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

\begin{document}
\centerline{MATH 136 -- Calculus 2}
\centerline{An Application of Integration By Parts:}
\centerline{Reduction Formulas}
\centerline{February 10, 2020}
\vskip 10pt
\noindent
{\it Background}
\vskip 10pt
There are a number of cases where repeated integration by parts can 
be used to completely evaluate integrals.  Instead of working these
out laboriously by hand every time, many mathematicians who need
to use these will use a \emph{reduction formula} that shows the 
general pattern of each application of the parts formula.  These
are usually obtained by consulting a compiled \emph{table of integrals}
rather than worked out by hand.  Here is an example:
$$\int x^n e^{ax}\ dx = \frac{x^n e^{ax}}{a} - \frac{n}{a}\int x^{n-1} e^{ax}\ dx$$
(This is derived by letting $u = x^n$ and $dv = e^{ax}\ dx$ as in an
example from the video on ``parts.'')  If we want to integrate $\int x^3 e^{4x}\ dx$, 
for instance, we apply this reduction formula three times with $n = 3$, then $n = 2$, 
then $n = 1$.  The power of $x$ is eventually reduced to a constant $1$.  
\begin{align*}
\int x^3 e^{4x}\ dx &= \frac{x^3 e^{4x}}{4} - \frac{3}{4} \int x^2 e^{4x}\ dx\\
&= \frac{x^3 e^{4x}}{4} - \frac{3}{4}\left(\frac{x^2e^{4x}}{4} - \frac{2}{4}\int x e^{4x}\ dx\right)\\
&= \frac{x^3 e^{4x}}{4} - \frac{3 x^2 e^{4x}}{16} + \frac{3}{8}\left(\frac{xe^{4x}}{4} - \frac{1}{4}\int e^{4x}\ dx\right)\\
&= \frac{x^3 e^{4x}}{4} - \frac{3 x^2 e^{4x}}{16} + \frac{3x e^{4x}}{32} - \frac{3e^{4x}}{128} + C.
\end{align*}

\vskip 10pt
\noindent
{\it Questions}

\begin{enumerate}
\item[(A)] Derive these reduction formulas for $n \ge 1$:
$$\int x^n \cos(ax)\ dx = \frac{x^n \sin(ax)}{a} - \frac{n}{a}\int x^{n-1}\sin(ax)\ dx$$
and 
$$\int x^n \sin(ax)\ dx = \frac{-x^n \cos(ax)}{a} + \frac{n}{a}\int x^{n-1}\cos(ax)\ dx$$
by integrating by parts (once each).
\item[(B)] Using the two reduction formulas from part (A) in sequence, integrate:
$$\int x^2 \cos(3x)\ dx$$
\item[(C)] Derive the following reduction formula by applying parts with $u = \sin^{n-1}(x)$
and $dv = \sin(x)\ dx$: 
$$\int \sin^n(x)\ dx = \frac{-\sin^{n-1}(x)\cos(x)}{n} + \frac{n-1}{n} \int \sin^{n-2}(x)\ dx.$$
\item[(D)]  Use the formula from (C) to compute $\int \sin^5(x)\ dx$.
\item[(E)]  Derive the following reduction formula for $n \ge 2$:
$$\int \tan^n(u)\ du = \frac{\tan^{n-1}(u)}{n-1} - \int \tan^{n-2}(u)\ du.$$
(Hint:  This does \emph{not} come from an integral by parts, 
but rather uses a ``clever'' application
of the trig identity $\tan^2(x) = \sec^2(x) - 1$.)
The case $n = 1$ is handled by a separate formula we have seen before ($u$-subsitution):
$$\int \tan(u)\ du = -\ln|\cos(u)| + C$$
\item[(F)] Use the formulas from (E) to integrate
$$\int \tan^5(x)\ dx$$
\end{enumerate}
\end{document}
