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\begin{document}

\noindent
p. 465

\smallskip
\noindent
\# 5)~~$y = {\rm ln} 3x, y^{'} = 1/x$

\smallskip
\noindent
\#11)~~$y = {\rm ln} (\theta + 1),
y^{'} = 1/(\theta + 1)$

\smallskip
\noindent
\#15)~~$y = t({\rm ln} t)^{2}, y^{'} =
({\rm ln} t)^{2} + 2 {\rm ln} t$

\smallskip
\noindent
\#19)~~$y = {\rm ln} t/ t, y^{'} =
1/t^{2} - {\rm ln} t/t^{2}$

\smallskip
\noindent
\# 21)~~$y = {\rm ln} x/(1 + {\rm ln}
x), y^{'} = 1/x(1 + {\rm ln} x) - {\rm
ln} x/x(1 + {\rm ln} x)^{2}$

\smallskip
\noindent
\# 25)~~$y = \theta (\sin ({\rm ln}
\theta)) + \cos ({\rm ln} \theta)),$
$$y^{'} = \sin ({\rm ln}
\theta) + \cos ({\rm ln} \theta)) +
\cos ({\rm ln}
\theta) - \sin ({\rm ln} \theta)$$

\smallskip
\noindent
\# 37)~~$y = \sqrt{x(x + 1)}$
$${\rm ln} y = {1 \over 2} \{{\rm ln} x
+ {\rm ln} (x + 1)\}$$
differentiation yields:
$$y^{'}/y = {1 \over 2} \{1/x + 1/(x +
1)\}$$
thus
$$y^{'} = {1 \over 2} y \{1/x + 1/(x +
1)\} = {1 \over 2}
\sqrt{x(x + 1)} \{1/x + 1/(x +
1)\}.$$

\smallskip
\noindent
\# 49)~~$$y = ({x(x - 2) \over x^{2} +
1})^{1/3}$$
hence
$${\rm ln} y = {1 \over 3} \{{\rm ln}
x + {\rm ln} (x - 2) - {\rm ln} (x^{2}
+ 1)\}$$
differentiation yields:
$$y^{'}/y = {1 \over 3} \{1/x
+ 1/(x - 2) - 1/(x^{2} + 1)\}$$
so
$$y^{'} = {1 \over 3} ({x(x - 2) \over
x^{2} + 1})^{1/3} \{1/x
+ 1/(x - 2) - 1/(x^{2} + 1)\}$$

\smallskip
\noindent
\# 53)~~$$\int {2y \over y^{2} - 25} dy
= ?$$
Let $u = y^{2} - 25$ then $du = 2y dy$
$$\int {2y \over y^{2} - 25} dy
= \int {du \over u} = \ln |u| + C =
\ln |y^{2} - 25| + C$$

\smallskip
\noindent
\# 53)~~$$\int_{1}^{2} {2 \ln x \over
x} dx = ?$$
Set $u = \ln x$ then $du = dx/x$
$$\int_{1}^{2} {2 \ln x \over
x} dx = 2 \int u du = u^{2} = (\ln
x)^{2} |_{1}^{2} = (\ln 2)^{2} - (\ln
1)^{2} = (\ln 2)^{2}$$

\smallskip
\noindent
\# 59)~~$$\int_{2}^{4} {dx \over x(\ln
x)^{2}} = ?$$
Let $u = \ln x$ then $du = dx /x$
\begin{eqnarray*}
\int_{2}^{4} {dx \over
x(\ln x)^{2}} = \int u^{-2} du = -
u^{-1} = - (\ln x)^{-1} |_{2}^{4} &=&
{1 \over \ln 4} - {1 \over \ln 2}\\
&=&
{1 \over 2\ln 2} - {1 \over \ln 2}\\
&=&
- {1 \over 2} {1 \over \ln 2}
\end{eqnarray*}

\smallskip
\noindent
\# 61)~~$$\int {3 \sec^{2} t \over 6 +
3 \tan t} dt = ?$$
Let $u = 6 + 3 tan t$ then $du = 3
sec^{2} t$ and
$$\int {3 \sec^{2} t \over 6 + 3
tan t} dt = \int {du \over u} = \ln
|u| + C = \ln |6 + 3 \tan t| + C$$

\bigskip
\noindent
p. 472

\smallskip
\noindent
\# 1a)~~$e^{\ln 7.2} = = 7.2$;
1b)~~$e^{-\ln x^{2}} = 1/x^{2}$;
1c)~~$e^{\ln x - \ln y} = e^{\ln
x}/e^{\ln y} = x/y.$

\smallskip
\noindent
\#3a)~~$$2 \ln \sqrt{e} = 2 \ln e^{1/2}
= 2.{1 \over 2} \ln e = 1.$$

\smallskip
\noindent
3b)~~$\ln (\ln e^{e}) = \ln (e \ln e)
= \ln e = 1$. 

\smallskip
\noindent
3c)~~$\ln (e^{-x^{2} - y^{2}}) = -
x^{2} - y^{2}.$

\smallskip
\noindent
5)~~$\ln y = 2t + 4, y = ?$
Take exponential on both sides, we get:
$$y = e^{2t + 4}.$$

\smallskip
\noindent
9)~~$\ln (y - 1) - \ln 2 = x + \ln x,
y = ?$
Take exponential on both sides:
$$e^{\ln (y - 1) - \ln 2} = e^{x + \ln
x}$$
which is equivalent to
$${y - 1 \over 2} = xe^{x}$$
hence
$$y = 2xe^{x} + 1.$$

\smallskip
\noindent
13a)~~$e^{-0.3t} = 27, t = ?$ Take
$\ln$ on both sides:
$$-0.3t = \ln 27, t = - (3\ln 3)/0.3 =
- 10 \ln 3.$$

\smallskip
\noindent
13b)~~$e^{kt} = 1/2$. Take $\ln$:
$$kt = - \ln 2, t = - \ln 2/k.$$

\smallskip
\noindent
13c)~~$e^{(\ln 0.2)t} = 0.4$. Simplify:
$$(0.2)^{t} = 0.4$$
then take $\ln$:
$$(\ln 0.2) t = \ln 0.4$$
so $t = \ln 0.2/\ln 0.4.$

\smallskip
\noindent
19)~~$y = e^{5-7x}$ then $y^{'} =
-7 e^{5-7x}.$

\smallskip
\noindent
23)~~$y = (x^{2} -2x + 2)e^{x}$ then
$y^{'} = (2x - 2)e^{x} + (x^{2} - 2x +
2)e^{x}$.

\smallskip
\noindent
27)~~$y = \cos (e^{-\theta^{2}})$ then
$y^{'} = - \sin (e^{-\theta^{2}})
e^{-\theta^{2}} (-2 \theta) = 2\theta
e^{-\theta^{2}} \cos
(e^{-\theta^{2}}).$

\smallskip
\noindent
37)~~$\ln y = e^{y} \sin x$ then
$$y^{'}/y = e^{y} y^{'} \sin x +
e^{y} \cos x$$
hence
$$y^{'} = ye^{y}y^{'} \sin x + ye^{y}
\cos x$$
which is equivalent to
$$y^{'}(1 - ye^{y} \sin x) = ye^{y}
\cos x.$$
Solving for $y^{'}$:
$$y^{'} = {ye^{y}
\cos x \over 1 - ye^{y} \sin x}.$$

\smallskip
\noindent
41)~~$\int (e^{3x} + 5e^{-x}) dx =
e^{3x}/3 - 5e^{-x}.$

\smallskip
\noindent
49)~~$$\int {e^{\sqrt r} \over \sqrt
r} dr = ?$$
Let $u = \sqrt r$ then $du = dr/2\sqrt
r$ so
$$\int {e^{\sqrt r} \over \sqrt
r} dr = 2 \int e^{u} du = 2 e^{u} + C
= 2 e^{\sqrt r} + C.$$

\smallskip
\noindent
55)~~$$\int_{0}^{\pi/4} (1+e^{\tan
\theta}) \sec^{2} \theta d\theta = ?$$
Let $u = \tan \theta$ then $d \theta =
\sec^{2} \theta$ hence:
\begin{eqnarray*}
\int_{0}^{\pi/4}
(1+e^{\tan
\theta}) \sec^{2} \theta d\theta &=&
\int (1+e^{u}) du \\
&=& u +
e^{u} \\
&=& \tan \theta + e^{\tan
\theta} |_{0}^{\pi/4}\\
&=& \tan \pi/4 + e^{\tan \pi/4} -
\tan 0 - e^{\tan 0}\\
&=& 1 + e - e^{0} = e
\end{eqnarray*}

\smallskip
\noindent
63)~~$y^{'} = e^{t} \sin (e^{t} - 2),
y(\ln 2) = 0$
Then
$$y = \int e^{t} \sin (e^{t} - 2) dt$$
set $u = e^{t} - 2$ then $du = e^{t}
dt$ and
$$y = \int e^{t} \sin (e^{t} - 2) dt
= \int \sin u du = - \cos u + C = -
\cos (e^{t} - 2) + C.$$
To find $C$ we use the initial
condition:
$$0 = y(\ln 2) = - \cos (e^{\ln 2} - 2)
+ C = - \cos 0 + C = - 1 + C$$
hence $C =1$ and $y = - \cos (e^{t} -
2) + 1.$

\smallskip
\noindent
65)~~$y^{''} = 2e^{-x}, y(0) = 1,
y^{'}(0) = 0$
First integration yields:
$$y^{'} = \int 2e^{-x} dx = -2e^{-x} +
C_{0}$$
Since $y^{'}(0) = 0$, we get:
$$0 = y^{'}(0) = - 2e^{-0} + C_{0} =
-2 + C_{0}$$
hence $C_{0} = 2$. Thus
$$y^{'} = 2 - 2e^{-x}$$
and integration yields:
$$y = \int (2 - 2e^{-x}) dx = 2x +
2e^{-x} + C_{1}.$$
Using the condition $y(0) = 1$ we get
$$1 = y(0) = 2 + C_{1}$$
hence $C_{1} = - 1$ and we have:
$$y = 2x +
2e^{-x} - 1.$$

\smallskip
\noindent
69)~~Find the absolute maximum valuse
of $f(x) = x^{2} \ln (1/x) = - x^{2}
\ln x.$

\smallskip
\noindent
Differentiation yields:
$$f^{'}(x) = - 2x \ln x - x^{2}/x
 = - 2x \ln x - x$$
Set $f^{'}(x) = - 2x \ln x - x = 0$
we get
$$x(- 2\ln x - 1) = 0$$
and so $x = 0, x = e^{-1/2}$ are the
critical points. Absolute maximum
occurs at $x = e^{-1/2}$ the absolute
maximum value is $- (e^{-1/2})^{2}
\ln (e^{-1/2}) = - e^{-1} (-1/2) =
1/2e.$

\bigskip
\noindent
p.480

\smallskip
\noindent
\#13)~~$y = 5^{\sqrt{x}}$,
$$\ln y = \sqrt{x} \ln 5, ~~y^{'}/y =
{\ln 5 \over 2\sqrt x}$$
hence
$$y^{'} = y{\ln 5 \over 2\sqrt x} =
5^{\sqrt{x}}{\ln 5 \over 2\sqrt x}.$$

\smallskip
\noindent
\#17)~~$y = (\cos \theta)^{\sqrt{2}}$,
$$\ln y = {\sqrt{2}} \ln (\cos
\theta),~~y^{'}/y = {\sqrt{2}} {1
\over \cos \theta} (- \sin \theta) =
- \sqrt{2} \tan \theta$$ 
so $$y^{'} = - \sqrt{2} \tan
\theta (\cos
\theta)^{\sqrt{2}}.$$

\smallskip
\noindent
\#33)~~$y = \log_{5} e^{x} = \ln e^{x}
/ \ln 5 = x/\ln 5$, hence
$$y^{'} = {1 \over \ln 5}.$$

\smallskip
\noindent
\#39)~~$y = (x + 1)^{x},~~\ln y = x
\ln (x + 1)$, hence
$$y^{'}/y = \ln (x + 1) + x/(x + 1)$$
so $$y = (x + 1)^{x} \{{x \over x + 1}
+ \ln (x + 1)\}.$$

\smallskip
\noindent
\#43)~~$y = (\sin x)^{x},~~\ln y = x
\ln \sin x$, hence
$$y^{'}/y = \ln \sin x + x {1 \over
\sin x} \cos x = \ln \sin x + x \tan
x$$
and so
$$y = (\sin x)^{x} \{\ln \sin x + x \tan
x\}.$$

\smallskip
\noindent
\#47)~~
$$\int 5^{x} dx = {5^{x} \over \ln 5}
+ C$$

\smallskip
\noindent
\#51)~~
$$\int_{1}^{\sqrt{2}} x2^{x^{2}} dx =
?$$
Let $u = x^{2}$ then $du = 2x dx$
\begin{eqnarray*}
\int_{1}^{\sqrt{2}}
x2^{x^{2}} dx = {1 \over 2} \int 2^{u}
du &=& {1
\over 2} {2^{u} \over \ln 2} \\
&=& {1 \over
2} {2^{x^{2}} \over \ln 2}
|_{1}^{\sqrt{2}}\\
&=& {1 \over
2\ln 2} (2^{2} - 2)\\
&=& {1 \over \ln 2}.
\end{eqnarray*}


\smallskip
\noindent
\#57)~~
$$\int 3x^{\sqrt{3}} dx =
{3 \over 11 + \sqrt{3}}
x^{1+\sqrt{3}}.$$

\smallskip
\noindent
\#65)~~
\begin{eqnarray*}
\int_{0}^{2} {\log_{2}
(x+2) \over x+2} &=& \int_{0}^{2} {\ln
(x+2)
\over (x+2)\ln 2}\\
&=&{1 \over \ln 2} \int_{0}^{2} {\ln
(x+2) \over x+2} dx 
\end{eqnarray*}
Let $u = \ln (x + 2)$ then $du =
dx/(x+2)$ and
\begin{eqnarray*}
\int_{0}^{2} {\log_{2} (x+2) \over
x+2} = {1 \over \ln 2} \int_{0}^{2} {\ln
(x+2) \over (x+2)} dx &=& {1 \over \ln 2} \int_{0}^{2} u
du = {u^{2} \over 2\ln 2} \\
&=& {(\ln (x+2))^{2} \over 2
\ln 2} |_{0}^{2}\\
&=& {(\ln 4)^{2} \over 2
\ln 2} - {(\ln 2)^{2} \over 2
\ln 2}\\
&=& {(2\ln 2)^{2} \over 2
\ln 2} - {(\ln 2)^{2} \over 2
\ln 2}\\
&=& 2\ln 2 - {\ln 2 \over 2}\\
&=& {3 \ln 2 \over 2}.
\end{eqnarray*}

\bigskip
\noindent
p.489

\smallskip
\noindent
\#3)~~$${dy \over dt} = - 0.6y, y(0) = 100$$
Then
$${dy \over y} = -0.6 dt$$
integration yields
$$\ln y = - 0.6 t + C$$
and $\ln 100 = -0.6(0) + C = C$.
Thus $\ln y = - 0.6 t + \ln 100.$ Taking exponential
yields:
$$y = 100e^{-0.6t}$$
For $t = 1$ we get
$$y = 100 e^{-0.6}.$$


\smallskip
\noindent
\#5)~~$${dL \over dx} = - kL$$
Intensity is halved at the depth of 18 feet.
Solving the equation yields:
$$L = L_{0} e^{-kx}$$
and
$${1 \over 2}L_{0} = L_{0} e^{-k18}$$
so
$$e^{-k18} = 1/2$$
and taking $\ln$ yields:
$$-18 k = - \ln 2,~~k = \ln 2/18$$
Thus
$$L = L_{0} e^{-(\ln 2)x/18}.$$
If $L(x) = L_{0}/10$ then
$$1/10 = e^{-(\ln 2)x/18}$$
thus
$${\ln 2 \over 18} x = \ln 10$$
and so $x = 18 \ln 10/\ln 2.$


\smallskip
\noindent
\#7)~~$2^{48}$

\smallskip
\noindent
\#13)~~$A(t) = A_{0}e^{.04t}$ thus
$$A(5) = A_{0}e^{.2}.$$
If $A(t) = 2A_{0}$ then
$$2 = e^{.04t},~\ln 2 = .04 t$$
so $t = \ln 2/.04.$ If $A(t) = 3A_{0}$ then
$$3 = e^{.04t},~\ln 3 = .04 t$$
so $t = \ln 3/.04.$

\bigskip
\noindent
p. 496

\smallskip
\noindent
\#7)~~
$$\lim_{t \rightarrow 0} {\sin t^{2} \over t} = \lim_{t
\rightarrow 0} {2t\cos t^{2} \over 1} = 0.$$


\smallskip
\noindent
\#13)~~
$$\lim_{\theta \rightarrow \pi/2} {1 - \sin \theta \over
1 + \cos 2\theta} =
\lim_{\theta \rightarrow \pi/2} {- \cos \theta \over -2
\sin 2\theta} =
\lim_{\theta \rightarrow \pi/2} {- \sin \theta \over -4
\cos 2\theta} = 0.$$

\smallskip
\noindent
\#15)~~
$$\lim_{x \rightarrow 0} {x^{2} \over
\ln (\sec x)} = \lim_{x \rightarrow 0} {2x \sec x \over
\tan x \sec x} = \lim_{x \rightarrow 0} {2x \over
\tan x} = \lim_{x \rightarrow 0} {2 \over
\sec^{2} x} = 2.$$

\smallskip
\noindent
\#19)~~
$$\lim_{x \rightarrow (\pi/2)^{-}} (x - {\pi \over 2})
\sec x = \lim_{x \rightarrow (\pi/2)^{-}} {(x - {\pi
\over 2}) \over 
\cos x} = \lim_{x \rightarrow (\pi/2)^{-}} {1 \over 
- \sin x} = - 1.$$

\smallskip
\noindent
\#27)~~
$$\lim_{x \rightarrow 0^{+}} {\ln (x^{2} + 2x)
\over
\ln x} = \lim_{x \rightarrow 0^{+}} {2(x + 1)
\over x^{2} + 2x} x = \lim_{x \rightarrow 0^{+}} {2(x + 1)
\over x + 2} = 1.$$

\smallskip
\noindent
\#31)~~
$$\lim_{x \rightarrow \infty} (\ln 2x -
\ln (x + 1)) = \lim_{x \rightarrow \infty} \ln {2x \over
x + 1} = \ln \lim_{x \rightarrow \infty} {2x \over
x + 1} = \ln \lim_{x \rightarrow \infty} \ln {2 \over
1} = \ln 2.$$

\smallskip
\noindent
\#37)~~
$$\lim_{x \rightarrow \infty} \int_{x}^{2x} {1 \over t}
dt = \lim_{x \rightarrow \infty} \ln t |_{x}^{2x} =
\lim_{x \rightarrow \infty} \ln {2x \over x} = \ln 2.$$

\smallskip
\noindent
\#42)~~
$$\lim_{x \rightarrow \infty} x^{2} e^{-x} = 
\lim_{x \rightarrow \infty} {x^{2} \over e^{x}} =
\lim_{x \rightarrow \infty} {2x \over e^{x}} = \lim_{x
\rightarrow \infty} {2 \over e^{x}} = 0.$$

\smallskip
\noindent
\#51)~~
$$\lim_{x \rightarrow 0^{+}} x^{x} = \lim_{x \rightarrow
0^{+}} e^{x \ln x}.$$
Now
$$\lim_{x \rightarrow
0^{+}} x \ln x = \lim_{x \rightarrow
0^{+}} {\ln x \over 1/x} = \lim_{x \rightarrow
0^{+}} - {1/x \over 1/x^{2}} = \lim_{x \rightarrow
0^{+}} - x = 0.$$
Thus
$$\lim_{x \rightarrow 0^{+}} x^{x} = \lim_{x \rightarrow
0^{+}} e^{x \ln x} = e^{0} = 1.$$

\noindent
\#53)~~\begin{eqnarray*}
\lim_{x \rightarrow \infty}
\sqrt{{9x + 1
\over x + 1}} = \lim_{x \rightarrow \infty} \sqrt{{(9x +
1)/x
\over (x + 1)/x}} &=& \lim_{x \rightarrow \infty} 
\sqrt{{(9 + 1/x)
\over (1 + 1/x)}} \\ &=&  
\sqrt{{(9 + 1/\infty)
\over (1 + 1/\infty)}} \\
&=& \sqrt{{9 \over 1}} = 3.
\end{eqnarray*}

\noindent
\#57)~~(a) is wrong and (b) is correct becuse
$$\lim_{x \rightarrow 3} {x - 3 \over x^{2} - 3}$$
is not of indeterminate form.

\noindent
\#63)~~$$\lim_{k \rightarrow \infty} (1 + {r \over
k})^{k} = \lim_{k \rightarrow \infty} e^{k \ln (1 +
r/k)}$$
and since
\begin{eqnarray*}
\lim_{k \rightarrow \infty} k \ln (1 +
{r \over k}) = \lim_{k \rightarrow \infty} \ln (1 +
{r \over k}) /(1/k) &=& \lim_{k \rightarrow \infty} {-
r/k^{2} \over 1 + r/k} (- k^{2}) \\
&=& 
\lim_{k \rightarrow \infty} {
r \over 1 + r/k} = r
\end{eqnarray*}
hence
$$\lim_{k \rightarrow \infty} (1 + {r \over
k})^{k} = \lim_{k \rightarrow \infty} e^{k \ln (1 +
r/k)} = e^{r}.$$

\bigskip
\noindent
p.503

\bigskip
\noindent
\#1) (Look at examples 1, 2, 3, 4 and 5) Slower : a), b),
c), e), h); Same rate: g); Faster d).

\bigskip
\noindent
\#3) c) ~Same rate:
$$\lim_{x \rightarrow \infty}
{\sqrt{x^{4} + x^{3}} \over x^{2}} = \lim_{x \rightarrow
\infty}
\sqrt{{x^{4} \over x^{4}} + {x^{3} \over x^{4}}} =
\lim_{x \rightarrow \infty} \sqrt{1 + {1 \over x}} =
1.$$
g)~Slower: $$\lim_{x \rightarrow \infty} {x^{3}e^{-x}
\over x^{2}} = \lim_{x \rightarrow \infty} {x \over
e^{x}} = 0.$$

\bigskip
\noindent
\# 23)~ $$\lim_{x \rightarrow \infty} {n \log_{2} n
\over n (\log_{2} n)^{2}} = \lim_{x \rightarrow \infty}
{1 \over \log_{2} n} = 0;$$
$$\lim_{x \rightarrow \infty} {n (\log_{2} n)^{2}
\over n^{3/2}} = \lim_{x \rightarrow \infty}
{(\log_{2} n)^{2} \over n^{1/2}} = 0.$$

\bigskip
\noindent
\# 23)~$$\int {dx \over \sqrt{9 - x^{2}}} = \sin^{-1} {x
\over 3}.$$

\bigskip
\noindent
\# 25)~$$\int {dx \over 17 +  x^{2}} = {1 \over
\sqrt{17}} \tan^{-1} {x
\over \sqrt{17}}.$$

\bigskip
\noindent
\# 27)~$$\int {dx \over x\sqrt{5x^{2} - 4}} = ?$$
Let $u = \sqrt{5}x$ then $du = \sqrt{5} dx$ then
$$\int {dx \over x\sqrt{5x^{2} - 4}} = \int {du \over
u \sqrt{u^{2} - 4}} = {1 \over
2} \sec^{-1} |{\sqrt{5}x \over 2}|$$

\bigskip
\noindent
\# 29)~$$\int_{0}^{1} {4ds \over \sqrt{4 - s^{2}}} =
4 \sin^{-1} {x \over 2} |_{0}^{1} = 4 \sin^{-1} {1\over
2}  -
4 \sin^{-1} 0 = 4\pi/6 = 2\pi/3.$$ 

\bigskip
\noindent
\# 31)~$$\int_{0}^{2} {dt \over 8 + 2t^{2}} = ?$$
Let $u = \sqrt{2} t$ then $du = \sqrt{2} dt$ hence
\begin{eqnarray*}
\int_{0}^{2} {dt \over 8 + 2t^{2}} = {1
\over
\sqrt{2}} \int {du \over 8 + u^{2}} = {1 \over
\sqrt{2}} {1 \over
2\sqrt{2}} \tan^{-1} {u \over 2\sqrt{2}} &=& {1 \over
4} \tan^{-1} {t \over 2} |_{0}^{2} \\ &=& {1 \over
4} \tan^{-1} 1  \\
&=& {\pi \over
16}
\end{eqnarray*}

\bigskip
\noindent
\# 33)~$$\int_{-1}^{-\sqrt{2}/2} {dy \over y
\sqrt{4y^{2} - 1}} = ?$$
Let $u = 2y$ then $du =
2 dy$ hence
\begin{eqnarray*}
\int_{-1}^{-\sqrt{2}/2} {dy \over y
\sqrt{4y^{2} - 1}} = \int {du \over u
\sqrt{u^{2} - 1}} &=& \sec^{-1} u \\
&=& \sec^{-1} |2y| |_{-1}^{-\sqrt{2}/2} \\ 
&=& \sec^{-1} \sqrt{2}  - \sec^{-1} 2\\
&=& {\pi \over
4} - {\pi \over
3}\\
&=& - {\pi \over
12}
\end{eqnarray*}

\noindent
\# 35)~$$\int {3dr \over 
\sqrt{1 - 4(r - 1)^{2}}} = ?$$
Let $u = 2(r - 1)$ then $du = 2 dr$ and
$$\int {3dr \over 
\sqrt{1 - 4(r - 1)^{2}}} = {3 \over 2} \int {du
\over 
\sqrt{1 - u^{2}}} = {3 \over 2} \sin^{-1} |u| =  {3
\over 2}
\sin^{-1} 2(r - 1) .$$


\end{document}