4.8: Integrals Involving Arctrig Functions
- Page ID
- 527
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Recall that
\[\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2},\nonumber \]
\[\dfrac{d}{dx} \sin^{-1} x = \dfrac{1}{\sqrt{1-x^2}},\nonumber \]
\[\dfrac{d}{dx} \sec^{-1} x = \dfrac{1}{x\sqrt{x^2-1}}.\nonumber \]
These three formulas immediately imply to integration:
\[ \int \dfrac{1}{1+x^2} dx = \tan^{-1} x + C,\nonumber \]
\[ \int \dfrac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + C,\nonumber \]
\[ \int \dfrac{1}{x\sqrt{x^2-1}} dx = \sec^{-1} x + C.\nonumber \]
Solve \( \int \dfrac{dx}{4+x^2}\).
Solution
\[ \int \dfrac{dx}{4+x^2} = \dfrac{1}{4} \int \dfrac{dx}{1+\left( \dfrac{x}{2} \right)^2}.\nonumber \]
Let \(u= \dfrac{x}{2}\) so \(du=\dfrac{1}{2}dx\)
and the integral becomes
\[\begin{align*} \dfrac{1}{2} \int \dfrac{du}{1+u^2} = \dfrac{1}{2} \tan^{-1} u + C &= \dfrac{1}{2} \tan^{-1} \left( \dfrac{x}{2} \right) + C. \end{align*} \]
Solve \( \int \dfrac{dx}{x \sqrt{x^2-4}} \).
Solution
\[\begin{align*} \int \dfrac{dx}{x \sqrt{x^2-4}} &= \int \dfrac{x\,dx}{x^2 \sqrt{4 (x^4/4 -1)}} \\ &= \dfrac{1}{4} \int \dfrac{x\,dx}{(x^2/2)\sqrt{(x^2/2)^2-1}} \end{align*} \]
which becomes
\[\begin{align*} \dfrac{1}{4} \int \dfrac{du}{u\sqrt{u^2-1}} &= \dfrac{1}{4} \sec^{-1} u + C \\ &= \dfrac{1}{4} \sec^{-1} \left(\dfrac{x^2}{2}\right) + C. \end{align*} \]
Solve \(\int \dfrac{2x\, dx}{x^2+6x+13}\).
Solution
\[\begin{align*} \int \dfrac{2x\, dx}{x^2+6x+13} &= \int \dfrac{2x\, dx}{(x+3)^2 + 4} \\ &=\dfrac{1}{2} \int \dfrac{x\, dx}{ \left(\dfrac{x+3}{2} \right)^2+1} \end{align*} \]
let \(u= \dfrac{x+3}{2}\) and \(du = \dfrac{1}{2} dx\) so \(x = 2u-3\).
\[\begin{align*} \int \dfrac{(2u-3)\, du}{u^2+1} &= \int \dfrac{2u\,du}{u^2+1}-3\int\dfrac{du}{u^2+1} \\ &= \ln \left| u^2+1 \right| -3\tan^{-1} u+C \\ &= \ln \left(\dfrac{ \left(\dfrac{x+3}{2} \right)^2}{4} +1\right) - 3 \tan^{-1}\dfrac{x+3}{2} + C \end{align*}.\nonumber \]
Contributors and Attributions
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.