7.6: Trigonometric Substitutions
- Page ID
- 155875
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Integrals containing one of the terms
\[\sqrt{a^{2}+x^{2}}, \quad \sqrt{a^{2}-x^{2}}, \quad \text { or } \quad \sqrt{x^{2}-a^{2}} \nonumber \]
can often be integrated by a trigonometric substitution. The idea is to take \(x, a\), and the square root as the three sides of a right triangle and use one of its acute angles as a new variable \(\theta\). The three kinds of trigonometric substitutions are shown in Figure 7.6.1. These figures do not have to be memorized. Just remember that the sides must be labeled so that
\[(\text { opposite })^{2}+(\text { adjacent })^{2}=(\text { hypotenuse })^{2} \nonumber \]
These substitutions frequently give an integral of powers of trigonometric functions discussed in the preceding section.
\(x=a \tan \theta\)
\(x=a \sin \theta\)
\(x=a \sec \theta\)
Figure 7:6.1
Find \(\int\left(a^{2}+x^{2}\right)^{-3 / 2} d x\).
Solution
Let \(\theta=\arctan (x / a)\). Then from Figure 7.6.2,
\[x=a \tan \theta, \quad d x=a \sec ^{2} \theta d \theta, \quad \sqrt{a^{2}+x^{2}}=a \sec \theta \nonumber \]
So
\[\begin{aligned} \int\left(a^{2}+x^{2}\right)^{-3 / 2} d x & =\int(a \sec \theta)^{-3} a \sec ^{2} \theta d \theta \\ & =\frac{1}{a^{2}} \int(\sec \theta)^{-1} d \theta=\frac{1}{a^{2}} \int \cos \theta d \theta \\ & =\frac{1}{a^{2}} \sin \theta+C=\frac{1}{a^{2}} \frac{\tan \theta}{\sec \theta}+C=\frac{x}{a^{2} \sqrt{a^{2}+x^{2}}}+C \end{aligned} \nonumber \]
Find \(\int \sqrt{x^{2}-a^{2}} d x\).
Solution
Let \(\theta=\operatorname{arcsec}(x / a)\) (Figure 7.6.3), so
\[x=a \sec \theta, \quad d x=a \tan \theta \sec \theta d \theta, \quad \sqrt{x^{2}-a^{2}}=a \tan \theta \nonumber \]
So
\[\begin{aligned} \int \sqrt{x^{2}-a^{2}} d x & =\int a \tan \theta a \tan \theta \sec \theta d \theta=a^{2} \int \tan ^{2} \theta \sec \theta d \theta \\ & =a^{2} \int\left(\sec ^{2} \theta-1\right) \sec \theta d \theta \\ & =a^{2} \int \sec ^{3} \theta d \theta-a^{2} \int \sec \theta d \theta \\ & =\left(\frac{1}{2} a^{2} \sec ^{2} \theta \sin \theta+\frac{1}{2} a^{2} \int \sec \theta d \theta\right)-a^{2} \int \sec \theta d \theta \\ & =\frac{1}{2} a^{2} \sec ^{2} \theta \sin \theta-\frac{1}{2} a^{2} \int \sec \theta d \theta \\ & =\frac{1}{2} x \sqrt{x^{2}-a^{2}}-\frac{1}{2} a^{2} \int \sec \theta d \theta \end{aligned} \nonumber \]
This is as far as we can go on this problem until we find out how to integrate \(\int \sec \theta d \theta\) in the next chapter.
\(\int \frac{1}{x^{2} \sqrt{a^{2}-x^{2}}} d x\). Let \(\theta=\arcsin (x / a)\) (Figure 7.6.4). Then
Solution
\[\begin{aligned} & x=a \sin \theta, \quad d x=a \cos \theta d \theta, \quad \sqrt{a^{2}-x^{2}}=a \cos \theta \\ & \int \frac{1}{x^{2} \sqrt{a^{2}-x^{2}}} d x=\int \frac{1}{a^{2} \sin ^{2} \theta a \cos \theta} a \cos \theta d \theta=\int \frac{1}{a^{2} \sin ^{2} \theta} d \theta \\ &=\frac{1}{a^{2}} \int \csc ^{2} \theta d \theta=-\frac{1}{a^{2}} \cot \theta+C \\ &=-\frac{1}{a^{2}} \frac{\sqrt{a^{2}-x^{2}}}{x}+C \end{aligned} \nonumber \]
\(\int \frac{\sqrt{x^{2}}-a^{2}}{x} d x\). Put \(\theta=\operatorname{arcsec}(x / a)\) (Figure 7.6.5). Then
Solution
\[\begin{aligned} & x=a \sec \theta, \quad d x=a \tan \theta \sec \theta d \theta, \quad \sqrt{x^{2}-a^{2}}=a \tan \theta \\ & \begin{aligned} \int \frac{\sqrt{x^{2}-a^{2}}}{x} d x & =\int \frac{a \tan \theta}{a \sec \theta} a \tan \theta \sec \theta d \theta=a \int \tan ^{2} \theta d \theta \\ \cdot & =a \int \sec ^{2} \theta d \theta-a \int d \theta=a \tan \theta-a \theta+C \\ & =\sqrt{x^{2}-a^{2}}-a \operatorname{arcsec}(x / a)+C \end{aligned} \end{aligned} \nonumber \]
Figure 7.6 .4
To keep track of a trigonometric substitution, it is a good idea to actually draw the triangle and label the sides.
The basic integrals:
Solution
(a)
\[\int \frac{1}{\sqrt{1-x^{2}}} d x=\arcsin x+C \nonumber \]
(b)
\[\int \frac{d x}{1+x^{2}} d x=\arctan x+C \nonumber \]
(c)
\[\int \frac{d x}{x \sqrt{x^{2}-1}}=\operatorname{arcsec} x+C, \quad x>1 \nonumber \]
can be evaluated very easily by a trigonometric substitution.
(a)
\[\int \frac{1}{\sqrt{1-x^{2}}} d x \nonumber \]
Let \(\theta=\arcsin x\) (Figure 7.6.6). Then \(x=\sin \theta, d x=\cos \theta d \theta, \sqrt{1-x^{2}}=\) \(\cos \theta\).
\[\begin{aligned} & \int \frac{1}{\sqrt{1-x^{2}}} d x=\int \frac{\cos \theta d \theta}{\cos \theta}=\int d \theta=\theta+C \\ & \int \frac{1}{\sqrt{1-x^{2}}} d x=\arcsin x+C \end{aligned} \nonumber \]
(b)
\[\int \frac{d x}{1+x^{2}} \nonumber \]
Let \(\theta=\arctan x\) (Figure 7.6.7). Then \(x=\tan \theta, d x=\sec ^{2} \theta d \theta, \sqrt{1+x^{2}}=\) \(\sec \theta\).
\[\begin{aligned} & \int \frac{d x}{1+x^{2}}=\int \frac{\sec ^{2} \theta}{\sec ^{2} \theta} d \theta=\int d \theta=\theta+C \\ & \int \frac{d x}{1+x^{2}}=\arctan x+C \end{aligned} \nonumber \]
(c)
\[\int \frac{d x}{x \sqrt{x^{2}-1}}, \quad x>1 \nonumber \]
Let \(\theta=\operatorname{arcsec} x\) (Figure 7.6.8). Then \(x=\sec \theta, d x=\tan \theta \sec \theta d \theta\), \(\sqrt{x^{2}-1}=\tan \theta\).
\[\begin{aligned} & \int \frac{d x}{x \sqrt{x^{2}-1}}=\int \frac{\tan \theta \sec \theta}{\sec \theta \tan \theta} d \theta=\int d \theta=\theta+C \\ & \int \frac{d x}{x \sqrt{x^{2}+1}}=\operatorname{arcsec} x+C, \quad x>1 \end{aligned} \nonumber \]
It is therefore more important to remember the method of trigonometric substitution than to remember the integration formulas (a), (b), (c).
PROBLEMS FOR SECTION 7.6
Draw the appropriate triangle and evaluate using trigonometric substitutions
\(1 \quad \int \frac{d x}{\sqrt{1}-4 x^{2}}\)
\(2 \quad \int \sqrt{a^{2}-x^{2}} d x\)
\(3 \quad \int \frac{x^{3} d x}{\sqrt{9+\overline{x^{2}}}}\)
\(4 \int \frac{\sqrt{x^{2}-1}}{x} d x\)
\(5 \int\left(4-x^{2}\right)^{-3 / 2} d x\)
\(6 \quad \int\left(1-3 x^{2}\right)^{3 / 2} d x\)
\(7 \int \frac{\sin \theta d \theta}{\sqrt{2-\cos ^{2} \theta}}\)
\(8 \quad \int \frac{d x}{\sqrt{x} \sqrt{1-x}}\)
\(9 \quad \int \frac{d x}{x^{2}\left(1+x^{2}\right)}\)
\(10 \int \frac{x^{4} d x}{9+x^{2}}\)
\(11 \int \frac{x^{2} d x}{\sqrt{4-x^{2}}}\)
\(12 \int \frac{\sqrt{9-4 x^{2}}}{x^{4}} d x\)
\(13 \int x^{2} \sqrt{1-x^{2}} d x\)
14 \(\int \sqrt{x} \sqrt{1-x} d x\)
15
\(\int \sqrt{4 x-x^{2}} d x\)
\(16 \int \frac{\sqrt{x}}{\sqrt{1-x}} d x\)
\(17 \int \frac{x^{3}}{\sqrt{4 x^{2}-1}} d x\)
\(18 \int-\frac{d x}{x^{3} \sqrt{x^{2}-3}}\)
\(19 \int \frac{x^{3}}{\sqrt{a^{2}-x^{2}}} d x\)
\(20 \int x^{3} \sqrt{1+a^{2} x^{2}} d x\)
\(21 \int \frac{\sqrt{x^{4}-1}}{x} d x\)
\(22 \int x \sqrt{1-x^{4}} d x\)
23 \(\int \frac{x^{3}}{\left(a^{2}+x^{2}\right)^{3 / 2}} d x\)
24
\[\int_{0}^{2} \sqrt{4-x^{2}} d x \nonumber \]
25 \(\int_{-1}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\)
\[\int_{0}^{4} \frac{d x}{\left(9+x^{2}\right)^{3 / 2}} \tag{26} \]
\(27 \quad \int_{0}^{\infty} \frac{d x}{\left(9+x^{2}\right)^{3 / 2}}\)
\[\int_{2}^{4} \frac{\sqrt{x^{2}-2}}{x} d x \tag{28} \]
\(29 \int_{2}^{x} \frac{\sqrt{x^{2}-2}}{x} d x\)
\(30 \quad \int_{0}^{x} \frac{x^{3}}{\sqrt{1+x^{2}}} d x\)
\(31 \int x \arcsin x d x\)
\(32 \int x \arccos x d x\)
\(33 \int x^{2} \arcsin x d x\)
\(34 \int x^{3} \arctan x d x\)
35 \(\int x^{-3} \arcsin x d x\)
\(36 \int x^{-3} \arctan x d x\)
37 Find the surface area generated by rotating the ellipse \(x^{2}+4 y^{2}=1\) about the \(x\)-axis.