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\begin{document}

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

February 5, 2003}
\end{center}


\begin{enumerate}

\item Find the integrals
	\begin{enumerate}
	\item  \[  \int \cos^3x \sin^6x \; dx \]
		\sol $\frac{1}{7}\sin^7x - \frac{1}{9}\sin^9x+C$
	\item  \[  \int \sin^2(3x) \; dx \]
		\sol $\frac{1}{2}- \frac{1}{12}\sin(6x) +C$
	\item  \[  \int \sec x \; dx \]
		\sol: This involves the following "trick:"
	\begin{eqnarray*}
	\int \sec x \left(\frac{\tan x + \sec x}{\tan x + \sec x} \right) \; dx 
	& = & \int \frac{\sec x\tan x + \sec^2 x}{\tan x + \sec x} \; dx 
	\end{eqnarray*}
	Now, you make the substitution $u=\sec x + \tan x$ and you should get:
	\[
	\int \sec x \; dx  = \ln |\sec x+ \tan x | +C
	\]

	\item  \[  \int \csc x \; dx \]
	\sol: (mimic the trick above) $-\ln|\csc x + \cot x| +C$.
	\end{enumerate}

\end{enumerate}

\newpage

Here is an example similar to what we were doing in class on February~3.
The point is to repeatedly use the half-angle identities (bottom of page 403) and keep simplifying algebraically.

\begin{eqnarray}
\int \cos^4x \sin^4x \; dx & = &
	\int (\cos^2x)^2 (\sin^2x)^2 \; dx \\
	& = & \int \left(\frac{1}{2}(1+\cos(2x))\right)^2 
		\left(\frac{1}{2}(1-\cos(2x))\right)^2 \; dx \\
	& = & \frac{1}{16} \int \left( (1+\cos(2x))(1-\cos(2x)) \right)^2 \; dx \\
	& = & \frac{1}{16} \int \left( 1-\cos^2(2x) \right)^2 \; dx \\
	& = & \frac{1}{16} \int 1-2\cos^2(2x)+\cos^4(2x) \; dx \\
	& = & \frac{1}{16} \int 1-2\left(\cos^2(2x)\right)+\left(\cos^2(2x)\right)^2 \; dx \\
	& = & \frac{1}{16} \int 1 - 2\left(\frac{1}{2}(1+\cos(4x))\right) + 		\left(\frac{1}{2}(1+\cos(4x))\right)^2 \; dx \\
	& = & \frac{1}{16} \int 1 - (1+\cos(4x)) 
		+ \frac{1}{4}(1+2\cos(4x)+\cos^2(4x)) \; dx \\
	& = & \frac{1}{16} \int \frac{1}{4} - \frac{1}{2}\cos(4x) 
		+ \frac{1}{4}\cos^2(4x) \; dx \\
	& = & \frac{1}{16} \int \frac{1}{4} - \frac{1}{2}\cos(4x) 
		+ \frac{1}{4}\left( \frac{1}{2}(1+ \cos(8x))\right) \; dx \\
	& = & \frac{1}{16} \int \frac{3}{8} - \frac{1}{2}\cos(4x) 
		+ \frac{1}{8} \cos(8x) \; dx \\
	& = & \int \frac{3}{128} - \frac{1}{32}\cos(4x) 
		+ \frac{1}{128} \cos(8x) \; dx \\
	& = & \frac{3}{128}x - \frac{1}{128}\sin(4x) 
		+ \frac{1}{1024} \sin(8x) + C
\end{eqnarray}

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