2.2: Trigonometric Substitution
- Page ID
- 522
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)When we have integrals that involve the square root term
\[\sqrt{a^2+x^2} \nonumber \]
we may be able to trigonometric substitution to solve the integral.
Solve
\[\int \sqrt{1-x^2}dx \nonumber \]
by substituting \(x=\sin \theta\) and \(dx=\cos \theta \, d\theta \).
The integrand then becomes
\[\begin{align} \sqrt{1-x^2} &= \sqrt{1-\sin^2 \theta} \\ &= \sqrt{\cos^2 \theta} \\ &= \cos \theta \end{align} \nonumber \]
We have
\[\begin{align} \int \sqrt{1-x^2}\; dx &= \int \cos\theta\cos\theta \;d\theta \\ &= \int \cos^2\theta \;d\theta \\ &= \int \Big( \dfrac{1}{2}+\dfrac{1}{2}\cos 2\theta \Big)\; d\theta \\ &= \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta +C \\ &= \dfrac{1}{2} \arcsin x +\dfrac{1}{2} \sin\theta\cos\theta+C \\ &= \dfrac{1}{2}\arcsin x +\dfrac{1}{2}x\sqrt{1-x^2}+C \end{align} \nonumber \]
- \[ \int \dfrac{\sqrt{1-x^2}}{x^4} dx \nonumber \]
- \[ \int \dfrac{1}{\sqrt{4-9x^2}}dx \nonumber \]
Two Key Formulas
From Trigonometry, we have the following two key formulas:
\[ \sec^2\, x = 1 + \tan^2\, x \nonumber \]
so
\[\sec x = \sqrt{1+\tan^2 x} \nonumber \]
and
\[ \tan^2\, x = \sec^2\, x - 1 \nonumber \]
so
\[ \tan x = \sqrt{\sec^2 \, x -1}. \nonumber \]
When we have integrals that involve any of the above square roots, we can use the appropriate substitution.
\[\begin{align} \int \dfrac{x^3}{\sqrt{1+x^2}}dx \\ x= \tan\theta, \; dx=\sec^2\theta \; d\theta \\ \sqrt{1+x^2}=\sqrt{1+\tan^2\theta}= \sqrt{\sec^2\theta} = \sec\theta \\ &= \int \dfrac{\tan^3\theta}{\sec\theta}\sec^2\theta \; d\theta \\ &= \int \tan^3\theta \sec\theta \; d\theta \\ &= \int \tan^2\theta \tan\theta \sec\theta \; d\theta \\ &= \int (\sec^2\theta-1)\sec\theta\tan\theta \; d\theta \\ u=\sec\theta, \; du=\sec\theta\tan\theta \; d\theta \\ &= \int (u^2-1) \; du \\ &=\dfrac{u^3}{3}-u+C \\ &= \dfrac{\sec^3\theta}{3}-\sec\theta+C \\ &= \dfrac{(1+x^2)^{\frac{3}{2}}}{3}-\sqrt{1+x^2}+C \end{align} \nonumber \]
- \[ \int \dfrac{x^3}{\sqrt{x^2-1}} \; dx \nonumber \]
- \[ \int \dfrac{x^2}{\sqrt{9+4x^2}} \; dx \nonumber \]
- \[ \int \dfrac{1}{\sqrt{x^2+2x}} \; dx \nonumber \]
Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.