Difference between revisions of "Trigonometric substitution"

 
(8 intermediate revisions by 6 users not shown)
Line 1: Line 1:
Trigonometric substitution is the technique of replacing unknown variables in equations with <math>\sin \theta\,</math> or <math>\cos {\theta}\,</math>
+
'''Trigonometric substitution''' is the technique of replacing [[variable]]s in equations with <math>\sin \theta\,</math> or <math>\cos {\theta}\,</math> or other [[function]]s from [[trigonometry]].
  
  This article is incomplete (did I need to tell you that?).
+
In [[calculus]], it is used to evaluate [[integral]]s of [[expression]]s such as <math>\sqrt{a^2+x^2},\sqrt{a^2-x^2}</math> or <math>\sqrt{x^2-a^2}</math>
 +
 
 +
== Examples ==
 +
=== <math>\sqrt{a^2+x^2}</math> ===
 +
To evaluate an expression such as <math>\int \sqrt{a^2+x^2}\,dx</math>, we make use of the identity <math>\tan^2x+1=\sec^2x</math>. Set <math>x=a\tan\theta</math> and the radical will go away. However, the <math>dx</math> will have to be changed in terms of <math>d\theta</math>: <math>dx=a\sec^2\theta</math> <math>d\theta</math>
 +
 
 +
=== <math>\sqrt{a^2-x^2}</math> ===
 +
Making use of the identity <math>\sin^2\theta+\cos^2\theta=1</math>, simply let <math>x=a\sin\theta</math>.
 +
 
 +
 
 +
=== <math>\sqrt{x^2-a^2}</math> ===
 +
Since <math>\sec^2(\theta)-1=\tan^2(\theta)</math>, let <math>x=a\sec\theta</math>.

Revision as of 07:24, 20 August 2019

Trigonometric substitution is the technique of replacing variables in equations with $\sin \theta\,$ or $\cos {\theta}\,$ or other functions from trigonometry.

In calculus, it is used to evaluate integrals of expressions such as $\sqrt{a^2+x^2},\sqrt{a^2-x^2}$ or $\sqrt{x^2-a^2}$

Examples

$\sqrt{a^2+x^2}$

To evaluate an expression such as $\int \sqrt{a^2+x^2}\,dx$, we make use of the identity $\tan^2x+1=\sec^2x$. Set $x=a\tan\theta$ and the radical will go away. However, the $dx$ will have to be changed in terms of $d\theta$: $dx=a\sec^2\theta$ $d\theta$

$\sqrt{a^2-x^2}$

Making use of the identity $\sin^2\theta+\cos^2\theta=1$, simply let $x=a\sin\theta$.


$\sqrt{x^2-a^2}$

Since $\sec^2(\theta)-1=\tan^2(\theta)$, let $x=a\sec\theta$.