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%TCIDATA{<META NAME="Title" CONTENT="Reference/Mathematics/Trigonometric Integrals">}
%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
%TCIDATA{LastRevised=Sat Mar 29 22:37:37 1997}
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\begin{document}


\section{Trigonometric Integrals}

\subsubsection{Trigonometric Integrals \label{trigonometric integrals}}

\begin{enumerate}
\item  T\QTR{BOXHEAD}{o evaluate $\tint \,\,\sin ^{m}x\cos \,^{n}x\,dx$}

\begin{enumerate}
\item  If the power of cosine is odd $(n=2k+1)$, save one cosine factor and
use $\cos ^{2}x=1-\sin ^{2}x$ to express the remaining factors in terms of
sine: 
\begin{eqnarray*}
\dint \,\sin ^{m}x\,\cos ^{2k+1}x\,dx &=&\dint \,\,\sin ^{m}x(\cos
^{2}x)^{k}\cos \,x\,dx \\
&=&\dint \,\,\sin ^{m}x(1-\sin ^{2}x)^{k}\cos \,x\,dx
\end{eqnarray*}
Then substitute $u=\sin \,x$.

\item  If the power of sine is odd $(m=2k+1)$, save one sine factor and use $%
\sin ^{2}x=1-\cos ^{2}x$ to express the remaining factors in terms of
cosine: 
\begin{eqnarray*}
\dint \,\,\sin ^{2k+1}x\,\cos ^{n}x\,dx &=&\dint \,\,(\sin ^{2}x)^{k}\cos
^{n}x\,\sin \,x\,dx \\
&=&\dint \,\,(1-\cos ^{2}x)^{k}\cos ^{n}x\,\sin \,x\,dx
\end{eqnarray*}
Then substitute $u=\cos \,x$.

\item  If the powers of both sine and cosine are even, use the half-angle
identities 
\[
\sin ^{2}x=\tfrac{1}{2}(1-\cos \,2x)\qquad \cos ^{2}x=\tfrac{1}{2}(1+\cos
\,2x)
\]
It is sometimes helpful to use the identity 
\[
\sin \,x\,\cos \,x=\tfrac{1}{2}\,\sin \,2x
\]
\end{enumerate}

\item  \QTR{BOXHEAD}{To evaluate }$\int \,\,\tan ^{m}x\,\sec ^{n}x\,dx$

\begin{enumerate}
\item  If the power of secant is even $(n=2k)$, save a factor of $\sec ^{2}x$
and use $\sec ^{2}x=1+\tan ^{2}x$ to express the remaining factors in terms
of $\tan x$: 
\begin{eqnarray*}
\dint \,\,\tan ^{m}x\sec ^{2k}x\,dx &=&\dint \,\,\tan ^{m}x(\sec
^{2}x)^{k-1}\sec ^{2}x\,dx \\
&=&\dint \,\,\tan ^{m}x(1+\tan ^{2}x)^{k-1}\sec ^{2}x\,dx
\end{eqnarray*}
Then substitute $u=\tan \,x$.

\item  If the power of tangent is odd $(m=2k+1)$, save a factor of $\sec
x\tan x$ and use $\tan ^{2}x=\sec ^{2}x-1$ to express the remaining factors
in terms of $\sec x$: 
\begin{eqnarray*}
\dint \,\,\tan ^{2k+1}x\,\sec ^{n}x\,dx &=&\dint \,\,(\tan ^{2}x)^{k}\sec
^{n-1}x\sec \,x\,\tan \,x\,dx \\
&=&\dint \,\,(\sec ^{2}x-1)^{k}\sec ^{n-1}x\,\sec \,x\,\tan \,x\,dx
\end{eqnarray*}
Then substitute $u=\sec x$.
\end{enumerate}
\end{enumerate}

\subsection{Related Reference Topics}

{\small \hyperref{Table of integrals}{}{}{TAB_integrals.tex}}

{\small \hyperref{Table of trigonometric functions and identities}{}{}{%
TAB_trigonometry.tex}}

\subsection{Related Scientific Notebook Topics}

{\small \hyperref{Calculus, change variable}{}{}{../../Help/DM7-5.tex#Change
of variables}}

{\small \hyperref{Indefinite integral}{}{}{../../Help/DM7-4.tex#Indefinite
integral}}

{\small \hyperref{Trigonometry}{}{}{../../Help/DM4.tex}}

\noindent \hrulefill

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