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\title{Math 166: Honors Calculus II\\
Exam I Solutions}
\date{February 18, 1999}

\begin{document}

\maketitle

\bigskip

Problems 1 and 2 are definitions and statements of theroems that
can be found in the text.

\nextproblem

3. Compute $\ddx\int_{-\log(x)}^{\log(x)} \sin(e^t)\,dt$

Let $A(x) = \int_a^x \sin(e^t) dt$. Then $A'(x) = \sin(e^x)$. Taking the
derivative of
$\int_{-\log(x)}^{\log(x)} \sin(e^t)\,dt = A(\log(x)) -
A(-\log(x))$ gives
\[
A'(\log(x))\frac{1}{x} -A'(-\log(x))(-\frac{1}{x}) =
\sin(e^{\log(x)})\frac{1}{x} + \sin(e^{-\log(x)})\frac{1}{x} =
\frac{\sin(x) + \sin(1/x)}{x}
\]

  a) Compute $\ddx \cos(x)^{\sin(x)}$.


\begin{eqnarray*}
\ddx \cos(x)^{\sin(x)} &=& \ddx \exp[\sin(x)\log(\cos(x))] =
\exp[\sin(x)\log(\cos(x))] [\cos(x)\log(x)-\frac{\sin^2(x)}{\cos(x)}]\\
&=& \cos(x)^{\sin(x)}[\cos(x)\log(x)-\sin(x)\tan(x)]
\end{eqnarray*}


  b) Prove $\ddx\arcsin(x) = \frac{1}{\sqrt{1-x^2}}$.

$y = \arcsin(x) \Leftrightarrow \sin(y) = x$. By the Theorem on Derivatives
of Inverse Functions,
\[
\ddx \arcsin(x) = \frac{]}{\sin'(y)} = \frac{1}{\cos(y)} =
\frac{1}{\sqrt{1 - \sin^2(y)}} = \frac{1}{\sqrt{1 - x^2}}
\]

\nextproblem

4. Compute the following integrals.


  a) $\int \frac{x-1}{\sqrt{x-2}}\,dx$

Let $u = x-2$. Then $x-1 = u+1$ and $du = dx$, so
\[
\int \frac{x-1}{\sqrt{x-2}}\,dx = \int \frac{u+1}{\sqrt{u}}\,du
= \int u^{1/2} + u^{-1/2}\,du = \frac{2}{3}u^{3/2} + 2 u^{1/2} + C
= \frac{2}{3}(x-2)^{3/2} + 2\sqrt{x-2} + C
\]

  b) $\int x^2 e^x\, dx$

Use integration by parts twice. The first time let $u = x^2$, $dv = e^x dx$,
$du = 2xdx$, $v = e^x$, so
\[
  \int x^2 e^x\, dx = x^2 e^x - 2 \int xe^x\,dx
\]
The second time let $u = x$, $dv = e^xdx$, $du = dx$, $v = e^x$, so
\[
x^2 e^x - 2 \int xe^x\,dx = x^2e^x - 2[ xe^x - \int e^x\,dx ]
= x^2 e^x - 2xe^x + 2e^x + C
\]

  c) $\int \log x\, dx$

Ue integration by parts. Let $u = \log(x)$, $dv = dx$, $du = \frac{1}{x}\,dx$,
$v = x$, so
\[
\int \log x\, dx = x\log(x) - \int x\frac{1}{x}\,dx = x\log(x) - x + C
\]

  d) $\int \frac{1}{(x^2+1)^2}\,dx$

Use a trig substitution, $\tan(u) = x$, $\sec^2(u)du = dx$,
$x^2 + 1 = \tan^2(u) + 1 = \sec^2(u)$, to get
\begin{eqnarray*}
\int \frac{1}{(x^2+1)^2}\,dx &=& \int \frac{1}{\sec^4(u)}\sec^2(u)\,du
 = \int \cos^2(u)\,du\\
 &=& \int \frac{1}{2} + \frac{1}{2}\cos(2u)\,du
 = \frac{1}{2}u + \frac{1}{4}\sin(2u) + C
 = \frac{1}{2}u + \frac{1}{2}\sin(u)\cos(u) + C\\
 &=& \frac{1}{2}\arctan(x) + \frac{1}{2}\frac{\tan(u)}{\sec^2(u)} + C
 = \frac{1}{2}\arctan(x) + \frac{x}{2(x^2+1)} + C
\end{eqnarray*}

  e) $\int \frac{1}{\sqrt{1 + x - x^2}}\,dx$

Complete the square, $1 + x - x^2 = \frac{5}{4} - (x-\frac{1}{2})^2$, and
use the standard formula
\[
\int \frac{1}{\sqrt{a^2 - u^2}}\,du = \arcsin\Big(\frac{u}{a}\Big) + C
\]
to obtain
\[
\int \frac{1}{\sqrt{1 + x - x^2}}\,dx
= \int\frac{1}{\sqrt{\frac{5}{4} - (x-\frac{1}{2})^2}}\,dx
=\arcsin\Big(\frac{2}{\sqrt{5}}(x - \frac{1}{2})\Big) + C
=\arcsin\Big(\frac{2x - 1}{\sqrt{5}}\Big) + C
\]

  f) $\int \frac{1}{x^2-5x+7}\,dx$

Complete the square again, $x^2 - 5x + 7 = (x - \frac{5}{2})^2 +
\frac{3}{4}$, and use the standard formula
\[
\int \frac{1}{u^2 + a^2}\,du = \frac{1}{a}\arctan\Big(\frac{u}{a}\Big) + C
\]
to obtain
\[
\int\frac{1}{x^2-5x+7}\,dx = \int\frac{1}{(x-\frac{5}{2})^2 + \frac{3}{4}}\,dx
= \frac{2}{\sqrt{3}}\arctan\Big(\frac{2}{\sqrt{3}}(x - \frac{5}{2})\Big) + C
= \frac{2}{\sqrt{3}}\arctan\Big(\frac{2x - 5}{\sqrt{3}}\Big) + C
\]

\nextproblem

5.  a) Give the partial fraction decomposition of
  $\frac{4x^3 - x^2 - 4x + 5}{(x-1)^2(x^2+1)}$

The decomposition should have the form
\[
\frac{4x^3 - x^2 - 4x + 5}{(x-1)^2(x^2+1)} =
\frac{A_1}{x-1} + \frac{A_2}{(x-1)^s} + \frac{Bx + C}{x^2 + 1}
\]
Multiplying through by the denominator on the left-hand side, we get
\[
4x^3 - x^2 - 4x + 5 =
A_1(x-1)(x^2+1) + A_2(x^2+1) + (Bx + C)(x^2-1)
\]
Plugging in $x = 1$ we see immediately that $A_2 = 2$. Equating the
coefficients of $x^k$, $k=3,2,1,0$, on both sides of this equation gives,
respectively,
\begin{eqnarray*}
4 &=& A_1 + B\\
-1 &=& -A_1 + A_2 - 2B + C\\
-4 &=& A_1 + B - 2C\\
5 &=& -A_1 + A_2 + C
\end{eqnarray*}
Subtract the second equation from the fourth to get $6 = 2B$, or $B = 3$.
The first equation then gives $A_1 = 4-B = 4 - 3 = 1$, and the fourth equation
gives $C = 5 + A_1 - A_2 =  5 + 1 - 2 = 4$.

\newpage

  b) Use a trignometric substitution to integrate
  $\int \frac{x}{\sqrt{x^2 - 6x}}\, dx$.

Complete the square, $x^2 - 6x = (x-3)^2 - 9$, and use the substitution
$3\sec(u) = x - 3$, $3\sec(u)\tan(u)du = dx$, 
$\sqrt{(x-3)^2 - 3^2} = 3\sqrt{\sec^2(u) - 1} = 3\tan(u)$ to obtain
\begin{eqnarray*}
\int \frac{x}{\sqrt{(x-3)^2 - 3^2}}\, dx &=&
\int \frac{3\sec(u)+3}{3\tan(u)}3\sec(u)\tan(u)\,du\\
&=& 3\int \sec^2(u) + \sec(u)\,du = 3\tan(u) + 3\log|\sec(u) + \tan(u)| + C\\
&=& \sqrt{x^2 - 6x} + 3\log\Big|\frac{x-3}{3} + \frac{\sqrt{x^2-6x}}{3}\Big| + C
\end{eqnarray*}

  c) Convert to an integral involving rational functions:
  $\int \frac{\sin x}{2 +  \cos x}\, dx$.

Substitute $u = \tan(x/2)$, $dx = \frac{2}{1+u^2}\,du$,
$\sin(x) = \frac{2u}{1+u^2}$ and $\cos(x) = \frac{1-u^2}{1+u^2}$ to get
\[
\int \frac{\sin x}{2 +  \cos x}\, dx =
\int \frac{\frac{2u}{1+u^2}}{\big(2+\frac{1-u^2}{1+u^2}\big)}
\frac{2}{(1+u^2)}\,du
= \int \frac{4u}{(1+u^2)(2(1+u^2) + 1 - u^2)}\,du
= \int \frac{4u}{(1+u^2)(3+u^2)}\,du
\]

\nextproblem

6. Find the Taylor polynomials.


  a) $T_{4n+1}\left(\frac{x}{1+x^4}\right)$

The algebraic identity $\frac{1}{1-u} = 1 + u + u^2 + \cdots + u^n +
\frac{u^{n+1}}{1-u}$ implies
\begin{eqnarray*}
\frac{x}{1+x^4} &=& x\left(\frac{1}{1 - (-x^4)}\right)
 = x\left(1 + (-x^4) + (-x^4)^2 + \cdots + (-x^4)^n +
\frac{(-x^4)^{n+1}}{1-(-x^4)}\right)\\
&=& x - x^5 + x^9 - \cdots + (-1)^nx^{4n+1} +
x^{4n+1}\frac{(-1)^{n+1}x^{4n}}{1+x^4}
\end{eqnarray*}
The last term is of the form $x^{4n+1}g(x)$ where $g(0) = 0$,
so our standard theorem about Taylor polynomials implies that
\[
T_{4n+1}\left(\frac{x}{1+x^4}\right)
= x - x^5 + x^9 - \cdots + (-1)^nx^{4n+1}
=\sum_{k=0}^{n} (-1)^k x^{4k+1}
\]

  b) $T_{2n}(\cos^2(x))$

Using the identity $\cos^2(x) = \frac{1}{2} + \frac{1}{2}\cos(2x)$ and
the linearity property of Taylor polynomials, we get
\[
T_{2n}(\cos^2(x)) = T_{2n}\Big(\frac{1}{2} + \frac{1}{2}\cos(2x)\Big)
=\frac{1}{2} + \frac{1}{2}T_{2n}(\cos(2x))
\]
We know the Taylor polynomial of cosine,
\[
T_{2n}(\cos(x)) = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \cdots +
\frac{(-1)^n}{(2n)!}x^{2n}
\]
so by the substitution property,
\[
T_{2n}(\cos(2x)) = 1 - \frac{1}{2!}(2x)^2 + \frac{1}{4!}(2x)^4 - \cdots +
\frac{(-1)^n}{(2n)!}(2x)^{2n}
\]
Therefore,
\begin{eqnarray*}
T_{2n}(\cos^2(x))
&=&\frac{1}{2} + \frac{1}{2}(1 - \frac{1}{2!}(2x)^2 + \frac{1}{4!}(2x)^4 -
\cdots + \frac{(-1)^n}{(2n)!}(2x)^{2n})\\
&=& 1 - \frac{2^1}{2!}x^2 + \frac{2^3}{4!}x^4 - \cdots +
\frac{(-1)^n2^{2n-1}}{(2n)!}x^{2n}\\
&=& 1 + \sum_{k=1}^n \frac{(-1)^k 2^{2k-1}}{(2k)!}x^{2k}
\end{eqnarray*}


\end{document}
