqnarray*}% \par Solution. This is another problem with unusual parts. \ We take $u=x^{2}$ and $dv=x\left( x^{2}+1\right) ^{1/2}dx.$ \ Then $du=2x\,dx$ and $v=\dfrac{1% }{3}\left( x^{2}+1\right) ^{3/2},$ where the last integral is found by substitution. \par Then \begin{eqnarray*} \int x^{3}(x^{2}+1)^{1/2}\ dx &=&x^{2}\left( \dfrac{1}{3}\left( x^{2}+1\right) ^{3/2}\right) -\dfrac{2}{3}\int x\left( x^{2}+1\right) ^{3/2}dx \\ &=&\dfrac{1}{3}x^{2}\left( x^{2}+1\right) ^{3/2}-\dfrac{2}{15}\left( x^{2}+1\right) ^{5/2}+C \end{eqnarray*}% where the last integral is determined using the substitution method with $% u=x^{2}+1$. \par {}} \qquad \FRAME{dtbpF}{4.2384in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0692in;depth 0pt;original-width 4.0205in;original-height 0.0138in;cropleft "0";croptop "0.6339";cropright "1";cropbottom "0.6334";filename 'maroon3.wmf';file-properties "XNPEU";}} Instructions \item $\int \ln x^{2}\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \ln x^{2}\,dx\;=\allowbreak 2x\ln x-2x+C$% \par \vspace{1pt} \par This is a trick problem. \ Here is how to do it. \ First $\ln x^{2}=2\ln x$, by the rules of logaritms. \ So, \[ \int \ln x^{2}\,dx=2\int \ln x\,dx \]% Let $u=\ln x,$ and $dv=dx.$ \ Then $du=\dfrac{1}{x}dx$ and $v=x.$ \ We have \begin{eqnarray*} 2\int \ln x\,dx &=&2\left( x\ln x-\int 1\,dx\right) \\ &=&2x\ln x-2x+C \end{eqnarray*}% \par {}} \item $\int \arcsin 2x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\dint \arcsin 2x\,dx=\allowbreak x\arcsin 2x+\frac{1}{2}% \sqrt{\left( 1-4x^{2}\right) }+C$% \par \emph{Solution.} Take \[ \begin{array}{ll} u=\arcsin 2x & dv=1\,dx=dx \\ du=\dfrac{2}{\sqrt{1-4x^{2}}}\,dx\qquad & v=x% \end{array}% \]% \par Thus \par \begin{center} $\dint \arcsin 2x\,dx=x\arcsin 2x-2\dint \dfrac{x}{\sqrt{1-4x^{2}}}\,dx$% \end{center} \par To compute the second integral, use the substitution $u=1-4x^{2}$. \ Then $% du=-8x\,dx$ and $x\,dx=-\dfrac{1}{8}\,du$. \ Thus, \begin{eqnarray*} \dint \dfrac{x}{\sqrt{1-4x^{2}}}\,dx &=&-\dfrac{1}{8}\dint \dfrac{1}{\sqrt{u}% }\,du \\ &=&-\dfrac{1}{4}\sqrt{u} \\ &=&-\dfrac{1}{4}\sqrt{1-4x^{2}} \end{eqnarray*}% Therefore, \par \begin{center} \begin{eqnarray*} \dint \arcsin 2x\,dx &=&x\arcsin 2x \\ &&+\dfrac{1}{2}\sqrt{1-4x^{2}}+C \end{eqnarray*}% $\;$% \end{center} \par $\;$Remember the ``replacement'' trick where we rename the multiple of $C$ to be $C$ itself. \par \vspace{1pt} \par {}} \item $\int \arctan 3x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \arctan 3x\,dx=\allowbreak x\arctan 3x-\frac{1}{6}% \ln \left( 1+9x^{2}\right) +C$% \par \emph{Solution.} Take \[ \begin{array}{ll} u=\arctan 3x & dv=1\,dx=dx \\ du=\dfrac{3}{1+9x^{2}}\,dx\qquad & v=x% \end{array}% \]% \par Thus \par \begin{center} $\int \arctan 3x\,dx=x\arctan 3x-\int \dfrac{3x}{1+9x^{2}}\,dx$% \end{center} \par To compute the second integral, use the substitution $u=1+9x^{2}$ \ Then $% du=18x\,dx$ and $3x\,dx=\dfrac{1}{6}\,du$. \ Thus, \begin{eqnarray*} \int \arctan 3x\,dx &=&x\arctan 3x-\int \dfrac{3x}{1+9x^{2}}\,dx \\ &=&x\arctan 3x-\dfrac{1}{6}\int \dfrac{1}{u}\,du \\ &=&x\arctan 3x-\dfrac{1}{6}\ln \left| 1+9x^{2}\right| +C \end{eqnarray*}% \par $\;$% \par {}} \item $\int \arctan 5x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int \arctan 5x\,dx=\allowbreak x\arctan 5x-\frac{1}{10}% \ln \left( 1+25x^{2}\right) +C$% \par \emph{Solution.} Take \[ \begin{array}{ll} u=\arctan 5x & dv=1\,dx=dx \\ du=\dfrac{5}{1+25x^{2}}\,dx\qquad & v=x% \end{array}% \]% \par Thus \par \begin{center} $\int \arctan 5x\,dx=x\arctan 5x-\int \dfrac{5x}{1+25x^{2}}\,dx$% \end{center} \par To compute the second integral, use the substitution $u=1+25x^{2}$ \ Then $% du=50x\,dx$ and $5x\,dx=\dfrac{1}{10}\,du$. \ Thus, \begin{eqnarray*} \int \arctan 3x\,dx &=&x\arctan 5x-\int \dfrac{5x}{1+25x^{2}}\,dx \\ &=&x\arctan 5x-\dfrac{1}{10}\int \dfrac{1}{u}\,du \\ &=&x\arctan 5x-\dfrac{1}{10}\ln \left| 1+25x^{2}\right| +C \end{eqnarray*}% \par $\;$% \par {}} \qquad \FRAME{dtbpF}{4.2384in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0692in;depth 0pt;original-width 4.0205in;original-height 0.0138in;cropleft "0";croptop "0.5504";cropright "1";cropbottom "0.55";filename 'maroon3.wmf';file-properties "XNPEU";}} Instructions \item $\int_{1}^{e}\ln x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\allowbreak \int_{1}^{e}\ln x\,dx=1$% \par \emph{Solution.} \ Let $u=\ln x,$ and $dv=dx.$ \ Then $du=\dfrac{1}{x}dx$ and $v=x.$ \ We have \begin{eqnarray*} \int_{1}^{e}\ln x\,dx &=&\left[ x\ln x\right] _{1}^{e}-\int_{1}^{e}1\,dx \\ &=&e\ln e-1\ln 1-\left. x\right| _{1}^{e} \\ &=&e-\left( e-1\right) =1 \end{eqnarray*}% Here we have used the facts that $\ln e=1$ and $\ \ln 1=0$. \par {}} \item $\int_{0}^{\pi /4}\arcsin 2x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{0}^{1/2}\arcsin 2x\,dx\;=\dfrac{\pi }{4}\allowbreak -\dfrac{1}{2}$% \par \emph{Solution.} Take \[ \begin{array}{ll} u=\arcsin 2x & dv=1\,dx=dx \\ du=\dfrac{2}{\sqrt{1-4x^{2}}}\,dx\qquad & v=x% \end{array}% \]% \par Thus \par \begin{center} $\dint_{0}^{1/2}\arcsin x\,dx=\left[ x\arcsin 2x\right] _{0}^{1/2}-2% \dint_{0}^{1/2}\dfrac{x}{\sqrt{1-4x^{2}}}\,dx$% \end{center} \par To compute the second integral, use the substitution $u=1-4x^{2}$. \ Then $% du=-8x\,dx$ and $x\,dx=-\dfrac{1}{8}\,du$. \ Thus, \begin{eqnarray*} \dint \dfrac{x}{\sqrt{1-4x^{2}}}\,dx &=&-\dfrac{1}{8}\dint \dfrac{1}{\sqrt{u}% }\,du \\ &=&-\dfrac{1}{4}\sqrt{u} \\ &=&-\dfrac{1}{4}\sqrt{1-4x^{2}} \end{eqnarray*}% Therefore, \par \begin{center} \begin{eqnarray*} \dint_{0}^{1/2}\arcsin x\,dx &=&\left[ x\arcsin 2x\right] _{0}^{1/2} \\ &&+\left[ \dfrac{1}{2}\sqrt{1-4x^{2}}\right] _{0}^{1/2} \\ &=&\dfrac{1}{2}\arcsin 1+\left( 0-\dfrac{1}{2}\right) \\ &=&\dfrac{\pi }{4}-\dfrac{1}{2} \end{eqnarray*}% $\;$% \end{center} \par {}} \item $\int_{-2}^{2}x\cos x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{-2}^{2}x\cos x\,dx=\allowbreak 0$% \par \emph{Solution.} \ Let $u=x,$ and $dv=\cos x\,dx.$ \ Then $du=dx$ and $% v=\sin x.$ \ We have \begin{eqnarray*} \int_{-2}^{2}x\cos x\,dx &=&\left[ x\sin x\right] _{-2}^{2}-\int_{-2}^{2}% \sin x\,dx \\ &=&2\sin 2-\left( -2\sin \left( -2\right) \right) +\cos |_{-2}^{2} \\ &=&2\sin 2-2\sin 2+(\cos 2-\cos \left( -2\right) \\ &=&0 \end{eqnarray*}% We have used the facts that $\sin \left( -x\right) =-\sin x$ for all $x$, and that $\cos \left( -x\right) $ $\cos x$ for all $x$.} \item $\int_{0}^{1}xe^{4x}\,dx$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{0}^{1}xe^{4x}\,dx=\allowbreak \frac{3}{16}e^{4}+% \frac{1}{16}$% \par \emph{Solution.} \ Let $u=x,$ and $dv=e^{4x}dx.$ \ Then $du=dx$ and $v=% \dfrac{1}{4}e^{4x}.$ \ Then \begin{eqnarray*} \int_{0}^{1}xe^{4x}\,dx &=&\left[ \dfrac{1}{4}xe^{4x}\right] _{0}^{1}-\dfrac{% 1}{4}\int_{0}^{1}e^{4x}\,dx \\ &=&\left[ \dfrac{1}{4}xe^{4x}\right] _{0}^{1}-\dfrac{1}{4}\left[ \dfrac{1}{4}% e^{4x}\right] _{0}^{1} \\ &=&\dfrac{1}{4}\allowbreak \left( e^{4}-0e^{0}\right) -\dfrac{1}{16}\left[ e^{4}-e^{0}\right] \\ &=&\frac{3}{16}e^{4}+\frac{1}{16} \end{eqnarray*}% Here we have used the fact that $e^{0}=1$. \par {}} \item $\int_{0}^{1}xe^{x}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{-1}^{1}xe^{x}\ dx=\allowbreak 2e^{-1}$} \item $\int_{1}^{2}x\ln |x|\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{1}^{2}x\ln \left| x\right| \ dx=\allowbreak 2\ln 2-% \frac{3}{4}$} \item $\int_{0}^{3}\ln |x+1|\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{0}^{3}\ln \left| x+1\right| \ dx=\allowbreak 8\ln 2-3$} \item $\int_{0}^{1}x\ln |x+1|\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{0}^{1}x\ln \left| x+1\right| \ dx=\allowbreak \frac{% 1}{4}$} \item $\int_{2}^{4}(x-2)e^{x+1}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{2}^{4}(x-2)e^{x+1}\ dx=\allowbreak e^{5}+e^{3}$} \item $\int_{0}^{1}3x\sqrt{2x+1}\ dx\;$\ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int_{0}^{1}3x\sqrt{2x+1}\ dx\;$} \qquad \FRAME{dtbpF}{4.2384in}{0.0692in}{0pt}{}{}{maroon.wmf}{\special% {language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0692in;depth 0pt;original-width 4.0205in;original-height 0.0138in;cropleft "0";croptop "0.5504";cropright "1";cropbottom "0.55";filename 'maroon3.wmf';file-properties "XNPEU";}} Compute the following integrals \item $\int x^{3}e^{x}\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{3}e^{x}\,dx=\allowbreak x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+C$% \par Solution. \ Let \[ \begin{array}{ll} u=x^{3} & dv=e^{x}dx \\ du=3x^{2}dx & v=e^{x}% \end{array}% \]% Hence \[ \int x^{3}e^{x}\,dx=x^{3}e^{x}-3\int x^{2}e^{x}\,dx \]% Now select the \ parts \[ \begin{array}{ll} u=x^{2} & dv=e^{x}dx \\ du=2x\,dx & v=e^{x}% \end{array}% \]% Hence \[ \int x^{3}e^{x}\,dx=x^{3}e^{x}-3\left( x^{2}e^{x}-2\int xe^{x}\,dx\right) \]% Finally select the parts \[ \begin{array}{ll} u=x^{{}} & dv=e^{x}dx \\ du=dx & v=e^{x}% \end{array}% \]% Hence \begin{eqnarray*} \int x^{3}e^{x}\,dx &=&x^{3}e^{x}-3x^{2}e^{x}+6\left( xe^{x}-\int e^{x}\,dx\right) \\ &=&x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+C \end{eqnarray*}% } \item $\int x^{2}\cos 2x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{2}\cos 2x\,dx=\allowbreak \frac{1}{2}x^{2}\sin 2x-% \frac{1}{4}\sin 2x+\frac{1}{2}x\cos 2x+C$% \par \emph{Solution.} \ Let \[ \begin{array}{ll} u=x^{2} & dv=\cos 2xdx \\ du=2xdx & v=\dfrac{1}{2}\sin 2x% \end{array}% \]% Hence \[ \int x^{2}\cos 2x\,dx=\dfrac{1}{2}x^{2}\sin 2x-\int x\sin 2x\,dx \]% Now select the \ parts \[ \begin{array}{ll} u=x^{{}} & dv=\sin 2xdx \\ du=dx & v=-\dfrac{1}{2}\cos 2x% \end{array}% \]% Hence \begin{eqnarray*} \int x^{2}\cos 2x\,dx &=&\dfrac{1}{2}x^{2}\sin 2x \\ &&-\left( -\dfrac{1}{2}x\cos 2x+\dfrac{1}{2}\int \cos 2x\,dx\right) \\ &=&\dfrac{1}{2}x^{2}\sin 2x+\dfrac{1}{2}x\cos 2x-\dfrac{1}{4}\sin 2x+C \end{eqnarray*}% \par {}} \item $\int x^{2}e^{13x}\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{2}e^{13x}\,dx=\allowbreak \frac{1}{13}% x^{2}e^{13x}-\frac{2}{13^{2}}xe^{13x}+\frac{2}{13^{3}}e^{13x}\allowbreak +C$% \par \emph{Solution.} \ Let \[ \begin{array}{ll} u=x^{2} & dv=e^{13x}dx \\ du=2xdx & v=\dfrac{1}{13}e^{x}% \end{array}% \]% Hence \[ \int x^{2}e^{13x}\,dx=\dfrac{1}{13}x^{2}e^{13x}-\dfrac{2}{13}\int xe^{13x}\,dx \]% Now select the \ parts \[ \begin{array}{ll} u=x & dv=e^{13x}dx \\ du=\,dx & v=\dfrac{1}{13}e^{13x}% \end{array}% \]% Hence \begin{eqnarray*} \int x^{2}e^{13x}\,dx &=&\dfrac{1}{13}x^{2}e^{13x}-\dfrac{2}{13}\left( \dfrac{1}{13}xe^{13x}-\dfrac{1}{13}\int e^{13x}\,dx\right) \\ &=&\dfrac{1}{13}x^{2}e^{13x}-\dfrac{2}{13^{2}}xe^{13x}+\dfrac{2}{13^{3}}% e^{13x}+C \end{eqnarray*}% } \item $\int x^{2}\sin 5x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{3}e^{x}\,dx=\allowbreak x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+C$% \par Solution. \ Let \[ \begin{array}{ll} u=x^{3} & dv=e^{x}dx \\ du=3x^{2}dx & v=e^{x}% \end{array}% \]% Hence \[ \int x^{3}e^{x}\,dx=x^{3}e^{x}-3\int x^{2}e^{x}\,dx \]% Now select the \ parts \[ \begin{array}{ll} u=x^{2} & dv=e^{x}dx \\ du=2x\,dx & v=e^{x}% \end{array}% \]% Hence \[ \int x^{3}e^{x}\,dx=x^{3}e^{x}-3\left( x^{2}e^{x}-2\int xe^{x}\,dx\right) \]% Finally select the parts \[ \begin{array}{ll} u=x^{{}} & dv=e^{x}dx \\ du=dx & v=e^{x}% \end{array}% \]% Hence \begin{eqnarray*} \int x^{3}e^{x}\,dx &=&x^{3}e^{x}-3x^{2}e^{x}+6\left( xe^{x}-\int e^{x}\,dx\right) \\ &=&x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+C \end{eqnarray*}% } \item $\int x^{3}e^{x}\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{3}e^{x}\,dx=\allowbreak x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+C$% \par Solution. \ Let \[ \begin{array}{ll} u=x^{3} & dv=e^{x}dx \\ du=3x^{2}dx & v=e^{x}% \end{array}% \]% Hence \[ \int x^{3}e^{x}\,dx=x^{3}e^{x}-3\int x^{2}e^{x}\,dx \]% Now select the \ parts \[ \begin{array}{ll} u=x^{2} & dv=e^{x}dx \\ du=2x\,dx & v=e^{x}% \end{array}% \]% Hence \[ \int x^{3}e^{x}\,dx=x^{3}e^{x}-3\left( x^{2}e^{x}-2\int xe^{x}\,dx\right) \]% Finally select the parts \[ \begin{array}{ll} u=x^{{}} & dv=e^{x}dx \\ du=dx & v=e^{x}% \end{array}% \]% Hence \begin{eqnarray*} \int x^{3}e^{x}\,dx &=&x^{3}e^{x}-3x^{2}e^{x}+6\left( xe^{x}-\int e^{x}\,dx\right) \\ &=&x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+C \end{eqnarray*}% } \item $\int (x-1)^{2}e^{-x}\ dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int (x-1)^{2}e^{-x}\ dx=\allowbreak -e^{-x}-x^{2}e^{-x}+C $% \par Solution. \ Let \[ \begin{array}{ll} u=\left( x-1\right) ^{2} & dv=e^{-x}\,dx \\ du=2\left( x-1\right) \,dx & v=-e^{-x}% \end{array}% \]% Hence \[ \int (x-1)^{2}e^{-x}\ dx=-\left( x-1\right) ^{2}e^{-x}+2\int \left( x-1\right) e^{-x}\,dx \]% Now select the \ parts \[ \begin{array}{ll} u=\left( x-1\right) & dv=e^{-x}dx \\ du=\,dx & v=-e^{-x}% \end{array}% \]% Hence \begin{eqnarray*} \int (x-1)^{2}e^{-x}\ dx &=&-\left( x-1\right) ^{2}e^{-x}+2\left( -\left( x-1\right) e^{-x}+\int e^{-x}\right) \\ &=&-\left( x-1\right) ^{2}e^{-x}-2\left( x-1\right) e^{-x}-2e^{-x}+C \end{eqnarray*}% \par {}} \item $\int x^{3}\sin 2x\,dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{3}\sin 2x\,dx=\allowbreak -\frac{1}{2}x^{3}\cos 2x+\frac{3}{4}x^{2}\sin 2x-\frac{3}{8}\sin 2x+\frac{3}{4}x\cos 2x\allowbreak +C$% \par Solution. \ Let \[ \begin{array}{ll} u=x^{3} & dv=\sin 2xdx \\ du=3x^{2}dx & v=-\dfrac{1}{2}\cos 2x% \end{array}% \]% Hence \[ \int x^{3}\sin 2x\,dx=-\dfrac{1}{2}\allowbreak x^{3}\cos 2x+\dfrac{3}{2}\int x^{2}\cos 2x\,dx \]% Now select the \ parts \[ \begin{array}{ll} u=x^{2} & dv=\cos 2xdx \\ du=2x\,dx & v=\dfrac{1}{2}\sin 2x\,dx% \end{array}% \]% and proceed to reduce the power function again. \par {}} \item For any real number $n$ show that \[ \int x^{n}\sin ax\,dx=-\dfrac{1}{a}x^{n}\cos ax+\dfrac{n}{a}\int x^{n-1}\cos ax\,dx \]% This formula is called a reduction formula for integrals.\thinspace $\,$% \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% Show that $\int x^{n}\sin ax\,dx=-\dfrac{1}{a}x^{n}\cos ax-\dfrac{n}{a}\int x^{n-1}\cos ax\,dx$% \par \emph{Solution.} Let $u=x^{n}$ and $dv=\sin ax\,dx.$ \ Then $du=nx^{n-1}$ and $v=-\dfrac{1}{a}\cos ax$. \ Applying integration by parts yields \par \[ \int x^{n}\sin ax\,dx=-\dfrac{1}{a}x^{n}\cos ax-\dfrac{n}{a}\int x^{n-1}\cos ax\,dx \]% \par which it what was to be shown$.$} \item Use the formula in the previous problem to reduce $\int x^{3}\sin 2xdx\,$by one power. \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{3}\sin 2xdx=\allowbreak -\frac{1}{2}x^{3}\cos 2x+% \frac{3}{4}x^{2}\sin 2x-\frac{3}{8}\sin 2x+\frac{3}{4}x\cos 2x\allowbreak +C$% \par We know that$\int x^{n}\sin 2xdx=-\dfrac{1}{a}x^{n}\cos ax+\dfrac{n}{a}\int x^{n-1}\cos ax\,dx$% \par \emph{Solution.} We have $a=2$ and $n=3.$ \ Then \[ \int x^{3}\sin 2xdx=-\dfrac{1}{2}x^{3}\cos 2x+\dfrac{3}{2}\int x^{2}\cos 2x\,dx \]% Note the constant of integration $C$ is not needed.} \item For any real number $n$ show that \[ \int x^{n}\cos ax\,dx=\dfrac{1}{a}x^{n}\sin ax-\dfrac{n}{a}\int x^{n-1}\sin ax\,dx \]% This formula is called a reduction formula for integrals.\thinspace $\,$% \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% Show that $\int x^{n}\sin ax\,dx=-\dfrac{1}{a}x^{n}\cos ax-\dfrac{n}{a}\int x^{n-1}\cos ax\,dx$% \par \emph{Solution.} Let $u=x^{n}$ and $dv=\sin ax\,dx.$ \ Then $du=nx^{n-1}$ and $v=-\dfrac{1}{a}\cos ax$. \ Applying integration by parts yields \par \[ \int x^{n}\sin ax\,dx=-\dfrac{1}{a}x^{n}\cos ax-\dfrac{n}{a}\int x^{n-1}\cos ax\,dx \]% \par which it what was to be shown$.$} \item Use both power-trig reduction formulas above to compute $\int x^{2}\sin 3x\,dx$ \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{2}\sin 3x\,dx\allowbreak +C=\allowbreak -\frac{1}{% 3}x^{2}\cos 3x+\frac{2}{27}\cos 3x+\frac{2}{9}x\sin 3x+C$% \par We know that$\int x^{n}\sin 2xdx=-\dfrac{1}{a}x^{n}\cos ax+\dfrac{n}{a}\int x^{n-1}\cos ax\,dx$ andwe know that $\int x^{n}\cos ax\,dx=\dfrac{1}{a}% x^{n}\sin ax-\dfrac{n}{a}\int x^{n-1}\sin ax\,dx$% \par \emph{Solution.} We have $a=3$ and $n=2.$ \ Then from the first reduction formula \[ \int x^{2}\sin 3xdx=-\dfrac{1}{3}x^{2}\cos 3x+\dfrac{2}{3}\int x\cos 2x\,dx \]% From the second reduction formula with $a=3$ and $n=1$% \[ \int x\cos 2x\,dx=\dfrac{1}{3}x\sin 3x=\dfrac{1}{3}\int \sin 3x\,dx \]% So \begin{eqnarray*} \int x^{2}\sin 3xdx &=&-\dfrac{1}{3}x^{2}\cos 3x+\dfrac{2}{3}\left( \dfrac{1% }{3}x\sin 3x=\dfrac{1}{3}\int \sin 3x\,dx\right) \\ &=&-\dfrac{1}{3}x^{2}\cos 3x+\dfrac{2}{9}x\sin 3x+\dfrac{2}{27}\cos 3x+C \end{eqnarray*}% \par {}} \item Use the formula in the previous problem to reduce $\int x^{3}\sin 2xdx\,$by one power. \dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int x^{3}\sin 2xdx=\allowbreak -\frac{1}{2}x^{3}\cos 2x+% \frac{3}{4}x^{2}\sin 2x-\frac{3}{8}\sin 2x+\frac{3}{4}x\cos 2x\allowbreak +C$% \par We know that$\int x^{n}\sin 2xdx=-\dfrac{1}{a}x^{n}\cos ax-\dfrac{n}{a}\int x^{n-1}\cos ax\,dx$% \par \emph{Solution.} We have $a=2$ and $n=3.$ \ Then \[ \int x^{3}\sin 2xdx=-\dfrac{1}{2}x^{3}\cos 2x-\dfrac{3}{2}\int x^{2}\cos 2x\,dx \]% Note the constant of integration $C$ is not needed.} \item For any real number $n$ show that $\int x^{n}e^{x}\,dx=\allowbreak x^{n}e^{x}-\left( n-1\right) \int x^{n-1}e^{x}\,dx$. $\ $This formula is called reduction formula for integrals.$\,$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% Show that $\int x^{n}e^{x}\,dx=\allowbreak x^{n}e^{x}-n\int x^{n-1}e^{x}\,dx$% \par \emph{Solution.} Let $u=x^{n}$ and $dv=e^{x}\,dx.$ \ Then $du=nx^{n-1}$ and $% v=e^{x}$. \ Applying integration by parts yields \par \[ \int x^{n}e^{x}\,dx=\allowbreak x^{n}e^{x}-n\int x^{n-1}e^{x}\,dx \]% \par which it what was to be shown$.$} \item Use the formula in the previous problem to determine $\int x^{5}e^{x}dx\;$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{% First $\int x^{5}e^{x}\,dx=\allowbreak x^{5}e^{x}-5\int x^{4}e^{x}\,dx$% \par Second $\int x^{4}e^{x}\,dx=x^{4}e^{x}-4\int x^{3}e^{x}\,dx$% \par Next $\int x^{3}e^{x}\,dx=x^{3}e^{x}-3\int x^{2}e^{x}\,dx$ and so on. \ Putting it altogether we get \par \begin{eqnarray*} \int x^{5}e^{x}\,dx &=&\allowbreak x^{5}e^{x}-5\left( x^{4}e^{x}-4\int x^{3}e^{x}\,dx\right) \\ &=&x^{5}e^{x}-5x^{4}e^{x}+20\int x^{3}e^{x}\,dx \\ &=&x^{5}e^{x}-5x^{4}e^{x}+20\left( x^{3}e^{x}-3\int x^{2}e^{x}\,dx\right) \end{eqnarray*}% \par Continuing we get the answer. \par \vspace{1pt} \[ \int x^{5}e^{x}\,dx \]% $=\allowbreak x^{5}e^{x}-5x^{4}e^{x}+20x^{3}e^{x}-60x^{2}e^{x}+\allowbreak 120xe^{x}-120e^{x}+C$% \par \vspace{1pt} \par \emph{Solution.} Let $u=x^{n}$ and $dv=e^{x}\,dx.$ \ Then $du=nx^{n-1}$ and $% v=e^{x}$. \ Applying integration by parts yields \par \[ \int x^{n}e^{x}\,dx=\allowbreak x^{n}e^{x}-n\int x^{n-1}e^{x}\,dx \]% \par which it what was to be shown$.$} \FRAME{dtbpF}{4.2384in}{0.0761in}{0pt}{}{}{maroon.wmf}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "PICT";valid_file "F";width 4.2384in;height 0.0761in;depth 0pt;original-width 290.5625pt;original-height 1pt;cropleft "0";croptop "0.7474";cropright "1";cropbottom "0.7469";filename 'maroon3.wmf';file-properties "XNPEU";}}\qquad Compute the following integrals \item $\int e^{2x}\cos x\,dx$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int e^{2x}\cos x\,dx=\allowbreak \dfrac{2}{5}e^{2x}\cos x+\dfrac{1}{5}e^{2x}\sin x+C$% \par \vspace{1pt} \par \emph{Solution}. First take the parts: \[ \begin{array}{ll} u=e^{2x} & dv=\cos x\,dx \\ du=2e^{2x\,}dx & v=\sin x% \end{array}% \]% $\;$% \par $\;$% \par Hence \par \[ \int e^{2x}\cos x\,dx=e^{2x}\sin x-2\int e^{2x}\sin x\,dx \]% Now take the ``parts'' \[ \begin{array}{ll} u=e^{2x} & dv=\sin x\,dx \\ du=2e^{2x\,}dx & v=-\cos x% \end{array}% \]% Hence \begin{eqnarray*} \int e^{2x}\cos x\,dx &=&e^{2x}\sin x-2\int e^{2x}\sin x\,dx \\ &=&e^{2x}\sin x \\ &&-2\left( -e^{2x}\cos x+2\int e^{2x}\cos \,dx\right) \end{eqnarray*}% Now collect terms and solve for $\int e^{2x}\cos \,dx$. \begin{eqnarray*} 5\int e^{2x}\cos \,dx &=&e^{2x}\sin x+2e^{2x}\cos x+C \\ \int e^{2x}\cos \,dx &=&\dfrac{1}{5}e^{2x}\sin x+\dfrac{2}{5}e^{2x}\cos x+C \end{eqnarray*}% Remember the ``replacement'' trick where we rename the multiple of $C$ to be $C$ itself.} \item $\int e^{-x}\cos x\,dx$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int e^{-x}\cos x\,dx=\allowbreak -\dfrac{1}{2}e^{-x}\cos x+\dfrac{1}{2}e^{-x}\sin x+C$% \par {} \par \emph{Solution}. First take the parts: \[ \begin{array}{ll} u=e^{-x} & dv=\cos x\,dx \\ du=-e^{-x\,}dx & v=\sin x% \end{array}% \]% $\;$% \par Hence \par \[ \int e^{-x}\cos x\,dx=-e^{-x}\sin x+\int e^{-x}\sin x\,dx \]% Now take the ``parts'' \[ \begin{array}{ll} u=e^{-x} & dv=\sin x\,dx \\ du=-e^{-x\,}dx & v=-\cos x% \end{array}% \]% Hence \begin{eqnarray*} \int e^{-x}\cos x\,dx &=&-e^{-x}\sin x+\int e^{-x}\sin x\,dx \\ &=&-e^{-x}\sin x \\ &&+\left( e^{-x}\cos x-\int e^{-x}\cos \,dx\right) \end{eqnarray*}% Now collect terms and solve for $\int e^{-x}\cos \,dx$. \begin{eqnarray*} 2\int e^{x}\cos \,dx &=&-e^{-x}\sin x+-e^{-x}\cos x+C \\ \int e^{-x}\cos \,dx &=&-\dfrac{1}{2}e^{-x}\sin x+\dfrac{1}{2}e^{-x}\cos x+C \end{eqnarray*}% Remember the ``replacement'' trick where we rename the multiple of $C$ to be $C$ itself.} \item $\int e^{3x}\sin 2x\,dx$\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int e^{3x}\sin 2x\,dx=\allowbreak -\dfrac{2}{13}% e^{3x}\cos 2x+\dfrac{3}{13}e^{3x}\sin 2x$% \par \emph{Solution}. First take the parts: \[ \begin{array}{ll} u=e^{3x} & dv=\sin 2x\,dx \\ du=3e^{3x\,}dx & v=-\dfrac{1}{2}\cos 2x% \end{array}% \]% $\;$% \par Hence \par \[ \int e^{3x}\sin 2x\,dx=-\dfrac{1}{2}e^{3x}\cos 2x+\dfrac{3}{2}\int e^{3x}\cos 2xdx \]% Now take the ``parts'' \[ \begin{array}{ll} u=e^{3x} & dv=\cos 2x\,dx \\ du=3e^{3x\,}dx & v=\dfrac{1}{2}\sin 2x% \end{array}% \]% Hence \begin{eqnarray*} \int e^{3x}\sin 2x\,dx &=&-\dfrac{1}{2}e^{3x}\cos 2x+\dfrac{3}{2}\int e^{3x}\cos 2xdx \\ &=&-\dfrac{1}{2}e^{3x}\cos 2x+\dfrac{3}{2}\left( \dfrac{1}{2}e^{3x}\sin 2x-% \dfrac{3}{2}\int e^{3x}\sin 2x\,dx\right) \end{eqnarray*}% Now collect terms and solve for $\int e^{3x}\sin 2x\,dx$. \[ \left( 1+\dfrac{9}{4}\right) \int e^{3x}\sin 2x\,dx=-\dfrac{1}{2}e^{3x}\cos 2x+\dfrac{3}{4}e^{3x}\sin 2x+C \]% \par \[ \int e^{3x}\sin 2x\,dx=-\dfrac{2}{13}e^{3x}\cos 2x+\dfrac{3}{13}e^{3x}\sin 2x+C \]% \par Remember the ``replacement'' trick where we rename the multiple of $C$ to be $C$ itself.} \item $\int xe^{x}\sin x\,dx$\ (This one is tough.)\dotfill \CustomNote[\hyperref{\TCIIcon{BITMAPSETAnswer}{0.1609in}{0.1487in}{0in}}{}{% }{}]{Margin Hint}{$\int xe^{x}\sin x\,dx=\allowbreak \left( -\frac{1}{2}x+% \frac{1}{2}\right) e^{x}\cos x+\frac{1}{2}xe^{x}\sin x$% \par {} \par \emph{Solution}. First take the parts: \[ \begin{array}{ll} u=x & dv=e^{x}\sin x\,dx \\ du=dx & v=\text{???}% \end{array}% \]% $\;$To find $v$ it is necessary to integration by parts twice as in a recurrence relation. \ In fact, this has already been done. \ It is \[ v=\int e^{x}\sin x\,dx \]% $=\allowbreak -\frac{1}{2}e^{x}\cos x+\frac{1}{2}e^{x}\sin x$% \par Hence \par \begin{eqnarray*} \int xe^{x}\sin x\,dx &=&xe^{x}\sin x \\ &&-\int \left( -\frac{1}{2}e^{x}\cos x+\frac{1}{2}e^{x}\sin x\right) dx \end{eqnarray*}% Each of the integrals above must be integrated by the recurrence method. \ This gives \[ \int \left( -\frac{1}{2}e^{x}\cos x+\frac{1}{2}e^{x}\sin x\right) dx \]% } \end{enumerate} \vspace{1pt} \vspace{0in} \end{document} %%%%%%%%%%%%%%%%%%%% End /document/lec_11_15_99.tex %%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%% Start /document/maroon3.wmf %%%%%%%%%%%%%%%%%%%%% WwlqZB@@W\ObOplAV~@zC@@@@@@OeF@@I@@@CLJB@@@A@x@@@@@@@P@@@@p@A`@@E@@@@l`@@@@ 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