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{{Wikiversity|Trigonometric Substitutions}}
{{Wikibooks|Calculus/Integration techniques/Trigonometric Substitution}}
{{calculus|expanded=Integral calculus}}
In [[mathematics]], '''trigonometric substitution''' is the substitution of trigonometric functions for other expressions. One may use the [[trigonometric identity|trigonometric identities]] to simplify certain [[integral]]s containing [[radical expression]]s:<ref>{{cite book | last=Stewart | first=James | authorlink=James Stewart (mathematician) | title=Calculus: Early Transcendentals |publisher=[[Brooks/Cole]] | edition=6th | year=2008 | isbn=0-495-01166-5}}</ref><ref>{{cite book | last1 = Thomas | first1 = George B. | last2=Weir | first2= Maurice D. | last3=Hass | first3=Joel | author3-link = Joel Hass | authorlink=George B. Thomas | title=Thomas' Calculus: Early Transcendentals | publisher=[[Addison-Wesley]] | year=2010 | edition=12th | isbn=0-321-58876-2}}</ref>

<blockquote>'''Substitution 1.''' If the integrand contains ''a''2 − ''x''2, let
: <math>x = a \sin(\theta)</math>
and use the [[list of trigonometric identities|identity]]
: <math>1-\sin^2(\theta) = \cos^2(\theta).</math></blockquote>

<blockquote>'''Substitution 2.''' If the integrand contains ''a''2 + ''x''2, let
: <math>x = a \tan(\theta)</math>
and use the identity
:<math>1+\tan^2(\theta) = \sec^2(\theta).</math></blockquote>

<blockquote>'''Substitution 3.''' If the integrand contains ''x''2 − ''a''2, let
:<math>x = a \sec(\theta)</math>
and use the identity
:<math>\sec^2(\theta)-1 = \tan^2(\theta).</math></blockquote>

==Examples==
=== Integrals containing ''a''2 − ''x''2 ===

In the integral

:<math>\int\frac{dx}{\sqrt{a^2-x^2}}</math>

we may use

:<math>x=a\sin(\theta),\quad dx=a\cos(\theta)\,d\theta, \quad \theta=\arcsin\left(\frac{x}{a}\right)</math>

:<math>\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)}} \\
&= \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2(1-\sin^2(\theta))}} \\
&= \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2\cos^2(\theta)}} \\
&= \int d\theta \\
&= \theta+C \\
&= \arcsin \left(\tfrac{x}{a}\right)+C
\end{align}</math>

Note that the above step requires that ''a'' > 0 and cos(θ) > 0; we can choose the ''a'' to be the positive square root of ''a''2; and we impose the restriction on θ to be −π/2 < θ < π/2 by using the [[arcsin]] function.

For a definite integral, one must figure out how the bounds of integration change. For example, as ''x'' goes from 0 to ''a''/2, then sin(θ) goes from 0 to 1/2, so θ goes from 0 to π/6. Then we have

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

Some care is needed when picking the bounds. The integration above requires that −π/2 < θ < π/2, so θ going from 0 to π/6 is the only choice. If we had missed this restriction, we might have picked θ to go from π to 5π/6, which would give us the negative of the result.

===Integrals containing ''a''2 + ''x''2===
In the integral

:<math>\int\frac{dx}{{a^2+x^2}}</math>

we may write

:<math>x=a\tan(\theta),\quad dx=a\sec^2(\theta)\,d\theta, \quad \theta=\arctan\left(\tfrac{x}{a}\right)</math>

so that the integral becomes

:<math>\begin{align}
\int\frac{dx}{{a^2+x^2}} &= \int\frac{a\sec^2(\theta)\,d\theta}{{a^2+a^2\tan^2(\theta)}} \\
&= \int\frac{a\sec^2(\theta)\,d\theta}{{a^2(1+\tan^2(\theta))}} \\
&= \int \frac{a\sec^2(\theta)\,d\theta}{{a^2\sec^2(\theta)}} \\
&= \int \frac{d\theta}{a} \\
&= \tfrac{\theta}{a}+C \\
&= \tfrac{1}{a} \arctan \left(\tfrac{x}{a}\right)+C
\end{align}</math>

(provided ''a'' ≠ 0).

===Integrals containing ''x''2 − ''a''2===
Integrals like

:<math>\int\frac{dx}{x^2 - a^2}</math>

should be done by [[partial fractions in integration|partial fractions]] rather than trigonometric substitutions. However, the integral

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

can be done by substitution:

:<math>x = a \sec(\theta),\quad dx = a \sec(\theta)\tan(\theta)\,d\theta, \quad \theta = \arcsec\left(\tfrac{x}{a}\right)</math>

:<math>\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>

We can then solve this using the formula for the [[integral of secant cubed]].

==Substitutions that eliminate trigonometric functions==
Substitution can be used to remove trigonometric functions. In particular, see [[Tangent half-angle substitution]].

For instance,

:<math>\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) \\
\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) \\
\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 (\tfrac{x}{2} \right ) \\
\int\frac{\cos x}{(1+\cos x)^3}\,dx &= \int\frac2{1+u^2}\frac{\frac{1-u^2}{1+u^2}}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\,du = \int \frac{1-u^2}{1+u^2}\,du
\end{align}</math>

==Hyperbolic functions==
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|accessdate=4 March 2013}}</ref>

In the integral <math>\int \frac{1}{\sqrt{a^2+x^2}}\,dx</math>, make the substitution <math>x=a\sinh{u}</math>, <math>dx=a\cosh{u}\,du</math>.

Then, using the identities <math>\cosh^2 (x) - \sinh^2 (x) = 1</math> and <math>\sinh^{-1}{x} = \ln(x + \sqrt{x^2 + 1})</math>,

<math>\begin{align}
\int \frac{1}{\sqrt{a^2+x^2}}\,dx &= \int \frac{a\cosh{u}}{\sqrt{a^2+a^2\sinh^2{u}}}\,du\\
&=\int \frac{a\cosh{u}}{a\sqrt{1+\sinh^2{u}}}\,du\\
&=\int \frac{a\cosh{u}}{a\cosh{u}}\,du\\
&=u+C\\
&=\sinh^{-1}{\frac{x}{a}}+C\\
&=\ln\left(\sqrt{\frac{x^2}{a^2} + 1} + \frac{x}{a}\right) + C\\
&=\ln\left(\frac{\sqrt{x^2+a^2} + x}{a}\right) + C
\end{align}</math>

==See also==
* [[Tangent half-angle formula]]

==References==
{{reflist}}

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

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