\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{amsmath}

\begin{document}
\centerline{MATH 136 -- Calculus 2}
\centerline{Practice Day on Trigonometric Integrals}
\centerline{February 11, 2020}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
Our approach to trigonometric integrals will be based in systematic
application of \emph{reduction formulas} like the ones we saw yesterday.  
When you do these 
and exams, the formulas on the sheet I distributed \emph{will be provided} for your
use in table format.  You will be responsible for deciding which formula
applies and for applying it correctly.  Here are some practice problems.
\vskip 10pt
\noindent
{\it Questions}
\vskip 10pt
For each of the following integrals, decide which of the reduction formula applies,
determine the appropriate $n$ (and $m$ in the last two).  Then apply the formula and
complete the computation.
\begin{enumerate}
\item $\displaystyle \int \sin^2(x)\ dx$
\item[] {\it Answer:}  Use SC1:
 $$\frac{x}{2} - \frac{\sin(x)\cos(x)}{2} + C$$
\item $\displaystyle \int \sec^3(x)\ dx$
\item[] {\it Answer:}  Use ST3 and ST4:
$$\frac{\sec(x)\tan(x)}{2} + \frac{1}{2}\ln|\sec(x) + \tan(x)| + C$$
\item $\displaystyle \int \cos^4(3x)\ dx$
\item[] {\it Answer:}  Use SC2 twice:
$$\frac{3x}{8} + \frac{\cos^3(3x)\sin(3x)}{12} + \frac{\cos(3x)\sin(3x)}{8} + C$$
\item $\displaystyle \int \tan^2(4x) \sec^3(4x)\ dx$ (Use $\tan^2(u) = 1 + \sec^2(u)$, then ST3.)
\item[] {\it Answer:}  Using the identity yields the equivalent sum of two integrals:
$$\int\sec^3(4x)\ dx + \int\sec^5(4x)\ dx$$
Using the reduction formulas and simplifying yields:
$$\frac{\sec^3(4x)\tan(4x)}{16} + \frac{7\sec(4x)\tan(4x)}{32} + \frac{7}{32}\ln|\sec(4x) + \tan(4x)| + C.$$
\item $\displaystyle \int \tan^3(5x)\sec^5(5x)\ dx$ (Convert to powers of $\sin(u)$ and $\cos(u)$.) 
\item[] {\it Answer:}  This can be done using SC4 or more cleverly by rewriting $\sin^3(5x)$
as $\sin(5x)\cdot \sin^2(5x) = \sin(5x)\cdot(1 - \cos^2(5x))$:
\begin{align*}
\int \frac{\sin^3(5x)}{\cos^8(5x)}\ dx &= \int \frac{\sin(5x)}{\cos^8(5x)}\ dx - 
\int\frac{\sin(5x)}{\cos^6(5x)}\ dx\\
&= \frac{-1}{35} \cos^{-7}(5x) + \frac{1}{25}\cos^{-5}(5x) + C
\end{align*}


\end{enumerate}
\end{document}

\begin{figure}
\begin{center}
\includegraphics[width=.8\linewidth]{needleregion.png}
\end{center}
\caption{The region for problem 4.}
\end{figure}

\end{document}



