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

Problem Set 4

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*}
\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
\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
\[
\frac{e^x}{2} \sqrt{1 + e^{2x}} + \frac{1}{2} \ln | e^x + \sqrt{1 + e^{2x}} | + C.
\]

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