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\leftline{Mathematics 132 -- Calculus for Physical and Life Sciences 2}
\leftline{Makeup Quiz Solutions -- February 29, 2008}
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Some possibly useful formulas:
$$\sin^2\theta + \cos^2\theta = 1, \qquad \tan^2\theta + 1 = \sec^2\theta, \qquad \sin(2\theta) = 2\sin\theta\cos\theta.$$
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(5) A)  Integrate:  $$\int {x^3 \over \sqrt{9 - x^2}}\, dx$$ 
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Because of the $9 - x^2$ under the radical sign, we use the $\sin$ substitution,
letting $x = 3\sin\theta$.  Then $dx = 3\cos\theta\, d\theta$ and we proceed as follows:
$$\eqalign{
\int {x^3 \over \sqrt{9 - x^2}}\, dx &= \int {(3\sin\theta)^3\over \sqrt{9 - (3\sin\theta)^2}}\, 3 \cos\theta\, d\theta\cr
&= \int {27 \sin^3\theta\cdot 3 \cos\theta\over \sqrt{9(1 - \sin^2\theta)}}\, d\theta\cr
&= \int {27 \sin^3\theta\cdot 3 \cos\theta\over \sqrt{9\cos^2\theta}}\, d\theta \quad {\rm (by\ the\ basic\ identity)}\cr
&= \int {27 \sin^3\theta\cdot 3 \cos\theta\over 3\cos\theta}\, d\theta \cr
&= 27 \int \sin^3\theta\, d\theta \quad {\rm (after\ cancelling)} \cr
&= 27 \int \sin^2\theta \cdot \sin\theta\, d\theta \quad {\rm (odd\ power\ strategy)}\cr
&= 27 \int (1 - \cos^2\theta) \sin\theta\, d\theta\cr
&= 27 \left(\int \sin\theta\, d\theta - \int \cos^2\theta\sin\theta\, d\theta\right) \quad {\rm (multiply\ out\ and\ split)}\cr
&= 27\left(-\cos\theta + {1\over 3}\cos^3\theta\right) + C}$$
The last line comes from applying the $u$-substitution $u = \cos\theta$ to the second integral 
(note that $du = -\sin\theta\, d\theta$).  Now we convert back to functions of $x$ from the 
substitution $x = 3\sin\theta$.  The reference triangle has opposite side $x$, hypotenuse $3$ and
adjacent side $\sqrt{9 - x^2}$.  Hence $\cos\theta = \sqrt{9 - x^2}/3$, and the integral equals
$$ = -9 \sqrt{9 - x^2} + {1\over 3}(9 - x^2)^{3/2} + C.$$
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(5) B)  Integrate:  $$\int {x^4 + 2x + 1\over x^3 + x}\, dx$$
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Since the degree of the top is larger, we have to divide the denominator into the 
numerator.  The quotient is $x$ and the remainder is $x^4 + 2x + 1 - x(x^3 + x) = -x^2 + 2x + 1$.
Therefore, 
$${x^4 + 2x + 1\over x^3 + x} = x + {-x^2 + 2x + 1\over x^3 + x} = x + {-x^2 + 2x + 1\over x(x^2 + 1)}.$$
We split the second term up into partial fractions like this:
$$\eqalign{
{-x^2 + 2x + 1\over x(x^2 + 1)} &= {A \over x} + {Bx + C\over x^2 + 1}, \quad {\rm (so\ after\ clearing\  denoms)}\cr
-x^2 + 2x + 1 &= A(x^2 + 1) + x(Bx + C)\cr
-x^2 + 2x + 1 &= (A + B)x^2 + Cx + A\cr}$$
By comparing coefficients $A = 1$, $C = 2$ and $A + B = -1$.  Hence $B = -2$.  This gives
$$
\eqalign{
x + {-x^2 + 2x + 1\over x(x^2 + 1)} &= x + {1\over x} + {-2x + 2\over x^2 + 1} \quad ({\rm so})\cr
\int x + {-x^2 + 2x + 1\over x(x^2 + 1)}\, dx &= \int x + {1\over x} + {-2x + 2\over x^2 + 1}\, dx\cr 
&= {x^2\over 2} + \ln|x| + \int {-2x\over x^2 + 1}\, dx + \int {2\over x^2 + 1}\, dx\cr
&= {x^2\over 2} + \ln|x| - \ln(x^2 + 1) + 2\tan^{-1}(x) + C\cr}$$
The third term here comes from the integral  $\int {-2x\over x^2 + 1}\, dx$, which is $-\int {du\over u}$
for $u = x^2 + 1$.  The last term is the basic inverse tangent integral.  

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