Editing Trigonometric substitution

{{short description|Technique of integral evaluation}}
{{Trigonometry}}
{{calculus|expanded=integral}}
In [[mathematics]], a '''trigonometric substitution''' replaces a [[trigonometric functions|trigonometric function]] for another expression. In [[calculus]], trigonometric substitutions are a technique for evaluating integrals. In this case, an expression involving a [[Radical expression|radical function]] 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> Like other methods of integration by substitution, when evaluating a definite integral, it may be simpler to completely deduce the antiderivative before applying the boundaries of integration.

==Case I: Integrands containing ''a''<sup>2</sup> − ''x''<sup>2</sup>==
Let <math>x = a \sin \theta,</math> and use the [[list of trigonometric identities|identity]] <math>1-\sin^2 \theta = \cos^2 \theta.</math>

===Examples of Case I===
[[File:Trig Sub Triangle 1.png|thumb|Geometric construction for Case I]]

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

<math display=block>\int\frac{dx}{\sqrt{a^2-x^2}},</math>

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

<math display=block>\int\sqrt{a^2-x^2}\,dx,</math>

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

<math display=block>\int_{-1}^1\sqrt{4-x^2}\,dx,</math>

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''<sup>2</sup> + ''x''<sup>2</sup>==
Let <math>x = a \tan \theta,</math> and use the identity <math>1+\tan^2 \theta = \sec^2 \theta.</math>

===Examples of Case II===
[[File:Trig Sub Triangle 2.png|thumb|Geometric construction for Case II]]
====Example 1====
In the integral

<math display=block>\int\frac{dx}{a^2+x^2}</math>

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

<math display=block>\int_0^1\frac{4\, dx}{1+x^2}\,</math>

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

<math display=block>\int\sqrt{a^2+x^2}\,{dx}</math>

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>

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

===Examples of Case III===
[[File:Trig Sub Triangle 3.png|thumb|Geometric construction for Case III]]

Integrals such as

<math display=block>\int\frac{dx}{x^2 - a^2}</math>

can also be evaluated by [[partial fractions in integration|partial fractions]] rather than trigonometric substitutions. However, the integral

<math display=block>\int\sqrt{x^2 - a^2}\, dx</math>

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 function]]s 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 [[Hyperbolic functions#Useful relations|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 [[Inverse hyperbolic functions#Definitions in terms of logarithms|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}}
* [[Integration by substitution]]
* [[Weierstrass substitution]]
* [[Euler substitution]]

==References==
{{reflist}}

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