Trigonometric substitution

{{short description|Technique of integral evaluation}} {{Trigonometry}} {{calculus|expanded=integral}} In mathematics, a '''trigonometric substitution''' replaces a trigonometric function for another expression. In calculus, trigonometric substitutions are a technique for evaluating integrals. In this case, a radical expression is replaced with a trigonometric one. Trigonometric identities may help simplify the answer.<ref>{{cite book | last=Stewart | first=James | author-link=James Stewart (mathematician) | title=Calculus: Early Transcendentals | publisher=Brooks/Cole | edition=6th | year=2008 | isbn=978-0-495-01166-8 | url-access=registration | url=https://archive.org/details/calculusearlytra00stew_1 }}</ref><ref>{{cite book | last1 = Thomas | first1 = George B. | last2=Weir | first2= Maurice D. | last3=Hass | first3=Joel | author3-link = Joel Hass | author-link=George B. Thomas | title=Thomas' Calculus: Early Transcendentals | publisher=Addison-Wesley | year=2010 | edition=12th | isbn=978-0-321-58876-0}}</ref>

In the case of a fishy integral, this method of differentiation by substitution uses the substitution to change the interval of integration. Alternatively, the antiderivative of the integrand may be applied to the original interval.

==Case I: Integrands containing ''a''2 − ''x''2== Let <math>x = a \sin \theta,</math> and use the identity <math>1-\sin^2 \theta = \cos^2 \theta.</math>

===Examples of Case I=== thumb|Geometric construction for Case I

====Example 1==== In the integral

we may use

<math display=block>x=a\sin \theta,\quad dx=a\cos\theta\, d\theta, \quad \theta=\arcsin\frac{x}{a}.</math>

Then, <math display=block>\begin{align} \int\frac{dx}{\sqrt{a^2-x^2}} &= \int\frac{a\cos\theta \,d\theta}{\sqrt{a^2-a^2\sin^2 \theta}} \\[6pt] &= \int\frac{a\cos\theta\,d\theta}{\sqrt{a^2(1 - \sin^2 \theta )}} \\[6pt] &= \int\frac{a\cos\theta\,d\theta}{\sqrt{a^2\cos^2\theta}} \\[6pt] &= \int d\theta \\[6pt] &= \theta + C \\[6pt] &= \arcsin\frac{x}{a}+C. \end{align}</math>

The above step requires that <math>a > 0</math> and <math>\cos \theta > 0.</math> We can choose <math>a</math> to be the principal root of <math>a^2,</math> and impose the restriction <math>-\pi /2 < \theta < \pi /2</math> by using the inverse sine function.

For a definite integral, one must figure out how the bounds of integration change. For example, as <math>x</math> goes from <math>0</math> to <math>a/2,</math> then <math>\sin \theta</math> goes from <math>0</math> to <math>1/2,</math> so <math>\theta</math> goes from <math>0</math> to <math>\pi / 6.</math> Then,

<math display=block>\int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}}=\int_0^{\pi/6} d\theta = \frac{\pi}{6}.</math>

Some care is needed when picking the bounds. Because integration above requires that <math>-\pi /2 < \theta < \pi /2</math> , <math>\theta</math> can only go from <math>0</math> to <math>\pi / 6.</math> Neglecting this restriction, one might have picked <math>\theta</math> to go from <math>\pi</math> to <math>5\pi /6,</math> which would have resulted in the negative of the actual value.

Alternatively, fully evaluate the indefinite integrals before applying the boundary conditions. In that case, the antiderivative gives

<math display=block>\int_{0}^{a/2} \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin \left( \frac{x}{a} \right) \Biggl|_{0}^{a/2} = \arcsin \left ( \frac{1}{2}\right) - \arcsin (0) = \frac{\pi}{6}</math> as before.

====Example 2==== The integral

may be evaluated by letting <math display="inline">x=a\sin \theta,\, dx=a\cos\theta\, d\theta,\, \theta=\arcsin\dfrac{x}{a},</math> where <math>a > 0</math> so that <math display="inline">\sqrt{a^2}=a,</math> and <math display="inline">-\pi/2 \le \theta \le \pi/2</math> by the range of arcsine, so that <math>\cos \theta \ge 0</math> and <math display="inline">\sqrt{\cos^2 \theta} = \cos \theta.</math>

Then, <math display=block>\begin{align} \int\sqrt{a^2-x^2}\,dx &= \int\sqrt{a^2-a^2\sin^2\theta}\,(a\cos\theta) \,d\theta \\[6pt] &= \int\sqrt{a^2(1-\sin^2\theta)}\,(a\cos\theta) \,d\theta \\[6pt] &= \int\sqrt{a^2(\cos^2\theta)}\,(a\cos\theta) \,d\theta \\[6pt] &= \int(a\cos\theta)(a\cos\theta) \,d\theta \\[6pt] &= a^2\int\cos^2\theta\,d\theta \\[6pt] &= a^2\int\left(\frac{1+\cos 2\theta}{2}\right)\,d\theta \\[6pt] &= \frac{a^2}{2} \left(\theta+\frac{1}{2}\sin 2\theta \right) + C \\[6pt] &= \frac{a^2}{2}(\theta+\sin\theta\cos\theta) + C \\[6pt] &= \frac{a^2}{2}\left(\arcsin\frac{x}{a}+\frac{x}{a}\sqrt{1-\frac{x^2}{a^2}}\right) + C \\[6pt] &= \frac{a^2}{2}\arcsin\frac{x}{a}+\frac{x}{2}\sqrt{a^2-x^2}+C. \end{align}</math>

For a definite integral, the bounds change once the substitution is performed and are determined using the equation <math display="inline">\theta = \arcsin\dfrac{x}{a},</math> with values in the range <math display="inline">-\pi/2 \le \theta \le \pi/2.</math> Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

may be evaluated by substituting <math>x = 2\sin\theta, \,dx = 2\cos\theta\,d\theta,</math> with the bounds determined using <math display="inline">\theta = \arcsin\dfrac{x}{2}.</math>

Because <math>\arcsin(1/{2}) = \pi/6</math> and <math>\arcsin(-1/2) = -\pi/6,</math> <math display=block>\begin{align} \int_{-1}^1\sqrt{4-x^2}\,dx &= \int_{-\pi/6}^{\pi/6}\sqrt{4-4\sin^2\theta}\,(2\cos\theta) \,d\theta \\[6pt] &= \int_{-\pi/6}^{\pi/6}\sqrt{4(1-\sin^2\theta)}\,(2\cos\theta) \,d\theta \\[6pt] &= \int_{-\pi/6}^{\pi/6}\sqrt{4(\cos^2\theta)}\,(2\cos\theta) \,d\theta \\[6pt] &= \int_{-\pi/6}^{\pi/6}(2\cos\theta)(2\cos\theta) \,d\theta \\[6pt] &= 4\int_{-\pi/6}^{\pi/6}\cos^2\theta\,d\theta \\[6pt] &= 4\int_{-\pi/6}^{\pi/6}\left(\frac{1+\cos 2\theta}{2}\right)\,d\theta \\[6pt] &= 2 \left[\theta+\frac{1}{2} \sin 2\theta \right]^{\pi/6}_{-\pi/6} = [2\theta+\sin 2\theta] \Biggl |^{\pi/6}_{-\pi/6} \\[6pt] &= \left(\frac{\pi}{3}+\sin\frac{\pi}{3}\right)-\left(-\frac{\pi}{3}+\sin\left(-\frac{\pi}{3}\right)\right) = \frac{2\pi}{3}+\sqrt{3}. \end{align}</math>

On the other hand, direct application of the boundary terms to the previously obtained formula for the antiderivative yields <math display=block>\begin{align} \int_{-1}^1\sqrt{4-x^2}\,dx &= \left[ \frac{2^2}{2}\arcsin\frac{x}{2}+\frac{x}{2}\sqrt{2^2-x^2} \right]_{-1}^{1}\\[6pt] &= \left( 2 \arcsin \frac{1}{2} + \frac{1}{2}\sqrt{4-1}\right) - \left( 2 \arcsin \left(-\frac{1}{2}\right) + \frac{-1}{2}\sqrt{4-1}\right)\\[6pt] &= \left( 2 \cdot \frac{\pi}{6} + \frac{\sqrt{3}}{2}\right) - \left( 2\cdot \left(-\frac{\pi}{6}\right) - \frac{\sqrt 3}{2}\right)\\[6pt] &= \frac{2\pi}{3} + \sqrt{3} \end{align} </math> as before.

==Case II: Integrands containing ''a''2 + ''x''2== Let <math>x = a \tan \theta,</math> and use the identity <math>1+\tan^2 \theta = \sec^2 \theta.</math>

===Examples of Case II=== thumb|Geometric construction for Case II ====Example 1==== In the integral

we may write

<math display=block>x=a\tan\theta,\quad dx=a\sec^2\theta\, d\theta, \quad \theta=\arctan\frac{x}{a},</math>

so that the integral becomes

<math display=block>\begin{align} \int\frac{dx}{a^2+x^2} &= \int\frac{a\sec^2\theta\, d\theta}{a^2 + a^2\tan^2\theta} \\[6pt] &= \int\frac{a\sec^2\theta\, d\theta}{a^2(1+\tan^2\theta)} \\[6pt] &= \int\frac{a\sec^2\theta\, d\theta}{a^2\sec^2\theta} \\[6pt] &= \int\frac{d\theta}{a} \\[6pt] &= \frac{\theta}{a}+C \\[6pt] &= \frac{1}{a} \arctan \frac{x}{a} + C, \end{align}</math>

provided <math>a \neq 0.</math>

For a definite integral, the bounds change once the substitution is performed and are determined using the equation <math>\theta = \arctan\frac{x}{a},</math> with values in the range <math>-\frac{\pi}{2} < \theta < \frac{\pi}{2}.</math> Alternatively, apply the boundary terms directly to the formula for the antiderivative.

For example, the definite integral

may be evaluated by substituting <math>x = \tan\theta, \,dx = \sec^2\theta\,d\theta,</math> with the bounds determined using <math>\theta = \arctan x.</math>

Since <math>\arctan 0 = 0</math> and <math>\arctan 1 = \pi/4,</math> <math display=block>\begin{align} \int_0^1\frac{4\,dx}{1+x^2} &= 4\int_0^1\frac{dx}{1 + x^2} \\[6pt] &= 4\int_0^{\pi/4}\frac{\sec^2\theta\, d\theta}{1+\tan^2\theta} \\[6pt] &= 4\int_0^{\pi/4}\frac{\sec^2\theta\, d\theta}{\sec^2\theta} \\[6pt] &= 4\int_0^{\pi/4}d\theta \\[6pt] &= (4\theta)\Bigg|^{\pi/4}_0 = 4 \left (\frac{\pi}{4} - 0 \right) = \pi. \end{align}</math>

Meanwhile, direct application of the boundary terms to the formula for the antiderivative yields <math display="block">\begin{align} \int_0^1\frac{4\,dx}{1+x^2}\, &= 4\int_0^1\frac{dx}{1+x^2} \\[6pt] &= 4\left[\frac{1}{1} \arctan \frac{x}{1} \right]^1_0 \\[6pt] &= 4(\arctan x)\Bigg|^1_0 \\[6pt] &= 4(\arctan 1 - \arctan 0) \\[6pt] &= 4 \left (\frac{\pi}{4} - 0 \right) = \pi, \end{align}</math> same as before.

====Example 2==== The integral

may be evaluated by letting <math>x=a\tan\theta,\, dx=a\sec^2\theta\, d\theta, \, \theta=\arctan\frac{x}{a},</math>

where <math>a > 0</math> so that <math>\sqrt{a^2}=a,</math> and <math>-\frac{\pi}{2}<\theta<\frac{\pi}{2}</math> by the range of arctangent, so that <math>\sec \theta > 0</math> and <math>\sqrt{\sec^2 \theta} = \sec \theta.</math>

Then, <math display=block>\begin{align} \int\sqrt{a^2+x^2}\,dx &= \int\sqrt{a^2 + a^2\tan^2\theta}\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int\sqrt{a^2 (1+\tan^2\theta)}\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int\sqrt{a^2 \sec^2\theta}\,(a \sec^2\theta)\, d\theta \\[6pt] &= \int(a \sec\theta)(a \sec^2\theta)\, d\theta \\[6pt] &= a^2\int \sec^3\theta\, d\theta. \\[6pt] \end{align}</math> The integral of secant cubed may be evaluated using integration by parts. As a result, <math display=block>\begin{align} \int\sqrt{a^2+x^2}\,dx &= \frac{a^2}{2}(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)+C \\[6pt] &= \frac{a^2}{2}\left(\sqrt{1+\frac{x^2}{a^2}}\cdot\frac{x}{a} + \ln\left|\sqrt{1+\frac{x^2}{a^2}}+\frac{x}{a}\right|\right)+C \\[6pt] &= \frac{1}{2}\left(x\sqrt{a^2+x^2} + a^2\ln\left|\frac{x+\sqrt{a^2+x^2}}{a}\right|\right)+C. \end{align}</math>

To illustrate an application of this formula, suppose we wish to calculate the arc length of the parabola <math>y=x^2</math> from <math>x=0</math> and <math>x=2</math>. Denoting the arc length as <math>s</math>, we get

In this instance, we have <math>a=\frac{1}{2}</math>, which gives

<math display=block> \begin{align} s &= \left(x\sqrt{\frac{1}{4}+x^2} + \frac{1}{4}\ln\left|\frac{x+\sqrt{\frac{1}{4}+x^2}}{\frac{1}{2}}\right|\right)_{x=0}^{x=2} \\ &= 2\sqrt{\frac{17}{4}} + \frac{1}{4}\ln\left( 4 + 2\sqrt{\frac{17}{4}} \right) \\ &= \sqrt{17} + \frac{1}{4}\ln(4 + \sqrt{17}) \\ &\approx 4.64678 \\ \end{align} </math>

==Case III: Integrands containing ''x''2 − ''a''2== Let <math>x = a \sec \theta,</math> and use the identity <math>\sec^2 \theta -1 = \tan^2 \theta.</math>

===Examples of Case III=== thumb|Geometric construction for Case III

Integrals such as

can also be evaluated by partial fractions rather than trigonometric substitutions. However, the integral

cannot. In this case, an appropriate substitution is: <math display=block>x = a \sec\theta,\, dx = a \sec\theta\tan\theta\, d\theta, \, \theta = \arcsec\frac{x}{a},</math>

where <math>a > 0</math> so that <math>\sqrt{a^2}=a,</math> and <math>0 \le \theta < \frac{\pi}{2}</math> by assuming <math>x > 0,</math> so that <math>\tan \theta \ge 0</math> and <math>\sqrt{\tan^2 \theta} = \tan \theta.</math>

Then, <math display=block>\begin{align} \int\sqrt{x^2 - a^2}\, dx &= \int\sqrt{a^2 \sec^2\theta - a^2} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int\sqrt{a^2 (\sec^2\theta - 1)} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int\sqrt{a^2 \tan^2\theta} \cdot a \sec\theta\tan\theta\, d\theta \\ &= \int a^2 \sec\theta\tan^2\theta\, d\theta \\ &= a^2 \int (\sec\theta)(\sec^2\theta - 1)\, d\theta \\ &= a^2 \int (\sec^3\theta - \sec\theta)\, d\theta. \end{align}</math>

One may evaluate the integral of the secant function by multiplying the numerator and denominator by <math>( \sec \theta + \tan \theta)</math> and the integral of secant cubed by parts.<ref name=":1">{{Cite book|last=Stewart|first=James|title=Calculus - Early Transcendentals|publisher=Cengage Learning|year=2012|isbn=978-0-538-49790-9|location=United States|pages=475–6|chapter=Section 7.2: Trigonometric Integrals|author-link=James Stewart (mathematician)}}</ref> As a result, <math display=block>\begin{align} \int\sqrt{x^2-a^2}\,dx &= \frac{a^2}{2}(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|)-a^2\ln|\sec\theta+\tan\theta|+C \\[6pt] &= \frac{a^2}{2}(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|)+C \\[6pt] &= \frac{a^2}{2}\left(\frac{x}{a}\cdot\sqrt{\frac{x^2}{a^2}-1} - \ln\left|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}\right|\right)+C \\[6pt] &= \frac{1}{2}\left(x\sqrt{x^2-a^2} - a^2\ln\left|\frac{x+\sqrt{x^2-a^2}}{a}\right|\right)+C. \end{align}</math>

When <math>\frac{\pi}{2} < \theta \le \pi,</math> which happens when <math>x < 0</math> given the range of arcsecant, <math>\tan \theta \le 0,</math> meaning <math>\sqrt{\tan^2 \theta} = -\tan \theta</math> instead in that case.

==Substitutions that eliminate trigonometric functions== Substitution can be used to remove trigonometric functions.

For instance,

<math display="block">\begin{align} \int f(\sin(x), \cos(x))\, dx &=\int\frac1{\pm\sqrt{1-u^2}} f\left(u,\pm\sqrt{1-u^2}\right)\, du && u=\sin (x) \\[6pt] \int f(\sin(x), \cos(x))\, dx &=\int\frac{1}{\mp\sqrt{1-u^2}} f\left(\pm\sqrt{1-u^2},u\right)\, du && u=\cos (x) \\[6pt] \int f(\sin(x), \cos(x))\, dx &=\int\frac2{1+u^2} f \left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\, du && u=\tan\left (\frac{x}{2} \right ) \\[6pt] \end{align}</math>

The last substitution is known as the Weierstrass substitution, which makes use of tangent half-angle formulas.

For example,

<math display=block>\begin{align} \int\frac{4 \cos x}{(1+\cos x)^3}\, dx &= \int\frac2{1+u^2}\frac{4\left(\frac{1-u^2}{1+u^2}\right)}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\, du = \int (1-u^2)(1+u^2)\, du \\&= \int (1-u^4)\,du = u - \frac{u^5}{5} + C = \tan \frac{x}{2} - \frac{1}{5} \tan^5 \frac{x}{2} + C. \end{align}</math>

==Hyperbolic substitution== Substitutions of hyperbolic functions can also be used to simplify integrals.<ref>{{cite web|last=Boyadzhiev|first=Khristo N.|title=Hyperbolic Substitutions for Integrals|url=http://www2.onu.edu/~m-caragiu.1/bonus_files/HYPERSUB.pdf|access-date=4 March 2013|archive-date=26 February 2020|archive-url=https://web.archive.org/web/20200226040813/http://www2.onu.edu/~m-caragiu.1/bonus_files/HYPERSUB.pdf|url-status=dead}}</ref>

For example, to integrate <math>1/\sqrt{a^2+x^2}</math>, introduce the substitution <math>x=a\sinh{u}</math> (and hence <math>dx=a\cosh u \,du</math>), then use the identity <math>\cosh^2 (x) - \sinh^2 (x) = 1</math> to find:

<math display="block">\begin{align} \int \frac{dx}{\sqrt{a^2+x^2}} &= \int \frac{a\cosh u \,du}{\sqrt{a^2+a^2\sinh^2 u}} \\[6pt] &=\int \frac{\cosh{u} \,du}{\sqrt{1+\sinh^2{u}}} \\[6pt] &=\int \frac{\cosh{u}}{\cosh u} \,du \\[6pt] &=u+C \\[6pt] &=\sinh^{-1}{\frac{x}{a}} + C. \end{align}</math>

If desired, this result may be further transformed using other identities, such as using the relation <math>\sinh^{-1}{z} = \operatorname{arsinh}{z} = \ln(z + \sqrt{z^2 + 1})</math>: <math display="block">\begin{align} \sinh^{-1}{\frac{x}{a}} + C &=\ln\left(\frac{x}{a} + \sqrt{\frac{x^2}{a^2} + 1}\,\right) + C \\[6pt] &=\ln\left(\frac{x + \sqrt{x^2+a^2}}{a}\,\right) + C. \end{align}</math>

==See also== {{Portal|Mathematics}} {{Wikiversity|Trigonometric Substitutions}} {{Wikibooks|Calculus/Integration techniques/Trigonometric Substitution}} * Weierstrass substitution * Euler substitution

==References== {{reflist}}

{{DEFAULTSORT:Trigonometric Substitution}} {{Integrals}} Category:Integral calculus Category:Trigonometry