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\noindent
p. 465

\smallskip
\noindent
\# 37)~~$$\int {dx \over 2 + (x - 1)^{2}} = ?$$
Let $u = x - 1$ then $du = dx$ hence
$$\int {dx \over 2 + (x - 1)^{2}} = \int {du \over 2 + u^{2}} =
{1 \over \sqrt{2}} \tan^{-1} {u \over \sqrt{2}} = {1 \over
\sqrt{2}} \tan^{-1} {(x - 1) \over \sqrt{2}}$$


\smallskip
\noindent
\# 39)~~$$\int {dx \over (2x - 1) \sqrt{(2x - 1)^{2} - 4}} = ?$$
Let $u = 2x - 1$ then $du = 2dx$ hence
\begin{eqnarray*}\int {dx \over (2x - 1) \sqrt{(2x - 1)^{2} -
4}} = {1 \over 2}\int {du \over u \sqrt{u^{2} - 4}} &=& {1 \over
2} {1 \over 2}
\sec^{-1} |{u
\over 2}| \\
&=& {1 \over 4} \sec^{-1} |{2x - 1 \over 2}|.
\end{eqnarray*}

\smallskip
\noindent
\# 41)~~$$\int_{-\pi/2}^{\pi/2} {2 \cos \theta d\theta \over 1 +
(\sin
\theta)^{2}} = ?$$
Let $u = \sin \theta$ then $du = \cos \theta d\theta$
hence
\begin{eqnarray*}
\int_{-\pi/2}^{\pi/2} {2 \cos \theta d\theta
\over 1 + (\sin
\theta)^{2}} = \int {2 du \over 1 + u^{2}} &=&
2 \tan^{-1} u \\
&=& 2 \tan^{-1} \sin \theta |_{-\pi/2}^{\pi/2}\\
&=& 2 \tan^{-1} 1 - 2 \tan^{-1} (-1) \\
&=& \pi/2 - (- \pi/2) = \pi.
\end{eqnarray*}

\smallskip
\noindent
\# 43)~~$$\int_{0}^{\ln \sqrt{3}} {e^{x} dx \over 1 + e^{2x}} =
?$$ Let $u = e^{x}$ then $du = e^{x} dx$
hence
\begin{eqnarray*}\int_{0}^{\ln \sqrt{3}} {e^{x} dx \over 1 +
e^{2x}} = \int {du \over 1 + u^{2}} = \tan^{-1} u &=& \tan^{-1}
e^{x} |_{0}^{\ln
\sqrt{3}} \\
&=& \tan^{-1} \sqrt{3} - \tan^{-1} \\
&=& \pi/3 - \pi/4 = \pi/12.
\end{eqnarray*}

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\noindent
\# 45)~~$$\int {y dy \over \sqrt{1 - y^{4}}} =
?$$ Let $u = y^{2}$ then $du = 2y dy$
hence
\begin{eqnarray*}\int {y dy \over \sqrt{1 - y^{4}}} = {1 \over
2} \int {du
\over \sqrt{1 - u^{2}}} = {1 \over
2} \sin^{-1} u = {1 \over
2} \sin^{-1} y^{2}.
\end{eqnarray*}

\smallskip
\noindent
\# 47)~~$$\int {dx \over \sqrt{- x^{2} + 4x - 3}} =
?$$
Note that $- x^{2} + 4x - 3 = 1 - (x - 2)^{2}$ then 
$$\int {dx \over \sqrt{- x^{2} + 4x - 3}} = \int {dx \over
\sqrt{1 - (x - 2)^{2}}}.$$
Let $u = (x - 2)$ then $du = dx$
hence
\begin{eqnarray*}\int {dx \over \sqrt{- x^{2} + 4x - 3}} = \int {dx \over
\sqrt{1 - (x - 2)^{2}}} = \int {du \over
\sqrt{1 - u^{2}}} &=& \sin^{-1} u \\ &=& \sin^{-1} (x - 2).
\end{eqnarray*}

\smallskip
\noindent
\# 51)~~$$\int {dy \over y^{2} - 2y + 5} =
?$$
Note that $y^{2} - 2y + 5 = (y - 1)^{2} + 4$ then 
$$\int {dy \over y^{2} - 2y + 5} = \int {dy \over (y - 1)^{2} +
4}.$$ Let $u = (y - 1)$ then $du = dy$ hence
\begin{eqnarray*}\int {dy \over y^{2} - 2y + 5} =\int
{dy \over (y - 1)^{2} +
4} =\int
{du \over u^{2} +
4} &=& {1 \over 2} \tan^{-1} u \\
&=& {1 \over 2} \tan^{-1} {y - 1
\over 2}.
\end{eqnarray*}

\smallskip
\noindent
\# 53)~~$$\int_{1}^{2} {8dx \over x^{2} - 2x + 2} =
?$$
Note that $x^{2} - 2x + 2 = (x - 1)^{2} + 1$ then 
$$\int_{1}^{2} {8dx \over x^{2} - 2x + 2} = \int_{1}^{2} {8dx
\over (x - 1)^{2} + 1}.$$ Let $u = (x - 1)$ then $du = dx$ hence
\begin{eqnarray*} \int_{1}^{2} {8dx \over x^{2} - 2x + 2} =
\int {8du \over u^{2} + 1} = 8 \tan^{-1} u &=& 8 \tan^{-1} (x -
1) |_{1}^{2} \\ &=& 8 \tan^{-1} 1 - 8 \tan^{-1} 0 \\
&=& 2\pi.
\end{eqnarray*}

\smallskip
\noindent
\# 55)~~$$\int {dx \over (x + 1) \sqrt{x^{2} + 2x}} =
?$$
Note that $x^{2} + 2x = (x + 1)^{2} - 1$ then 
$$\int {dx \over (x + 1) \sqrt{x^{2} + 2x}} = \int
{dx \over (x + 1) \sqrt{(x + 1)^{2} - 1}}.$$ Let $u = (x + 1)$
then
$du = dx$ hence
\begin{eqnarray*} \int {dx \over (x + 1) \sqrt{x^{2} + 2x}} =
\int {du \over u \sqrt{u^{2} - 1}} = \sec^{-1} |u| = \sec^{-1}
|x + 1|.
\end{eqnarray*}

\smallskip
\noindent
\# 57)~~$$\int {e^{\sin^{-1}x}dx \over \sqrt{1 - x^{2}}} = ?$$
Let $u = e^{\sin^{-1}x}$
then
$$du = {e^{\sin^{-1}x} \over \sqrt{1 - x^{2}}}dx$$ hence
\begin{eqnarray*} \int {e^{\sin^{-1}x}dx \over \sqrt{1 - x^{2}}}
=
\int du = u = e^{\sin^{-1}x}.
\end{eqnarray*}

\smallskip
\noindent
\# 59)~~$$\int {(\sin^{-1}x)^{2}dx \over \sqrt{1 - x^{2}}} = ?$$
Let $u = \sin^{-1}x$
then
$$du = {1 \over \sqrt{1 - x^{2}}}dx$$ hence
\begin{eqnarray*} \int {(\sin^{-1}x)^{2}dx \over \sqrt{1 -
x^{2}}} =
\int u^{2} du = {u^{3} \over 3} = {(\sin^{-1}x)^{3} \over 3}.
\end{eqnarray*}

\smallskip
\noindent
\# 61)~~$$\int {dy \over (\tan^{-1} y) (1 + y^{2})} = ?$$
Let $u = \tan^{-1}y$
then
$$du = {1 \over 1 + y^{2}}dy$$ hence
\begin{eqnarray*} \int {dy \over (\tan^{-1} y) (1 + y^{2})} =
\int {du  \over u} = \ln |u| = \ln |\tan^{-1} y|.
\end{eqnarray*}

\smallskip
\noindent
\# 63)~~$$\int {\sec^{2}(\sec^{-1} x) dx \over x \sqrt{x^{2} -1}}
= ?$$ Let
$u =
\sec^{-1}x$ then
$$du = {1 \over x \sqrt{x^{2} -1}}dx$$ hence
\begin{eqnarray*} \int {\sec^{2}(\sec^{-1} x) dx \over x
\sqrt{x^{2} -1}} =
\int \sec^{2} u du = \tan u = \tan |\sec^{-1}x|.
\end{eqnarray*}

\smallskip
\noindent
\# 65)~~$$\lim_{x \rightarrow 0} {\sin^{-1} 5x \over x} = \lim_{x
\rightarrow 0} {5 \over \sqrt{1 - 25x^{2}}} = 5.$$

\smallskip
\noindent
\# 73)~~$$y^{'} = {1 \over \sqrt{1 - x^{2}}},~y(0) = 0$$
Thus
$$y = \int {1 \over \sqrt{1 - x^{2}}} dx = \sin^{-1} x + C$$
and $0 = y(0) = \sin^{-1} 0 + C = C$ so $y = \sin^{-1} x$.


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