\magnification=\magstep1
\def\t{\theta}
\def\st{\sin(\theta)}
\def\ct{\cos(\theta)}
\def\dt{\ d\theta}
\centerline{Mathematics 132 -- Calculus for Physical and Life Science 2}
\centerline{Solutions for Discussion 3 -- Choosing an Integration Method}
\centerline{February 21, 2005}
\bigskip
\noindent
A)  This can be done either by trigonometric substitution, 
or integration by parts.  Using the substitution $t = 4\st$
(so $dt = 4\ct\dt$), 
$$\eqalign{
\int {t^3\over \sqrt{16 - t^2}}\ dt 
&= \int {(4\st)^3\cdot 4\ct\dt\over \sqrt{16(1 - (\st)^2}}\cr
&= \int {64 \sin^3(\theta) 4\ct \dt\over 4\ct}\cr
&= 64\int \sin^3(\t)\dt\cr
&= 64\left({-1\over 3}\sin^2(\t)\ct - {2\over 3}\ct\right) + C  \qquad {\rm\ (using\ \# 17\ in\ table)}\cr
&= 64\left({-1\over 3}\left({t\over 4}\right)^2\cdot {\sqrt{16-t^2}\over 4} - {2\over 3}{\sqrt{16-t^2}\over 4}\right) + C\cr
&= {-t^2\over 3}\sqrt{16-t^2} -{32\over 3}\sqrt{16-t^2} +C \cr}$$
(Note: This one can also be done by parts with $u = t^2$!)
\bigskip
\noindent
B)  Rewrite the function algebraically first, splitting into a sum 
of two terms and cancelling powers of $x$:
$$\int{x + x^{5/2}\over x^2}\ dx = 
\int {1\over x} + x^{1/2}\ dx = \ln|x| + {2\over 3} x^{3/2} + C.$$
\bigskip
\noindent
C)  Note that ${d\over dx}\sqrt{x^2 + 3} = {x \over \sqrt{x^2 + 3}}$.
So the form of this integral is $\int \sin(u)\ du$ if we make the 
substitution $u = \sqrt{x^2 + 3}$.  We have
$$\int \sin(\sqrt{x^2 + 3}) {x \over \sqrt{x^2 + 3}}\ dx 
= \int\sin(u)\ du = -\cos(u) + C = -\cos(\sqrt{x^2 + 3}) + C.$$
\bigskip
\noindent
D)  Start by applying integration by parts, with $u = \arcsin(v)$
and $dw = v^2\ dv$.  Then 
$$
\int v^2 \arcsin(v)\ dv = {v^3\over 3}\arcsin(v) - {1\over 3} \int {v^3\over 
\sqrt{1 - v^2}}\ dv.$$
The remaining integral now has almost the same form as the integral 
in question A (except the quadratic inside the square
root is $1 - v^2$, not $16 - v^2$.  So we can finish with a trigonometric
substitution $v = \st$, $dv = \ct\dt$.  We get
$$\eqalign{
&= {v^3\over 3}\arcsin(v) - {1\over 3}\int\sin^3(\t)\dt\cr
&= {v^3\over 3}\arcsin(v) - {1\over 3}\left({-1\over 3}\sin^2(\t)\ct - {2\over 3}\ct\right) + C\cr
&= {v^3\over 3}\arcsin(v) + {1\over 9}v^2\sqrt{1-v^2} + {2\over 9}\sqrt{1-v^2}+C\cr}$$
\bigskip
\noindent
E)  Note that ${d\over dx}\sin(x^2) = \cos(x^2)\cdot 2x$.  Hence
after the substitution $u = \sin(x^2)$, this integral becomes
$$\int_{u=0}^{u=1} u^5\ du = \left {u^6\over 12}\right|_0^1 = {1\over 12}.$$
Note:  It is {\it not correct} to write the limits of integration
on the substituted integral as the original $0$ and $\sqrt{\pi/2}$.
Those are $x$-values, not $u$-values.  The corresponding values
of $u$ are found with the substitution equation:
$$\eqalign{x = 0 &\leftrightarrow u = \sin(0^2) = 0\cr
           x = \sqrt{\pi/2}  &\leftrightarrow u = \sin((\sqrt{\pi/2})^2) = 
           \sin(\pi/2) = 1\cr}$$
\bigskip
\noindent
F)  Complete the square, substitute $w = x - 2$,
and use the table of integrals (\# 31 and \# 29):
$$\eqalign{\int \sqrt{x^2 - 4x}\ dx &= \int\sqrt{(x-2)^2 - 4}\ dx\cr
&= {x-2\over 2}\sqrt{x^2-4x} - 2\ln|x-2 + \sqrt{x^2 - 4x}| + C\cr}$$
\bigskip
\noindent
G)  This one is done by the method of partial fractions.  The 
denominator factors as $u(u^2 + 8u + 17)$.  So we have 
$${u + 1\over u(u^2 + 8u + 17)} = {A \over u} + {B u + C\over u^2 + 8u + 17}$$
For some $A,B,C$.  To find them, we clear denominators to see:
$$u + 1 = A(u^2 + 8u + 17) + (B u + C)u = (A+B)u^2 + (8A + C)u + 17A.$$
Then equating coefficients,
$A+B =0, 8A + C = 1, 17A = 1$.  Therefore,
$$A = {1\over 17}, \qquad B = {-1\over 17}, \qquad C = {9\over 17}$$
So,
$$\int {u + 1\over u(u^2 + 8u + 17)}\ du = 
{1\over 17} \int {du\over u} + {1\over 17} \int {-u + 9\over u^2 + 8u + 17}\ du
\leqno(1)$$
The first term here is easy; we just get ${1\over 17}\ln|u|$.  For the other
term, we complete the square downstairs:  $u^2 + 8u + 17 = (u+4)^2 + 1$,
and substitute $w = u + 4$ (so $u = w - 4$, and $dw = du$):
$${1\over 17} \int {-u + 9\over u^2 + 8u + 17}\ du = 
{1\over 17} \int {-w + 13\over w^2 + 1}\ dw$$
At this point, we can use the table, or just split up the integral
into two parts:
$$\eqalign{{1\over 17} \int {-w + 13\over w^2 + 1}\ dw &= 
{1\over 17} \int {-w\over w^2 + 1}\ dw + 
{13\over 17} \int {dw \over w^2 + 1}\cr
&= -{1\over 34} \ln|w^2 + 1| + {13\over 17}\arctan(w) + C\cr}$$
Substituting $w = x+4$ back in, and adding the first term above, we
get the final answer:
$$\int {u + 1\over u(u^2 + 8u + 17)}\ du = 
{1\over 17}\ln|u| - {1\over 34} \ln|x^2 + 8x + 17| + {13\over 17}\arctan(x+4) + C$$

\bye