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\noindent MTH 154 \hfill Helson---Additional 
Problems \hfill Spring 09

\noindent
Prof. Cowen
\vskip 18 pt

\noindent
{\bf Chapter 5.5}
\vskip 12 pt

\noindent 5.5.1.  For which $p$ does the improper integral $\displaystyle 
{\int_{4}^{\infty}
\frac {1}{x (\ln x)^p} \, dx}$ converge/diverge?
\vskip 12 pt

\noindent 5.5.2.  For which $p$ does the improper integral $\displaystyle {\int_{4}^{\infty}
\frac {1}{x (\ln x)(\ln \ln x)^p} \, dx}$ converge/diverge?
\vskip 12 pt

\noindent 5.5.3.  Does $\displaystyle {\int_{10}^{\infty} \frac {1}{\sqrt{x^{3} -
5x}} \,dx}$  converge or diverge? Why?
\vskip 12 pt

\noindent 5.5.4.  Does $\displaystyle {\int_{0}^{1} \frac {1}{\sqrt{3x - x^{5}}}
\, dx}$ converge or diverge?  Why?  
\vskip 12 pt

\noindent 5.5.5.  Does $\displaystyle {\int_{1}^{\infty}
\frac {x}{\sqrt{x^{4} -1}} \, dx}$ converge or diverge? Why?
\vskip 12 pt

\noindent
{\bf Chapter 6.1}
\vskip 12 pt

\noindent 6.1.1. Let $a_n = 1 + \frac {1}{2} + ... + \frac {1}{n} - \ln n$ . Show that 
$a_n$ is decreasing. 

\noindent Hint: $\ln (n +1) - \ln n = \int_{n}^{n + 1} \frac {1}{x} \, dx$ . 
Why is this greater than or equal to  $\frac {1}{n + 1}$? 
\vskip 12 pt
       
\noindent 6.1.2. Show that $a_n$ is bounded below. Hint: 
$1 + \frac {1}{2} + ... + \frac {1}{n - 1} \ge \int_{1}^{n} \frac {1}{x} \,
dx$ . 
\vskip 12 pt

Note: The limit of $a_{n}$ as $n \to \infty$ is the Euler constant
$\gamma = .57721 56649  \ldots$.  \vskip 12 pt

\noindent 6.1.3.  Let  $f(x)$  and  $g(x)$  be positive functions which are
continuous on the interval $(a,b\,]$.  
Let  $L = \lim_{x \to a^{+}} \frac {f(x)}{g(x)}$.  State and prove a theorem
relating the improper integrals $\int_{a}^{b} f(x) \, dx$ and
$\int_{a}^{b} g(x) \, dx$ when  $L = 0$.  {\it Hint: This is similar
to the case where $L \ne 0, \infty$, but $L = 0$ means
that $f(x)$  is much smaller than $g(x)$  near  $a$.}
\vskip 12 pt

\noindent 6.1.4.  Same question as problem 6.1.3 but for $L = \infty$.
{\it Hint: $L = \infty$ means
that $f(x)$  is much larger than $g(x)$  near  $a$.}
\vskip 12 pt

\noindent 6.1.5.  We showed in class that an increasing sequence 
bounded above has a limit. Use this result to show that a decreasing 
sequnce bounded below has a limit.
\vskip 12 pt

\noindent 6.1.6.  We showed in class that a continuous function on a 
closed finite interval is bounded above. Use this result to prove 
that continuous function on a closed finite interval is bounded below.
\vskip 12 pt

\noindent 6.1.7.  Let  $f(x)$ be a continuous function defined
on a closed finite interval
$[a,b\,]$.  Prove that the range of  $f$  is a closed finite interval
$[c,d\,]$.  {\it Hint:  draw the graph of a general continuous function. 
What are $c$ and $d$? Now prove the result.}
\vskip 12 pt

\noindent 6.1.8.  Draw the graph of a continous function defined on an open finite
interval $(a, b)$ such that the range is a closed finite interval 
$[c,d\,]$.
\vskip 12 pt


\noindent
{\bf Chapter 6.4}
\vskip 12 pt

\noindent 6.4.1.  Use the Taylor approximation to find $e^{.5}$ to eight 
decimal places. Include an analysis of how large $n$ has to be. You 
may use a calculator or computer to do the arithmetic. Check your 
answer against the actual value for $e^{.5}$ given by your calculator.
\vskip 12 pt

\noindent 6.4.2. Same question for $\sin 0.2$.
\vskip 12 pt

\noindent 6.4.3. We showed that $\displaystyle {1/\ln \frac {1}{x}}$ goes to 0 slower than  
$x^{p}$ for any real number $p > 0$, in the sense that 
\[
\lim_{x \to 0^{+}} \frac {x^{p}}{1/\ln \frac {1}{x}} = 0.
\] 

Give an example of a function $f(x)$ defined for $x > 0$ such that 
$f(x)$ goes to $0$ as $x \to 0^{+}$, but $f(x)$ goes to $0$ slower than  
$\displaystyle {\left(1/\ln \frac {1}{x}\right)^{p}}$ for any real number $p > 0$.

Prove that your example actually does what it is supposed to do.
\vskip 12 pt

\noindent 6.4.4.  (Continuation of 6.4.3) Give an example of a function
$g(x)$ defined for $x > 0$ such that $g(x)$ goes to $0$ as $x \to
0^{+}$, but $g(x)$ goes to $0$ slower than $\left(f(x)\right)^{p}$ for any
real number $p > 0$.

Prove that your example actually does what it is supposed to do.
\vskip 12 pt

\noindent 6.4.5.  (Continuation of 6.4.4) Give an example of a sequence of 
functions $f_{1}(x), f_{2}(x), \ldots $ defined for $x > 0$ such that 
$f_{1}(x) = x$; 
for each $n$,
$f_{n}(x)$ goes to $0$ as $x \to
0^{+}$; but $f_{n+1}(x)$ goes to $0$ slower than $\left(f_{n}(x)\right)^{p}$
for any real number $p > 0$.

Do not give a proof.
\vskip 12 pt


\noindent 6.4.6. Prove that $e$ is irrational as follows:

If $e$ is rational, then $e = m/n$ where $m$ and $n$ are positive 
integers with no common factors. Note that $n \ge 2$, since 
otherwise $n =1$ and $e = m$; this would imply that $e$ is an
integer, which we know is not true by calculation. 

Use Taylor's  Theorem to show that

\[ 
0 < e - \{1 + 1/1! + \ldots + 1/n! \} < 
3/(n+1)!.
\]

Now let $r = n! (m/n - \{1 + 1/1! + \ldots + 1/n! \})$. Show that $r$ 
is an integer and that $0 < r < 3/(n+1)$.  Why is this a contradiction?
{\em Note: to prove that $\pi$ is irrational is infinitely harder.}
\vskip 12 pt

\noindent 6.4.7.  We showed in class that: if $f(x)$ is differentiable at
$a$; if $L(x) = f(a) + f'(a)(x-a)$ is the linearization; and if $R(x)
= f(x) - L(x)$, then 
\[
\lim_{x->a} \frac {R(x)}{x-a} = 0.
\]
Let $\tilde
L(x)$ be any linear function of $x$, and let $\tilde R(x) = f(x) - 
\tilde L(x)$. If 
\[
\lim_{x->a} \frac {\tilde R(x)}{x-a} = 0,
\]
 show 
that $\tilde L(x) = L(x)$. Hint: write $\tilde L(x)$ in the form 
$L(x) = b_{0} + b_{1}(x-a)$.
\vskip 12 pt
\newpage

\noindent
{\bf Chapter 6.5}
\vskip 12 pt

\noindent 6.5.1.  Let $f$ and $g$ be continuous on $[a, b\,]$  and 
differentiable on $(a, b)$, with  $g'(x) \ne 0$  for all  $x$  in $(a, b)$.
\vskip 12 pt

\noindent i. Prove that $g(b) \ne g(a)$.
\vskip 12 pt

\noindent ii. Let 
\[h(x) = \left ( g(x) - g(a)\right) \left( f(b) - f(a)\right) - 
\left( f(x) - f(a)\right) \left( g(b) - g(a)\right).
\]
\noindent Show that $h(a) = h(b) = 0$.
\vskip 12 pt

\noindent iii. Use $h(x)$ from ii. to prove there is a  $c$ in $(a, b)$ such
that
\[
\frac {f(b) - f(a)} {g(b) - g(a)} = \frac {f'(c)} {g'(c)}
\]
\noindent Hint:  prove that $h'(c) = 0$ for some $c$ in $(a, b)$.
\vskip 12 pt

\noindent 6.5.2.  $\displaystyle {\lim_{x \to 1} \frac {\ln x}{1-x}}$
\vskip 12 pt

\noindent 6.5.3.  $\displaystyle {\lim_{x \to 0} \frac {e^{x}-1}{\ln (1+x)}}$
\vskip 12 pt

\noindent 6.5.4.  $\displaystyle {\lim_{x \to 1} \frac {\ln (x +2)}{1+x}}$
\vskip 12 pt

\noindent 6.5.5.  $\displaystyle {\lim_{x \to 0} (e^{x} + x)^{1/x}}$
\vskip 12 pt

\noindent 6.5.6.  $\displaystyle {\lim_{x \to \infty} x^{2} \sin 1/x}$
\vskip 12 pt

\noindent 6.5.7.  $\displaystyle {\lim_{n \to \infty} (n^{2} + n)^{1/n}}$
\vskip 12pt
\noindent Hint: if $f(x) \to L$ as $x \to \infty$ then 
$f(n)  \to L$ as $n \to \infty$. So switch the variable to $x$, find 
the limit, then switch back to  $n$ (note: it doesn't make sense to 
take derivative with respect to $n$, but it does make sense to take 
derivative with respect to $x$.)
\vskip 12 pt

\noindent 6.5.8.  $\displaystyle {\lim_{x \to 0^{+}} \frac {x^{p}} {1/\ln x}}$  
\qquad for $p > 0$.
\vskip 12pt
\noindent Hint: let $y = 1/x$ and express the limit in terms of  $y$
\vskip 12pt

\noindent 6.5.9. Prove the following version of L'Hospital's rule: if
$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) =0$ and
$\displaystyle{\lim_{x \to \infty} f'(x)/g'(x)}$ exists, then
$\displaystyle {\lim _{x \to \infty} f(x)/g(x) = \lim _{x \to \infty}
f'(x)/g'(x)}$. 
\vskip 12 pt

\noindent Hint:  let $y = 1/x$, so $y \to 0^{+}$ as $x \to \infty$. 
Define $F(y) = f(1/y)$ and 
$G(y) = g(1/y)$ for $y > 0$. Apply L'Hospital's rule to $F(y)/G(y)$. 
What is the relationship between $F'(y)/G'(y)$ and $f'(1/y)/g'(1/y)$?
\vskip 12 pt

\noindent 6.5.10. Here is another version of 
L'Hospital's rule, with a more complicated proof (an expanded 
version of the proof in Helson, Problem 2 p. 149): If 
$\lim_{x \to a} f(x) = \lim_{x \to a} g(x) =\infty$ and $\lim _{x \to 
a} f'(x)/g'(x)$ exists, then
$\lim_{x \to a} f(x)/g(x) = \lim _{x \to a} f'(x)/g'(x)$.

\noindent Note: you may use a part of the problem, even if you cannot 
prove it, in subsequent parts of the problem.
\vskip 12 pt

\noindent i. We show the rule is true for $\lim_{x \to a^{+}}$. Assume
that $f(x)$,$g(x)$,$f'(x)$, and $g'(x)$ are defined on $(a, b]$, for
some $b > a$. (At this point it really doesn't matter what $b$ is, as
long as $b > a$, but we will see below that we have to control how 
close $b$ is to $a$ in order to make everything work.) If $a < x < b$, then
\[
\frac {f(b) - f(x)} {g(b) - g(x)} = \frac {f'(c_{x})} {g'(c_{x})}
\]
for some  $c_{x}$, $x < c_{x} < b$. Now divide the numerator
and denominator of the left side of 
the equation above by $g(x)$ and show that
\[
\frac {f(x)}{g(x)} = \frac {f(b)}{g(x)} - \frac {g(b)}{g(x)}
\frac {f'(c_{x})}{g'(c_{x})} + \frac {f'(c_{x})} {g'(c_{x})}\, . 
\]
Why can we assume that $f(x) > 0$ and $g(x) > 0$ for
$a < x \le b$?
\smallskip

\noindent ii.  Let $\lim _{x \to a^{+}} f'(x)/g'(x) = L$.  We want to
show that $\lim _{x \to a^{+}} f(x)/g(x) = L$.  That is, given
$\varepsilon > 0$, we need to show there exists $\delta > 0$ such that
$\displaystyle{\left|\frac {f(x)}{g(x)} - L\right| < \varepsilon}$ whenever $0 < x - a < \delta$.

Using part i), show that
\[
\left|\frac {f(x)}{g(x)} - L\right| \le \frac {f(b)}{g(x)} + \frac
{g(b)}{g(x)} \left|\frac {f'(c_{x})}{g'(c_{x})}\right| + \left|\frac
{f'(c_{x})} {g'(c_{x})} - L\right|,
\]
if $a < x < b$.
\smallskip

\noindent iii. Show that if a function has a limit as $x \to a^{+}$, 
then the function is bounded for $x$ near $a$ and greater than $a$.
\smallskip

\noindent iv. Use parts ii and iii to show that if $b$ is close 
enough to $a$, then
\[
\left|\frac {f(x)}{g(x)} - L\right| \le \frac {f(b)}{g(x)} + \frac
{g(b)}{g(x)} K + \left|\frac {f'(c_{x})} {g'(c_{x})} - L\right|,
\]
for some constant $K > 0$, whenever $a < x < b$.
\smallskip

\noindent v.  By the definition of limit, given $\varepsilon > 0$,
there exists $\delta > 0$ such that $\displaystyle{|\frac
{f'(x)}{g'(x)} - L| < \frac {\varepsilon}{2}}$ whenever $0 < x - a <
\delta$.  Show there exists $\delta_{1}$, $0 < \delta_{1} \le \delta$, 
such that if $b - a < \delta_{1}$, then
\[
\left|\frac {f(x)}{g(x)} - L\right| < \frac {f(b) + Kg(b)}{g(x)} +
\frac {\varepsilon}{2},
\]
whenever $a < x < b$.
\smallskip

\noindent vi.  Use v and the fact that $g(x) \to \infty$ as $x \to 
a^{+}$ to prove there exists $\delta_{2}$, $0 < \delta_{2} \le 
\delta_{1}$,  such that
if $0 < x - a < \delta_{2}$, then $\displaystyle{\left|\frac
{f(x)}{g(x)} - L\right| < \varepsilon}$. This shows that $\displaystyle{\lim_{x \to 
a^{+}} \frac {f(x)}{g(x)} = L}$. 
\vskip 12 pt

\noindent
{\bf Chapter 7.1}
\vskip 12 pt

\noindent {\em In the next three problems, assume that you have never 
heard of $\log$ or $\ln$---so these functions are ineligible for any 
use whatsoever.}
\vskip 12 pt

\noindent 7.1.1. Let $f(x)$ be a function defined for all $x > 0$ 
such that $\displaystyle{f'(x) = 1/x}$. Let $a$ be a positive 
constant and let $g(x) := f(ax)$. Find $g'(x)$ and simplify your answer.
\vskip 12 pt

\noindent 7.1.2. Find a positive integer $n$ such that when you chop
the interval from 1 to 2.8 into $n$ equal pieces and approximate
$\displaystyle{\int_{1}^{2.8} \frac {dx}{x}}$ by the Riemann sum using
right endpoints (i.e. by the lower sum), then you get a sufficiently 
accurate approximation to show $\displaystyle{\int_{1}^{2.8} \frac {dx}{x} > 1}$. You
may use a spreadsheet or a computer algebra system (such as
matlab/maple/mathematica). Include a printout verifying your answer.
\vskip 12 pt

\noindent 7.1.3. Find a positive integer $n$ such that when you chop
the interval from 1 to 2.7 into $n$ equal pieces and approximate
$\displaystyle{\int_{1}^{2.7} \frac {dx}{x}}$ by the Riemann sum using
left endpoints (i.e. by the upper sum), then you get a sufficiently 
accurate approximation to show $\displaystyle{\int_{1}^{2.7} \frac 
{dx}{x} < 1}$ You may use a spreadsheet or a computer algebra system 
(such as matlab/maple/mathematica). 
Include a printout verifying your answer.
\vskip 12 pt

\noindent 7.1.4. Let $b > 0$ be a real number. Use the definition of $b^{x}$ to show:

\noindent 1) $b^{s + t} = b^{s}b^{t}$ for all $s$ and $t$ real.

\noindent 2) $b^{s - t} = b^{s}/b^{t}$ for all $s$ and $t$ real.

\noindent 3) $\log b^{s} = s \log b$ for all $s$ real.

\noindent 4) $b^{st} = (b^{s})^{t}$ for all  $s$ and $t$ real.
\vskip 12 pt

\noindent 7.1.5. Let $b > 0$ be a real number. Use the definition of 
$b^{x}$ to show:
\smallskip

\noindent 1) $\displaystyle{\frac {db^{x}}{dx} = b^{x} \, \log b}$.
\medskip

\noindent 2) $\displaystyle{\int b^{x} \, dx = \frac {b^{x}}{\log b} + C}$.
\vskip 12 pt

\noindent Use logarithmic differentiation to find the derivative with 
respect to $x$ of:
\vskip 12 pt

\noindent 7.1.6. $\displaystyle {\frac {(x + 1)^{3} (x^{2} + 2)^{4} 
(x^{3} + 3)^{5}}{(x^{4} + 4)^{6} (x^{5} + 5)^{7} (x^{6} + 6)^{8}}}$
\vskip 12 pt

\noindent 7.1.7. $\displaystyle {\sqrt {(x^{2} + x + 2)^{7}(x^{5} + 9x^{3} + 
8)^{11} }}$
\vskip 12 pt

\noindent In problems 7.1.8-7.1.12, find the derivative with respect 
to $x$. 
Note: 
\begin{align}
\frac {d \tan^{-1} x}{dx} &= \frac {1}{1 + x^{2}} \\
\frac {d \sin^{-1} x}{dx} &= \frac {1}{\sqrt {1 - x^{2}}}
\end{align}
\vskip 12 pt

\noindent 7.1.8. $\displaystyle {\left( \sqrt {2} \right)^{x}}$
\vskip 12 pt

\noindent 7.1.9. $\displaystyle {\left( \tan x \right)^{\ln x}}$
\vskip 12 pt

\noindent 7.1.10. $\displaystyle {\pi^{x} \tan^{-1} x}$
\vskip 12 pt

\noindent 7.1.11. $\displaystyle {e^{x^{2} \sin^{-1} x}}$
\vskip 12 pt

\noindent 7.1.12. $\displaystyle {\log_{10} \left(x e^{x} + 1 \right)}$
\vskip 12 pt

\noindent Find the following integrals:
\vskip 12 pt

\noindent 7.1.13. $\displaystyle {\int \left(\sin 4x\right) 5^{\cos 4x} 
\, dx}$
\vskip 12 pt

\noindent 7.1.14. $\displaystyle {\int \frac {x \log_{2} \left(x^{2} + 
1\right)}{x^{2} + 1} 
\, dx}$
\vskip 12 pt

\noindent 7.1.15. $\displaystyle {\int \frac {\pi^{\log_{10}x}} {x}
\, dx}$
\vskip 12 pt

\noindent
{\bf Chapter 8.1}
\vskip 12 pt

\noindent 8.1.1.  Let $1,1,2,3,5,8,\ldots$ be the Fibonacci sequence. 
Find the first six terms of a series $a_{1} + a_{2} + \ldots$ such
that the first six partial sums $S_{1},\ldots,S_{6}$ for the series match the first
six terms of the Fibonacci sequence. That is, 
$S_{1} = 1,S_{2} = 1,S_{3} = 2,S_{4} = 3,S_{5} = 5,S_{6} = 8$. 
\vskip 12 pt

\noindent 8.1.2. Let $s_{1}, s_{2}, \ldots $ be any sequence. Find a series 
$a_{1} + a_{2} + \ldots$ such that the partial sum $S_{n}$ for the 
series satisfies $S_{n} = s_{n}$ for all $n$. {\em This shows that 
partial sums can exhibit the behavior of arbitrary sequences.}
\vskip 12 pt

\noindent 8.1.3. A series $a_{1} + a_{2} + \ldots$ is called a {\em 
telescoping series} if $a_{1} = b_{1} - b_{2}, a_{2} = b_{2} - b_{3}, 
\ldots$. For example, $\sum_{n=1}^{\infty} \frac {1}{n} - \frac {1}{n + 1}$ is a 
telescoping series.
Show that a telescoping series converges if and only if 
$b_{n} \to L$ as $n \to \infty.$ What is the sum of the series if it 
does converge?
\vskip 12 pt

\noindent 8.1.4. Find $\sum_{n=2}^{\infty} 1/(n^{2} - 1)$. {\em Hint: use 
partial fractions to show the series is somethng like a telescoping 
series.}
\vskip 12 pt

\noindent 8.1.5. Does $\displaystyle {\sum_{n=5}^{\infty} \frac {n}{n^{3} - 7n^{2} + 4}}$ converge or 
diverge? Give complete reasons.
\vskip 12 pt

\noindent 8.1.6. Does $\displaystyle {\sum_{n=5}^{\infty} \frac {5^{n} - 2n}{8^{n} + 
4n^{2}}}$ converge or diverge? Give complete reasons.
\vskip 12 pt

\noindent 8.1.7. $1 + 3/2 + 9/4 + 27/8 + \ldots = ?$
\vskip 12 pt

\noindent
{\bf Chapter 9.2}
\vskip 12 pt

\noindent In problems 9.2.1 - 9.2.4, find the radius of convergence and
check convergence at the endpoints for each of the following: \vskip
12 pt

\noindent 9.2.1. $\displaystyle {\sum_{n=1}^{\infty} \frac {x^{n}}{n}}$
\vskip 12 pt

\noindent 9.2.2.  $\displaystyle {\sum_{n=1}^{\infty} \frac {(3x - 5)^{n}}{n^{2}}}$
\vskip 12 pt

\noindent 9.2.3.  $\displaystyle {\sum_{n=1}^{\infty} \frac {8^{n} (x - 1)^{3n}}{n}}$
\vskip 12 pt

\noindent 9.2.4.  $\displaystyle {\sum_{n=1}^{\infty} \frac {8^{n} 
(4x + 1)^{2n + 1}}{\sqrt n}}$
\vskip 12 pt

\noindent 9.2.5.  Estimate $\displaystyle {\int_{0}^{2} e^{x^{2}} \, dx}$ correct to 6 
decimal places. How do you know your estimate is correct?
\vskip 12 pt



\noindent
{\bf Chapter 10.2}
\vskip 12 pt

\noindent 10.2.1.  $\displaystyle {\int \csc x \, dx}$
\vskip 12 pt

\noindent 10.2.2.  $\displaystyle {\int \cos^{5} x \, dx}$
\vskip 12 pt

\noindent 10.2.3.  $\displaystyle {\int \tan^{2} x \, dx}$
\vskip 12 pt

\noindent 10.2.4.  $\displaystyle {\int \sec^{4} x \, dx}$
\vskip 12 pt

\noindent 10.2.5.  $\displaystyle {\int \cos^{4} x \, dx}$
\vskip 12 pt

\noindent 10.2.6.  $\displaystyle {\int \frac {dx} {\sqrt {1 - 4x^2}}}$
\vskip 12 pt

\noindent 10.2.7.  $\displaystyle {\int \frac {dx} {\sqrt {9x^2 - 1}}}$
\vskip 12 pt

\noindent 10.2.8.  $\displaystyle {\int \frac {x^2 \, dx} {\sqrt {9 - x^2}}}$
\vskip 12 pt

\noindent 10.2.9.  $\displaystyle {\int \frac {x^3 \, dx} {\sqrt {x^2 + 1}}}$
\vskip 12 pt

\noindent 10.2.10. $\displaystyle {\int \frac {dx} {\sqrt {2x - x^2}}}$
\vskip 12 pt

\noindent 10.2.11. $\displaystyle {\int \frac {dx} {4x^2 + 4x + 2}}$
\vskip 12 pt

\noindent 10.2.12. $\displaystyle {\int \frac {dx} {\sqrt {2 - 5 x^2}}}$
\vskip 12 pt

\noindent 10.2.13.  $\displaystyle {\int \frac {dx} {1 + \sin x}}$  
{\em Hint: multiply numerator and denominator by } $1 - \sin x$
\vskip 12 pt

\noindent 10.2.14. Show that
\begin{align} \sin 2\theta &= \frac{2 \tan \theta}{1 + \tan^2 \theta}\\
\cos 2\theta &= \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}\\
d \theta &= \frac{d \tan \theta}{1 + \tan^2 \theta}
\end{align}
\vskip 12 pt

\noindent 10.2.15. Let \[z = \tan \frac {x}{2}\] 
Show that 
\begin{align} \sin x &= \frac{2 z}{1 + z^2}\\
\cos x &= \frac{1 - z^2}{1 + z^2}\\
d x &= \frac{2 d z}{1 + z^2}
\end{align}
Hint: let $\theta = x/2$ and apply problem 10.2.14.
\vskip 12 pt

\noindent 10.2.16.  Use 10.2.15 to find $\displaystyle {\int \frac {dx} {2 + \sin x}}$
\vskip 12 pt

\noindent
{\bf Chapter 10.3}
\vskip 12 pt

\noindent 10.3.1.  $\displaystyle {\int \frac {x+7}{x^{2} - 6x + 5} \, 
dx}$
\vskip 12 pt

\noindent 10.3.2.  $\displaystyle {\int \frac {3x+1}{x^{2} - 10x + 25} 
\, dx}$
\vskip 12 pt

\noindent 10.3.3.  $\displaystyle {\int \frac {x + 2}{x^{3} - 3x^{2}} 
\, dx}$
\vskip 12 pt

\noindent 10.3.4.  $\displaystyle {\int \frac {1}{(x + 1)(x^{2} + 4)} 
\, dx}$
\vskip 12 pt

\noindent 10.3.5.  $\displaystyle {\int \frac {x^{5}}{x^{2} - 4} 
\, dx}$
\vskip 12 pt

\noindent 10.3.6.  $\displaystyle {\int \frac {e^{x}}{e^{2x} + 4e^{x} + 3} 
\, dx}$
\vskip 12 pt


\end{document}