Skip to main content Contents Prev Up Next \(\newcommand{\Z}{{\mathbb Z}}
\newcommand{\Q}{{\mathbb Q}}
\newcommand{\R}{{\mathbb R}}
\newcommand{\C}{{\mathbb C}}
\newcommand{\T}{{\mathbb T}}
\newcommand{\F}{{\mathbb F}}
\newcommand{\PP}{{\mathbb P}}
\newcommand{\HH}{{\mathbb H}}
\newcommand{\compose}{\circ}
\newcommand{\bolda}{{\mathbf a}}
\newcommand{\boldb}{{\mathbf b}}
\newcommand{\boldc}{{\mathbf c}}
\newcommand{\boldd}{{\mathbf d}}
\newcommand{\bolde}{{\mathbf e}}
\newcommand{\boldi}{{\mathbf i}}
\newcommand{\boldj}{{\mathbf j}}
\newcommand{\boldk}{{\mathbf k}}
\newcommand{\boldn}{{\mathbf n}}
\newcommand{\boldp}{{\mathbf p}}
\newcommand{\boldq}{{\mathbf q}}
\newcommand{\boldr}{{\mathbf r}}
\newcommand{\bolds}{{\mathbf s}}
\newcommand{\boldt}{{\mathbf t}}
\newcommand{\boldu}{{\mathbf u}}
\newcommand{\boldv}{{\mathbf v}}
\newcommand{\boldw}{{\mathbf w}}
\newcommand{\boldx}{{\mathbf x}}
\newcommand{\boldy}{{\mathbf y}}
\newcommand{\boldz}{{\mathbf z}}
\newcommand{\boldzero}{{\mathbf 0}}
\newcommand{\boldmod}{\boldsymbol{ \bmod }}
\newcommand{\boldT}{{\mathbf T}}
\newcommand{\boldN}{{\mathbf N}}
\newcommand{\boldB}{{\mathbf B}}
\newcommand{\boldF}{{\mathbf F}}
\newcommand{\boldS}{{\mathbf S}}
\newcommand{\boldE}{{\mathbf E}}
\newcommand{\boldG}{{\mathbf G}}
\newcommand{\boldK}{{\mathbf K}}
\newcommand{\boldL}{{\mathbf L}}
\DeclareMathOperator{\lns}{lns}
\DeclareMathOperator{\lcm}{lcm}
\DeclareMathOperator{\Span}{span}
\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\NS}{null}
\DeclareMathOperator{\RS}{row}
\DeclareMathOperator{\CS}{col}
\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\range}{range}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\nullity}{nullity}
\DeclareMathOperator{\sign}{sign}
\DeclareMathOperator{\Fix}{Fix}
\DeclareMathOperator{\Aff}{Aff}
\DeclareMathOperator{\Frac}{Frac}
\DeclareMathOperator{\Ann}{Ann}
\DeclareMathOperator{\Tor}{Tor}
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\mdeg}{mdeg}
\DeclareMathOperator{\Lt}{Lt}
\DeclareMathOperator{\Lc}{Lc}
\DeclareMathOperator{\disc}{disc}
\DeclareMathOperator{\Frob}{Frob}
\DeclareMathOperator{\adj}{adj}
\DeclareMathOperator{\curl}{curl}
\DeclareMathOperator{\grad}{grad}
\DeclareMathOperator{\diver}{div}
\DeclareMathOperator{\flux}{flux}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\ab}{ab}
\newcommand{\surjects}{\twoheadrightarrow}
\newcommand{\injects}{\hookrightarrow}
\newcommand{\bijects}{\leftrightarrow}
\newcommand{\isomto}{\overset{\sim}{\rightarrow}}
\newcommand{\floor}[1]{\lfloor#1\rfloor}
\newcommand{\ceiling}[1]{\left\lceil#1\right\rceil}
\newcommand{\mclass}[2][m]{[#2]_{#1}}
\newcommand{\val}[2][]{\left\lvert #2\right\rvert_{#1}}
\newcommand{\abs}[2][]{\left\lvert #2\right\rvert_{#1}}
\newcommand{\valuation}[2][]{\left\lvert #2\right\rvert_{#1}}
\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\newcommand{\anpoly}{a_nx^n+a_{n-1}x^{n-1}\cdots +a_1x+a_0}
\newcommand{\anmonic}{x^n+a_{n-1}x^{n-1}\cdots +a_1x+a_0}
\newcommand{\bmpoly}{b_mx^m+b_{m-1}x^{m-1}\cdots +b_1x+b_0}
\newcommand{\pder}[2]{\frac{\partial#1}{\partial#2}}
\newcommand{\normalin}{\trianglelefteq}
\newcommand{\angvec}[1]{\langle #1\rangle}
\newcommand{\varpoly}[2]{#1_{#2}x^{#2}+#1_{#2-1}x^{#2-1}\cdots +#1_1x+#1_0}
\newcommand{\varpower}[1][a]{#1_0+#1_1x+#1_1x^2+\cdots}
\newcommand{\limasto}[2][x]{\lim_{#1\rightarrow #2}}
\newenvironment{linsys}[2][m]{
\begin{array}[#1]{@{}*{#2}{rc}r@{}}
}{
\end{array}}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Section 1.19 Trigonometric integrals
We now develop an integration technique for two very particular types of integrals:
\begin{align*}
\int \sin^m x\cos^n x\, dx \amp \amp \int \tan^m x\sec^n x\, dx \text{.}
\end{align*}
Why these particular pairings of products of functions? In a nutshell, because one of the following four choices of substitution will often come in handy:
\begin{align*}
u\amp =\sin x \amp u\amp =\tan x \\
du\amp =\cos x\, dx \amp du\amp =\sec^2 x\\
\amp \\
u\amp =\cos x \amp u\amp =\sec x \\
du\amp =-\sin x\, dx \amp du\amp =\sec x\tan x\text{.}
\end{align*}
Procedure 1.19.2 and
Procedure 1.19.3 articulate in greater detail when and where such substitutions will be useful. The basic principle is that a given substitution as above will help, when after “peeling off” part of the integrand to account for the
\(du\text{,}\) the remaining part of the integrand can be expressed completely in terms of
\(u\text{.}\)
Theorem 1.19.1 . Trigonometric identities.
The following identities hold for all real numbers \(\theta\) and \(\phi\) for which the given expression is defined.
Pythagorean identities.
\begin{align*}
\sin^2\theta+\cos^2\theta \amp = 1 \amp \tan^2\theta+1\amp = \sec^2\theta
\end{align*}
Sum-to-product and double-angle formulas.
\begin{align*}
\sin(\theta+\phi) \amp =\sin\theta\cos\phi+\cos\theta\sin\phi
\amp \sin(\theta-\phi) \amp =\sin\theta\cos\phi-\cos\theta\sin\phi \\
\cos(\theta+\phi)\amp =\cos\theta\cos\phi-\sin\theta\sin\phi
\amp \cos(\theta-\phi)\amp =\cos\theta\cos\phi+\sin\theta\sin\phi \\
\sin(2\theta)\amp =2\sin\theta\cos\theta \\
\cos(2\theta) \amp = \cos^2\theta-\sin^2\theta \text{.}
\end{align*}
Product-to-sum and square formulas.
\begin{align*}
\sin\theta\sin\phi \amp = \frac{1}{2}(\cos(\theta-\phi)-\cos(\theta+\phi)) \\
\cos\theta\cos\phi \amp =\frac{1}{2}(\cos(\theta-\phi)+\cos(\theta+\phi)) \\
\sin\theta\cos\phi \amp = \frac{1}{2}(\sin(\theta-\phi)+\sin(\theta+\phi))\\
\sin^2\theta \amp = \frac{1}{2}(1-\cos 2\theta) \\
\cos^2\theta\amp = \frac{1}{2}(1+\cos 2\theta)\text{.}
\end{align*}
Procedure 1.19.2 . Integrating \(\sin^m x\cos^n x\) .
Let \(m\) and \(n\) be nonnegative integers. When computing
\begin{equation*}
\int \sin^m x\cos^n x\, dx
\end{equation*}
use the following strategies.
Odd sine power.
If \(m=2k+1\) is odd, write
\begin{equation*}
\int \sin^m x\cos^n x\, dx=\int (1-\cos^2x)^k\cos^n x\sin x\, dx
\end{equation*}
and use the substitution \(u=\cos x, du=-\sin x\, dx\text{.}\)
Odd cosine power.
If \(n=2k+1\) is odd, write
\begin{equation*}
\int \sin^m x\cos^n x\, dx=\int \sin^m x(1-\sin^2x)^k\cos x \, dx
\end{equation*}
and use the substitution \(u=\sin x, du=\cos x\, dx\text{.}\)
Even powers. If \(m\) and \(n\) are both even use \(\displaystyle\sin^2 x=\frac{1-\cos 2x}{2}\) and \(\displaystyle\cos^2 x=\frac{1+\cos 2x}{2}\) to reduce to a lower power of \(\cos 2x\text{.}\)
Procedure 1.19.3 . Integrating \(\tan^m x\sec^n x\) .
Let \(m\) and \(n\) be nonnegative integers. When computing
\begin{equation*}
\int \tan^m x\sec^n x \, dx
\end{equation*}
use the following strategies.
Odd tangent power.
If \(m=2k+1\) is odd and \(n\) is positive, write
\begin{equation*}
\int \tan^m x\sec^n x \, dx=\int (\sec^2 x-1)^k\sec^{n-1} x\sec x\tan x \, dx
\end{equation*}
and use the substitution \(u=\sec x, du=\sec x\tan x\, dx\text{.}\)
Even secant power.
If \(n=2k\) is even and positive, write
\begin{equation*}
\int \tan^m x\sec^n x \, dx=\int (\tan^2 x+1)^{k-1}\tan^m x\sec^2 x\, dx
\end{equation*}
and use the substitution \(u=\tan x, du=\sec^2 x\, dx\text{.}\)
Even tangent power, odd secant power. If \(m\) is even and \(n\) is odd, express everything in terms of \(\sec x\) and possibly use integration by parts.
Tangent power. If \(n=0\text{,}\) use the identity \(\tan^2 x=\sec^2 x-1\) and strategies from the previous cases.
Example 1.19.4 . Odd sine power.
Compute \(\displaystyle \int \sin^3x \cos^2 x\, dx\text{.}\)
Solution .
Peel off a factor of \(\sin x\) and use the substitution \(u=\cos x\text{,}\) \(du=-\sin x\text{:}\)
\begin{align*}
\int \sin^3x \cos^2 x\, dx \amp = \int \sin^2x \cos^2 x \sin x\, dx \\
\amp = \int (1-\cos^2x) \cos^2 \sin x\, dx\\
\amp = -\int (1-u^2)u^2\, du \amp (u=\cos x, du=-\sin x dx)\\
\amp = -\frac{1}{3}u^3+\frac{1}{5}u^5+C\\
\amp = -\frac{1}{3}\cos ^3 x+\frac{1}{5}\cos^5 x+C
\end{align*}
Example 1.19.5 . Even powers.
Compute \(\displaystyle \int \sin^2 x\cos^4 x\, dx\text{.}\)
Solution .
Both powers are even. We use the square identity to reduce powers across the board:
\begin{align*}
\int \sin^2 x\cos^4 x\, dx \amp = \int \frac{1}{2}(1-\cos 2x)\left(\frac{1}{2}(1+\cos 2x)\right)^2\, dx \\
\amp = \frac{1}{8}\int (1-\cos 2x)(1+\cos 2x)^2\, dx\\
\amp = \frac{1}{8}\int 1+\cos 2x-\cos^2 2x-\cos^3 2x\, dx\\
\amp = \frac{1}{8}\int 1-\cos^2 2x+\sin^2 x\cos x\, dx \amp (\cos^3 x=(1-\sin^2 x)\cos x)\\
\amp =\frac{1}{8}\int \frac{1}{2}-\frac{1}{4}\cos 4x+\underset{u^2}{\underbrace{\sin^2 2x}}\, \underset{\frac{du}{2}}{\underbrace{\cos 2x\, dx}} \amp (\cos 2x=\frac{1}{2}(1+\cos 4x))\\
\amp = \frac{1}{16}x-\frac{1}{64}\sin 4x+\frac{1}{48}\sin^3 2x+C \amp (u=\sin 2x, du=2\cos 2x dx)\text{.}
\end{align*}
Example 1.19.6 . Even secant power.
Compute \(\displaystyle \int \sec^4 x\, dx\text{.}\)
Solution .
Peel off a factor of \(\sec^2 x\) and use the substitution \(u=\tan x\text{,}\) \(du=\sec^2 x dx\text{:}\)
\begin{align*}
\int \sec^4 x\, dx\amp = \int (\tan^2 x+1)\, \underset{u}{\underbrace{\sec^2 x\, dx}}\\
\amp = \frac{1}{3}\tan^3 x+\tan x+C \amp (u=\tan x, du=\sec^2 x\, dx)\text{.}
\end{align*}
Example 1.19.7 . Odd tangent power.
Compute \(\displaystyle \int \tan^5 x\sec^7 x\, dx\text{.}\)
Solution .
Peel off a factor of \(\tan x\sec x\) and use the substitution \(u=\sec x\text{,}\) \(du=\sec x\tan x\text{:}\)
\begin{align*}
\int \tan^5 x\sec^7 x\, dx \amp = \int (\sec^2-1)^2\sec^6 x \, \underset{du}{\underbrace{\sec x\tan x\, dx}}\\
\amp = \int u^{10}-2u^{8}+u^6\, du \amp (u=\sec x, du=\sec x\tan x dx)\\
\amp = \frac{1}{11}\sec^{11}x-\frac{2}{9}\sec^9 x+\frac{1}{7}\sec x+C\text{.}
\end{align*}
Example 1.19.8 . Even tangent power, odd secant power.
Compute \(\displaystyle \int \sec^3 x\, dx\text{.}\)
Solution .
Not much we can do here besides integration by parts:
\begin{align*}
\int \sec^3 x\, dx \amp = \sec x\tan x-\int \sec x\tan^2 x\, dx \amp (u=\sec x, v'=\sec^2 x)\\
\amp = \sec x\tan x-int \sec^3 x-\sec x\, dx \amp (\tan^2 x=\sec^2 x-1) \\
\amp = \sec x\tan x+\ln\abs{\sec x+\tan x}-\int \sec^3 x\, dx\text{.}
\end{align*}
We’ve shown that
\begin{equation*}
\int \sec^3 x\, dx=\sec x\tan x+\ln\abs{\sec x+\tan x}-\int \sec^3 x\, dx\text{.}
\end{equation*}
Doing some algebra and solving for \(\int \sec^3 x\, dx\text{,}\) we conclude that
\begin{equation*}
\int \sec^3 x\, dx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln\abs{\sec x+\tan x}+C. \text{.}
\end{equation*}
Example 1.19.9 . Tangent power.
Compute \(\displaystyle \int \tan^5 x \, dx\text{.}\)
Solution .
\begin{equation*}
\tan^5 x=(\tan^2 x)^2\tan x=(\sec^4 x-2\sec^2 x+1)\tan x\text{,}
\end{equation*}
from which we conclude
\begin{align*}
\int \tan^5 x\, dx \amp = \int \sec^4 x\tan x-2\sec^2 x\tan x+\tan x\, dx \\
\amp = \int \underset{u^3}{\underbrace{\sec^3 x}}\, \underset{du}{\underbrace{\sec x\tan x}}-2\underset{u}{\underbrace{\sec x}} \, \underset{du}{\underbrace{\sec x\tan x}}+\tan x\, dx \\
\amp =\frac{1}{4}\sec^4 x-\sec^2 x-\ln\abs{\cos x}+C\text{.}
\end{align*}
