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\begin{center}
Math 132: Calculus for the Physical \& Life Sciences 2

Problem Set 4 -- Answers

%Due Friday, February 15, 2008, at the beginning of class.
\end{center}


\vskip 10pt
%\noindent
%General Directions:  You must show all work for credit on these
%problems.

\begin{enumerate}
\item 
%Use trigonometric substitutions (possibly after completing
%the square) to evaluate each of the following
%integrals.

\begin{enumerate}
\item[a.]  
%${\displaystyle \int \frac{1}{\sqrt{9 + 4x^2}}\ dx}$
%\vskip 8pt
%\noindent
%{\it Solution:}  The integral is $\frac{1}{3} \int \frac{dx}{\sqrt{1 + (2x/3)^2}}$.
%Let $2x/3 = \tan\theta$, so $\sqrt{1+(2x/3)^2} = \sec\theta$ and $dx = (3/2) \sec^2(\theta)\, d\theta$.
%The integrand becomes
%\begin{eqnarray*}
%\frac{1}{2} \int \frac{\sec^2\theta\ d\theta}{\sec\theta}
%&=& \frac{1}{2} \int \sec\theta\ d\theta\\
%&=& \frac{1}{2} \ln|\sec\theta + \tan\theta| + C\\
%&=& \frac{1}{2} \ln|\sqrt{9 + 4x^2}/3 + 2x/3| + C,\\
%\end{eqnarray*}
%which can also be written as 
\[
\frac{1}{2} \ln|\sqrt{9 + 4x^2} + 2x| + C\\
\]
%(by incorporating $-\ln(3)/2$ into the constant of integration).

%\vskip 10pt
%\item[b.]  ${\displaystyle \int \frac{\sqrt{x^2 - 1}}{x^2}\ dx}$
%\vskip 8pt
%\noindent
%{\it Solution:}  Let $x = \sec\theta$, so $dx = \sec\theta\tan\theta\, d\theta$.
%The integrand becomes
%\begin{eqnarray*}
%\int \frac{\tan\theta \sec\theta\tan\theta\, d\theta}{\sec^2\theta}
%&=& \int \frac{\tan^2\theta \, d\theta}{\sec\theta}\\
%&=& \int \frac{\sec^2\theta - 1\, d\theta}{\sec\theta}\\
%&=& \int \sec\theta - \cos\theta\, d\theta\\
%&=& \ln|\sec\theta + \tan\theta| - \sin\theta + C\\
%&=& \ln|x + \sqrt{x^2 - 1}| - \frac{\sqrt{x^2 - 1}}{x} + C
%\end{eqnarray*}
%\vskip 8pt
\item[c.]  
%${\displaystyle \int \frac{1}{\sqrt{3 - 2x - x^2}}\ dx}$
%\vskip 10pt
%\noindent
%{\it Solution:}  Completing the square, $3 - 2x - x^2 = 4 - (x+1)^2$.
%So we let $x + 1 = 2\sin\theta$ and $dx = 2\cos\theta\, d\theta$.
%The integrand becomes
%\begin{eqnarray*}
%\int \frac{2\cos\theta\, d\theta}{2\cos\theta} & = & \int\,  d\theta\\
%&=& \theta + C\\
%&=& \sin^{-1}\left(\frac{x+1}{2}\right) + C.
%\end{eqnarray*}
\[
\sin^{-1}\left(\frac{x+1}{2}\right) + C.
\]
\end{enumerate}

\item  
%Integrate using the partial fraction technique (divide the 
%denominator polynomial into the top first when necessary).

\begin{enumerate}
%\item[a.]  ${\displaystyle \int \frac{1}{x^2 + x - 6}\ dx}$
%\vskip 10pt
%\noindent
%{\it Solution:}
%The denominator factors as $(x + 3)(x - 2)$, so the partial fractions are
%\[
%\frac{1}{x^2 + x - 6} = \frac{A}{x + 3} + \frac{B}{x-2}.
%\]
%This gives $1 = A(x - 2) + B(x + 3)$, so $B = 1/5$ and $A = -1/5$.
%Then 
%\[
%\int \frac{1}{x^2 + x - 6}\, dx = \int \frac{-1/5}{x + 3} + \frac{1/5}{x - 2}\ dx = 
%\frac{-1}{5}\ln|x + 3| + \frac{1}{5} \ln|x - 2| +C
%\]
%This could also be written as $\frac{1}{5}\ln\left|\frac{x-2}{x+3}\right| + C$
%using properties of logarithms. 
%\vskip 8pt
\item[b.]  
%${\displaystyle \int \frac{x^4}{x^2 + 4}\ dx}$
%\vskip 10pt
%Since the degree of the numerator is larger than the degree of the denominator,
%we divide first, yielding $x^4 = (x^2-4)(x^2 + 4) + 16$.  Therefore
%\begin{eqnarray*}
%\int \frac{x^4}{x^2 + 4}\ dx
%& = & \int x^2 - 4\ dx + \int \frac{16\ dx}{x^2 + 4}\\
%& = & \frac{x^3}{3} - 4x + 8\tan^{-1}(x/2) + C
%\end{eqnarray*}  
%\vskip 8pt
\[
\frac{x^3}{3} - 4x + 8\tan^{-1}(x/2) + C.
\]
%\item[c.]  ${\displaystyle \int \frac{x + 10}{2x^2 + 5x - 3}\ dx}$
%\vskip 10pt
%\noindent
%{\it Solution:} The denominator factors as $(2x-1)(x+3)$, so the 
%partial fractions are
%\[
%\frac{x + 10}{(2x-1)(x+3)} = \frac{A}{2x - 1} + \frac{B}{x+3}.
%\]
%This gives $x + 10 = A(x+3) + B(2x-1)$, so $A = 3$ and $B = -1$.
%Then 
%\[
%\int \frac{x+10}{(2x-1)(x+3)}\, dx = \int \frac{3}{2x-1} + \frac{-1}{x+3}\ dx = 
%\frac{3}{2}\ln|2x - 1| - \ln|x + 3| + C.
%\]

%\vskip 8pt
\item[d.]  
%${\displaystyle \int \frac{x^2+1}{x^3 + 2x^2 + x}\ dx}$

%\vskip 10pt
%\noindent
%{\it Solution:}
%The denominator factors as $x(x^2 + 2x + 1) = x(x+1)^2$.  Because
%of the repeated linear factor, the partial fractions here will be
%\[
%\frac{x^2+1}{x^3 + 2x^2 + x} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2}.
%\]
%After clearing denominators,
%\[
%x^2 + 1 = A(x+1)^2 + Bx(x+1) + Cx.
%\]
%Hence $A = 1$, $B = 0$ and $C = -2$.  
%So the integral is 
%\[
%\int \frac{1}{x} - \frac{2}{(x+1)^2}\ dx = 
\[
\ln|x| + \frac{2}{x+1} + C.
\]
%\]
%\vskip 8pt
%\item[e.]  ${\displaystyle \int \frac{x + 4}{x^3 + 4x}\ dx}$
%\vskip 10pt
%\noindent
%{\it Solution:}
%The denominator factors as $x(x^2 + 4)$.  Because of the quadratic factor,
%the partial fractions look like
%\[
%\frac{x + 4}{x^3 + 4x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4},
%\]
%so after clearing denominators,
%\[
%x + 4 = A(x^2 + 4) + (Bx + C)x.
%\]
%This gives $C = 1$, $A = 1$ and $B = -1$.  So
%\[
%\int \frac{x + 4}{x^3 + 4x} = \int \frac{1}{x} + \frac{-x + 1}{x^2 + 4}\ dx
%= \ln|x| - \frac{1}{2}\ln(x^2 + 4) + \frac{1}{2}\tan^{-1}(x/2) + C.
%\]
\end{enumerate}

\item  
%Integrate using the table of integrals in Stewart, and preliminary
%substitutions or algebraic manipulation as necessary.  In each case,
%state clearly the number of the table entry you are using.

\begin{enumerate}
%\vskip 8pt
\item[a.]   
%${\displaystyle \int \frac{x^2}{\sqrt{x^2 + 25}}\ dx}$
%\vskip 10pt
%{\it Solution:}  
%Use number 26 in the table with $a = 5$.  The answer is 
\[
\frac{x}{2}\sqrt{x^2 + 25} - \frac{25}{2}\ln|x + \sqrt{25 + x^2}| + C.
\]
%\vskip 8pt
%\item[b.]   ${\displaystyle \int \sqrt{2x - x^2}\ dx}$
%\vskip 10pt
%\noindent
%{\it Solution:}  Complete the square inside the radical:
%$\sqrt{2x - x^2} = \sqrt{1 - (x - 1)^2}$.  So after the substitution 
%$u = x - 1$, this is $\int \sqrt{1 - u^2}\ du$, which is number 30 in 
%the table with $a = 1$.  The integral is
%\[
%\frac{x-1}{2}\sqrt{1 - (x-1)^2} + \frac{1}{2}\sin^{-1}(x-1) + C.
%\]
%\vskip 8pt
\item[c.]   
%${\displaystyle \int \frac{1}{\sqrt{e^{2x} - 1}}\, dx}$
%\vskip 10pt
%\noindent
%{\it Solution:}
%First substitute $u = e^x$, so $du = e^x dx$ and $dx = du/e^x = du/u$.
%The form is $\int \frac{1}{u\sqrt{u^2 - 1}}\, du$, which is number 18 in 
%the table with $a = 1$.  The answer is
\[
\sec^{-1}(e^x) + C.
\]
%\vskip 8pt
%\item[d.]   ${\displaystyle \int \frac{\cos(x)}{\sqrt{4 - \sin^2(x)}}\ dx}$

%\vskip 10pt
%\noindent
%{\it Solution:}  First substitute $u = \sin x$, so $du = \cos x\, dx$.
%The form is $\int \frac{1}{\sqrt{4 - u^2}}\, du$, which is number 16 in the table with
%$a = 2$.  The integral is 
%\[
%\sin^{-1}\left(\frac{\sin(x)}{2}\right) + C.
%\] 
%\vskip 8pt
\item[e.]   
%${\displaystyle \int \frac{\sec^2(x)}{\tan^2(x) + 2\tan(x) + 2}\ dx}$
%\vskip 10pt
%\noindent
%{\it Solution:}  The substitution $u = \tan(x)$, $du = \sec^2(x)\, dx$ converts this to 
%$\int \frac{du}{u^2 + 2u + 2} = \int \frac{du}{(u+1)^2 + 1}$, which is in the
%basic arctangent integral form for $v = u+1$ (number 17 in the table).  Answer
%is 
\[
\tan^{-1}(\tan(x) + 1) + C.
\]
%\vskip 8pt
%\item[f.]   ${\displaystyle \int e^x \sqrt{1 + e^{2x}}\ dx}$
%\vskip 10pt
%\noindent
%{\it Solution:}  Substitute $u = e^x$, then $du = e^x dx$, so the form
%is $\int \sqrt{1 + u^2}\ du$, which is number 21 in the table with $a = 1$.
%Answer is
%\[
%e^x \sqrt{1 + e^{2x}} + \frac{1}{2} \ln | e^x + \sqrt{1 + e^{2x}} | + C.
%\]

\end{enumerate}
\end{enumerate}
\end{document}
