7.5: Integrals of Powers of Trigonometric Functions
- Page ID
- 155873
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)It is often possible to transform an integral into one of the forms
\[\int \sin ^{n} u d u, \quad \int \cos ^{n} u d u, \quad \int \tan ^{n} u d u, \quad \text { etc. } \nonumber \]
These integrals can be evaluated by means of reduction formulas, which express the integral of the \(n\)th power of a trigonometric function in terms of the \((n-2)\) nd power. The easiest reduction formulas to prove are those for the tangent and cotangent, so we shall give them first.
Let \(n \neq 1\). Then
- \(\int \tan ^{n} x d x=\frac{\tan ^{n-1} x}{n-1}-\int \tan ^{n-2} x d x\).
- \(\int \cot ^{n} x d x=-\frac{\cot ^{n-1} x}{n-1}-\int \cot ^{n-2} x d x\).
PROOF
We recall that
\[\begin{aligned} & \tan ^{2} x=\sec ^{2} x-1, \quad d(\tan x)=\sec ^{2} x d x \\ & \text { Then } \int \tan ^{n} x d x=\int \tan ^{n-2} x \tan ^{2} x d x=\int \tan ^{n-2} x\left(\sec ^{2} x-1\right) d x \\ & =\int \tan ^{n-2} x \sec ^{2} x d x-\int \tan ^{n-2} x d x \\ & =\int \tan ^{n-2} x d(\tan x)-\int \tan ^{n-2} x d x \\ & =\frac{\tan ^{n-1} x}{n-1}-\int \tan ^{n-2} x d x \end{aligned} \nonumber \]
These reduction formulas are true for any rational number \(n \neq 1\). They are most useful, however, when \(n\) is a positive integer.
\(\int \tan ^{2} x d x=\frac{\tan x}{1}-\int \tan ^{\circ} x d x=\tan x-x+C\).
Solution
EXAMPLE
\( \int \tan ^{4} x d x=\frac{\tan ^{3} x}{3}-\int \tan ^{2} x d x=\frac{\tan ^{3} x}{3}-\tan x+x+C\).
Solution
Add example text here.
\(\int \tan ^{3} x d x=\frac{\tan ^{2} x}{2}-\int \tan x d x\)
Solution
We will evaluate \(\int \tan x d x\) in the next chapter.
Each time we use the reduction formula the exponent in the integral goes down by two. By repeated use of the reduction formulas we can integrate any even power of \(\tan x\) or \(\cot x\). We can also work the integral of any odd power of \(\tan x\) or \(\cot x\) down to an expression involving \(\int \tan x\) or \(\int \cot x\).
The reduction formulas for the other trigonometric functions are obtained by using integration by parts.
Let \(n \neq 0\). Then
- \(\int \sin ^{n} x d x=-\frac{1}{n} \sin ^{n-1} x \cos x+\frac{n-1}{n} \int \sin ^{n-2} x d x\).
- \(\int \cos ^{n} x d x=\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} \int \cos ^{n-2} x d x\).
PROOF
(i) Break the term \(\sin ^{n} x d x\) into two parts,
\[\sin ^{n} x d x=\sin ^{n-1} x(\sin x d x) \nonumber \]
We shall let
\[u=\sin ^{n-1} x, \quad v=-\cos x \nonumber \]
\[d u=(n-1) \sin ^{n-2} x \cos x d x, \quad d v=\sin x d x \nonumber \]
and use integration by parts. Then
\[\begin{aligned} \int \sin ^{n} x d x & =\int u d v=u v-\int v d u \\ & =-\sin ^{n-1} x \cos x-\int(n-1)(-\cos x) \sin ^{n-2} x \cos x d x \\ & =-\sin ^{n-1} x \cos x+(n-1) \int \sin ^{n-2} x\left(1-\sin ^{2} x\right) d x \\ & =-\sin ^{n-1} x \cos x+(n-1) \int \sin ^{n-2} x d x-(n-1) \int \sin ^{n} x d x \end{aligned} \nonumber \]
We find that \(\int \sin ^{n} x d x\) appears on both sides of the equation, and we solve for it,
\[\begin{aligned} & n \int \sin ^{n} x d x=-\sin ^{n-1} x \cos x+(n-1) \int \sin ^{n-2} x d x \\ & \int \sin ^{n} x d x=-\frac{1}{n} \sin ^{n-1} x \cos x+\frac{n-1}{n} \int \sin ^{n-2} x d x \end{aligned} \nonumber \]
We already know the integrals
\[\int \sin x d x=-\cos x+C, \quad \int \cos x d x=\sin x+C \nonumber \]
We can use the reduction formulas to integrate any positive power of \(\sin x\) or \(\cos x\). Again, the formulas are true where \(n\) is any rational number, \(n \neq 0\).
\(4 \int \sin ^{2} x d x=-\frac{1}{2} \sin x \cos x+\frac{1}{2} \int d x=-\frac{1}{2} \sin x \cos x+\frac{1}{2} x+C\).
Solution
\[\int \cos ^{2} x d x=\frac{1}{2} \cos x \sin x+\frac{1}{2} \int d x=\frac{1}{2} \cos x \sin x+\frac{1}{2} x+C \nonumber \]
\(5 \int \cos ^{3} x d x=\frac{1}{3} \cos ^{2} x \sin x+\frac{2}{3} \int \cos x d x\)
Solution
\[=\frac{1}{3} \cos ^{2} x \sin x+\frac{2}{3} \sin x+C \text {. } \nonumber \]
Let \(m \neq 1\). Then
- \(\int \sec ^{m} x d x=\frac{1}{m-1} \sec ^{m-1} x \sin x+\frac{m-2}{m-1} \int \sec ^{m-2} x d x\).
- \(\int \csc ^{m} x d x=-\frac{1}{m-1} \csc ^{m-1} x \cos x+\frac{m-2}{m-1} \int \csc ^{m-2} x d x\).
PROOF
(ii) This can be done by integration by parts, but it is easier to use Theorem 2. Let \(n=2-m\). For \(m \neq 2, n \neq 0\) and Theorem 2 gives
\[\begin{aligned} & \int \sin ^{2-m} x d x=-\frac{1}{2-m} \sin ^{1-m} x \cos x+\frac{1-m}{2-m} \int \sin ^{-m} x d x \\ & \int \csc ^{m-2} x d x=\frac{1}{m-2} \csc ^{m-1} x \cos x+\frac{m-1}{m-2} \int \csc ^{m} x d x \end{aligned} \nonumber \]
whence
\[\int \csc ^{m} x d x=-\frac{1}{m-1} \csc ^{m-1} x \cos x+\frac{m-2}{m-1} \int \csc ^{m-2} x d x \nonumber \]
For \(m=2\) the formula is already known,
\[\int \csc ^{2} x d x=-\cot x+C=-\csc x \cos x+C \nonumber \]
These reduction formulas can be used to integrate any even power of sec \(x\) or \(\csc x\), and to get the integral of any odd power of \(\sec x\) or \(\csc x\) in terms of \(\int \sec x\) or \(\int \csc x\). We shall find \(\int \sec x\) and \(\int \csc x\) in the next chapter.
\(\int \sec ^{3} x d x=\frac{1}{2} \sec ^{2} x \sin x+\frac{1}{2} \int \sec x d x\).
Solution
EXAMPLE \(7 \int \sec ^{4} x d x=\frac{1}{3} \sec ^{3} x \sin x+\frac{2}{3} \int \sec ^{2} x d x\)
\(\int \sec ^{4} x d x=\frac{1}{3} \sec ^{3} x \sin x+\frac{2}{3} \int \sec ^{2} x d x\)
Solution
\[=\frac{1}{3} \sec ^{3} x \sin x+\frac{2}{3} \tan x+C \nonumber \]
By using the identity \(\sin ^{2} x+\cos ^{2} x=1\) we can evaluate any integral of the form \(\int \sin ^{m} x \cos ^{n} x d x\) where \(m\) and \(n\) are positive integers. If either \(m\) or \(n\) is odd we let \(u=\sin x\) or \(u=\cos x\) and transform the integral into a polynomial in \(u\).
\(\int \sin ^{4} x \cos ^{3} x d x\). Let \(u=\sin x, d u=\cos x d x\).
Solution
\[\begin{aligned} \int \sin ^{4} x \cos ^{3} x d x & =\int u^{4}\left(1-u^{2}\right) d u \\ & =\frac{1}{5} u^{5}-\frac{1}{7} u^{7}+C \\ & =\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+C . \end{aligned} \nonumber \]
This method also works for an odd power of \(\sin x\) times any power of \(\cos x\), and vice versa.
\(\int \sqrt{\cos x} \sin ^{3} x d x\). Let \(u=\cos x, d u=-\sin x d x\).
Solution
\[\begin{aligned} \int \sqrt{\cos x} \sin ^{3} x d x & =\int \sqrt{u}\left(1-u^{2}\right)(-1) d u \\ & =\int-u^{1 / 2}+u^{5,2} d u=-\frac{2}{3} u^{3 / 2}+\frac{2}{7} u^{7 / 2}+C \\ & =-\frac{2}{3}(\cos x)^{3 / 2}+\frac{2}{7}(\cos x)^{7,2}+C . \end{aligned} \nonumber \]
\(\int \sin ^{5} x d x\). Let \(u=\cos x, d u=-\sin x d x\).
Solution
\[\begin{aligned} \int \sin ^{5} x d x & =\int\left(1-u^{2}\right)^{2}(-1) d u \\ & =-\int\left(1-2 u^{2}+u^{4}\right) d u=-u+\frac{2}{3} u^{3}-\frac{1}{5} u^{5}+C \\ & =-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+C \end{aligned} \nonumber \]
If \(m\) and \(n\) are both even, the integral \(\int \sin ^{m} x \cos ^{n} x d x\) can be transformed into the integral of a sum of even powers of \(\sin x\). Then the reduction formula can be used.
\(\int \sin ^{4} x \cos ^{4} x d x=\int \sin ^{4} x\left(1-\sin ^{2} x\right)^{2} d x\)
Solution
\[=\int \sin ^{4} x-2 \sin ^{6} x+\sin ^{8} x d x \nonumber \]
We can also evaluate integrals of the form
\[\begin{aligned} & \int \tan ^{m} x \sec ^{n} x d x \\ & \int \cot ^{m} x \csc ^{n} x d x \end{aligned} \nonumber \]
When \(m\) is even use \(\tan ^{2} x=\sec ^{2} x-1\).
Solution
\[\begin{aligned} \int \tan ^{4} x \sec x d x & =\int\left(\sec ^{2} x-1\right)^{2} \sec x d x \\ & =\int \sec ^{5} x-2 \sec ^{3} x+\sec x d x \end{aligned} \nonumber \]
Now the reduction formula for \(\sec x\) can be used.
When \(m\) is odd use the new variable \(u=\sec x\) or \(u=-\csc x\).
Solution
\[\begin{aligned} \int \cot ^{3} x \csc ^{3} x d x & =\int \cot ^{2} x \csc ^{2} x(\cot x \csc x d x) \\ & =\int\left(u^{2}-1\right) u^{2} d u=\frac{u^{5}}{5}-\frac{u^{3}}{3}+C \\ & =-\frac{\csc ^{5} x}{5}+\frac{\csc ^{3} x}{3}+C . \end{aligned} \nonumber \]
PROBLEMS FOR SECTION 7.5
Evaluate the integrals in Problems 1-32.
\(1 \quad \int \frac{\sin ^{3} t}{\cos ^{2} t} d t\)
\(2 \quad \int \sin ^{2}(2 t) d t\)
\(3 \quad \int \cot ^{2} x d x\)
\(4 \quad \int \sin ^{3}(5 u) d u\)
\(5 \quad \int \cos ^{4} \dot{x} d x\)
\(6 \quad \int \frac{1}{\sin ^{4} x} d x\)
\(7 \quad \int \tan ^{3} x \sec ^{4} x d x\)
\(8 \int \tan ^{6} \theta d \theta\)
\(9 \quad \int \sin ^{2} x \cos ^{3} x d x\)
\(10 \int \cot \theta \csc ^{2} \theta d \theta\)
\(11 \int \cot ^{2} \theta \csc ^{2} \theta d \theta\)
\(12 \int \sin x(\cos x)^{3 / 2} d x\)
\(13 \int(\tan x)^{3 / 2} \sec ^{4} x d x\)
\(14 \int \sec ^{4}(3 u-1) d u\)
15
\(\int \sec ^{2} \theta \csc ^{2} \theta d \theta\)
\(16 \int \frac{\sin ^{2} \theta}{1-\cos \theta} d \theta\)
\(17 \int \frac{1-\cos \theta}{\sin ^{2} \theta} d \theta\)
\(18 \quad \int_{0}^{\pi / 2} \sin ^{3} x \cos x d x\)
\(19 \int_{0}^{\pi / 3} \tan ^{3} \theta \sec \theta d \theta\)
\(20 \quad \int_{0}^{\pi \cdot 2} \sqrt{\cos x} \sin x d x\)
\(21 \quad \int_{0}^{\pi / 4} \tan ^{4} x d x\)
\(22 \int_{0}^{\pi / 2} \tan ^{2} x d x\)
\(23 \quad \int_{0}^{\pi} \sin ^{4} \theta d \theta\) 24 \(\int \sin ^{3}(2 u) \cos ^{3}(2 u) d u\)
25 \(\int \frac{\cos ^{2} \sqrt{x}}{\sqrt{x}} d x\) \(\int x \tan \left(x^{2}\right) \sec ^{2}\left(x^{2}\right) d x\)
27
\(\int x \sin ^{3} x d x\)
28
\[\int x \tan ^{3} x \sec ^{2} x d x \nonumber \]
29
\(\int x \sin ^{2} x \cos x d x\)
30
\(\int \sin ^{6} \theta \cos ^{5} \theta d \theta\)
\(31 \int \tan ^{4} \theta \sec ^{6} \theta d \theta\)
\[\int \sin ^{2} x \cos ^{2} x d x \tag{32} \]
In Problems 33-39, express the given integral in terms of
\(\int \tan x d x, \int \cot x d x, \int \sec x d x, \int \csc x d x\)
33
\(\int \sec ^{3} x d x\)
34
\(\int \cot ^{3} x d x\)
35
\(\int \tan ^{2} x \sec x d x\)
36
\(\int \csc ^{5} x d x\)
37
\(\int \cot ^{2} x \csc ^{3} x d x\)
38
\[\int \tan ^{4} x \sec x d x \tag{39} \]
\(\int \frac{\sin x+\cos x}{\sin x \cos x} d x\)
40 Check the reduction formula for \(\int \sin ^{n} x d x\) by differentiating both sides of the equation. Do the same for \(\int \tan ^{n} x d x\) and \(\int \sec ^{n} x d x\).
41 Find a reduction formula for \(\int x^{n} \sin x d x\) using integration by parts.
42 Find the volume of the solid generated by rotating the region under the curve \(y=\sin ^{2} x\) \(0 \leq x \leq \pi\), about (a) the \(x\)-axis, (b) the \(y\)-axis.
43 Find the volume of the solid generated by rotating the region under the curve \(y=\sin x \cos x, 0 \leq x \leq \pi / 2\), about (a) the \(x\)-axis, (b) the \(y\)-axis.