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\begin{center}
{\Large Warm-Up Problems and Lecture Problems

Solutions

February 7, 2003}
\end{center}


\begin{enumerate}

\item For each of the following integrals, determine the correct choice of trigonometric substitution.  (Use the chart handed out last time.)
	\begin{enumerate}
	\item  \[ \int \frac{1}{x^2\sqrt{x^2-25}} \; dx \]
		\\ \sol $x=5 \sec \theta$
	\item  \[ \int \frac{1}{(25+16x^2)^{3/2}} \; dx \]
		\\ \sol $x=\frac{5}{4}\tan \theta$
	\item  \[ \int \frac{\sqrt{5-x^2}}{x} \; dx \]
		\\ \sol $x=\sqrt{5}\sin \theta$
	\end{enumerate}

\item  For the substitutions made in the previous problem, convert the following expressions into expressions of $x$.  For example, if your substitution was $x=\sin \theta$ and you are asked to put $\sin^2\theta$ into an expression of $x$, you would write $\sin^2\theta = x^2$.

Remember to draw the appropriate triangle!
	\begin{enumerate}
	\item  $\theta+\cos \theta -\tan \theta$
		\\ \sol $\cos^{-1}(5/x) + (5/x) - \frac{1}{5}\sqrt{x^2-25}$
	\item  $\theta+\cos \theta -\tan \theta$
		\\ \sol $\tan^{-1}(4x/5) +\frac{5}{\sqrt{25+16x^2}} - \frac{4x}{5}$
	\item  $\theta+\cos \theta -\tan \theta$
		\\ \sol $\sin^{-1}(x/\sqrt{5}) + \frac{1}{\sqrt{5}}\sqrt{5-x^2} - \frac{5}{\sqrt{5-x^2}}$
	\end{enumerate}

\newpage

\item  Write the following fractions in the correct ``partial fraction decomposition form:''
	\begin{enumerate}
	\item  \[ \frac{x^2+3x}{(x^2+x+1)(x-2)} \]
		\\ \sol \[ \frac{x^2+3x}{(x^2+x+1)(x-2)} = \frac{Ax+B}{(x^2+x+1)} + 
				\frac{C}{(x-2)} \]
	\item  \[ \frac{5x^3-1}{(2x+3)^3(x+4)} \]
		\\ \sol \[ \frac{A}{(2x+3)} + \frac{B}{(2x+3)^2} + \frac{C}{(2x+3)^3} + 
			\frac{D}{(x+4)} \]
	\end{enumerate}

\newpage

\item  Solve for the constants in the following partial fraction decomposition:
	\[
	\frac{5}{(2x+1)(x-2)} = \frac{A}{2x+1} + \frac{B}{x-2}
	\]
	\\ \sol: ``Clear'' the denominators:
	\[
	5 = A(x-2) + B(2x+1)
	\]
	Now collect powers of $x$:
	\[
	5 = (A+2B)x + (-2A+B)
	\]
	Now set up equations to solve:
	\begin{eqnarray*}
	A+2B & = & 0 \\
	-2A+B & = & 5
	\end{eqnarray*}
	Now solve the equations and get: $A=-2$ and $B=1$

\newpage

\item  Here is an integral that I have done the partial fraction decomposition.
Do the rest of the work and compute the antiderivative.
(You might want to perform the partial fraction decomposition at home later.)
\[
\int \frac{5x^3-3x^2+2x-1}{x^4+x^2} \; dx 
	= \int \left( \frac{2}{x} - \frac{1}{x^2} + \frac{3x-2}{x^2+1} \right) \; dx 
\]
\\ \sol  The main idea is to split up the integral:
\begin{eqnarray*}
\int \frac{5x^3-3x^2+2x-1}{x^4+x^2} \; dx 
	& = & \int \left( \frac{2}{x} - \frac{1}{x^2} + \frac{3x-2}{x^2+1} \right) \; dx \\
	& = & 
	\int \frac{2}{x} \; dx - \int \frac{1}{x^2}\; dx + \int \frac{3x-2}{x^2+1} \; dx \\
	& = & 2\ln|x| + \frac{1}{x} + 3\int \frac{x}{x^2+1} \; dx 
		- 2\int \frac{1}{x^2+1} \; dx
\end{eqnarray*}
We now use a $u$-substitution ($u=x^2+1$) to do the second to last integral and the last integral is $\tan^{-1}x$:
\begin{eqnarray*}
\int \frac{5x^3-3x^2+2x-1}{x^4+x^2} \; dx 
	& = & 2\ln|x| + \frac{1}{x} + \frac{3}{2}\ln(x^2+1) -2 \tan^{-1}x + C
\end{eqnarray*}



\end{enumerate}

\end{document}