Methods of integration - Revision history

Karajou: Reverted edits by BertJonson64 (talk) to last revision by FredericBernard

2020-10-15T21:36:06Z

Reverted edits by BertJonson64 (talk) to last revision by FredericBernard

BertJonson64: new integration aspects

2020-10-15T19:24:11Z

new integration aspects

FredericBernard: /* Algebraic Substitution */ Tidy up formatting

2018-12-19T18:06:57Z

Algebraic Substitution: Tidy up formatting

FredericBernard: /* Trigonometric Substitution */ Tidy up formatting

2018-12-19T17:58:51Z

Trigonometric Substitution: Tidy up formatting

FredericBernard: /* See also */ Add Table of Indefinite Integrals

2018-12-19T17:50:48Z

See also: Add Table of Indefinite Integrals

DavidB4-bot: /* Reverse Chain Rule */Spelling, Grammar, and General Cleanup, typos fixed: immeadiately → immediately

2017-09-08T13:00:52Z

Reverse Chain Rule: Spelling, Grammar, and General Cleanup, typos fixed: immeadiately → immediately

FredericBernard: Added a section on the reverse chain rule

2017-08-28T12:18:50Z

Added a section on the reverse chain rule

DavidB4-bot: /* top */clean up & uniformity

2016-07-13T15:58:53Z

top: clean up & uniformity

DavidB4-bot: Spelling, Grammar, and General Cleanup, typos fixed: subsitution → substitution, Therefore → Therefore,

2016-07-01T13:17:50Z

Spelling, Grammar, and General Cleanup, typos fixed: subsitution → substitution, Therefore → Therefore,

DrewJ: Undo revision 808256 by CPdefeated (Talk)

2010-08-20T19:16:21Z

Undo revision 808256 by CPdefeated (Talk)

@@ Line 97: / Line 97: @@
 ==Trigonometric Substitution==
-'''Integration by Trigonometric Substitution''' is a technique to facilitate the integration of a rational expression by substituting a more complicated radical expression with a trigonometric expression. This is the least Christian technique of integration and should be avoided at all costs. Atheists invented the sine function to force conservatives learning math to write the word ''sin''.
+'''Integration by Trigonometric Substitution''' is a technique to facilitate the integration of a rational expression by substituting a more complicated radical expression with a trigonometric expression.
 Given an integral

@@ Line 97: / Line 97: @@
 ==Trigonometric Substitution==
-'''Integration by Trigonometric Substitution''' is a technique to facilitate the integration of a rational expression by substituting a more complicated radical expression with a trigonometric expression.
+'''Integration by Trigonometric Substitution''' is a technique to facilitate the integration of a rational expression by substituting a more complicated radical expression with a trigonometric expression. This is the least Christian technique of integration and should be avoided at all costs. Atheists invented the sine function to force conservatives learning math to write the word ''sin''.
 Given an integral

@@ Line 37: / Line 37: @@
 Given an integral
-:<big><math>\int\frac{2x}{x^2+3}dx</math></big>
+:<math>\int\frac{2x}{x^2+3}dx</math>
-We can substitute the term :<big><math>x^2+3</math></big> with a u.  Giving us
+We can substitute the term ''x''<sup>2</sup>+3 with a ''u''.  Giving us
-:<big><math>u=x^2+3</math></big>
+:<math>u=x^2+3</math>
-We then take the derivative of u with respect to x,
+We then take the derivative of ''u'' with respect to ''x'',
-:<big><math>\frac{du}{dx}=2x</math></big>
+:<math>\frac{du}{dx}=2x</math>
-We then set the terms equal to du,
+We then set the terms equal to d''u'',
-:<big><math>du=2xdx</math></big>
+:<math>du=2xdx</math>
 Now we are ready to rewrite the integral,
-:<big><math>\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du</math></big>
+:<math>\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du</math>
-We can rewrite the integral this way due to the substitution of the x terms with the u terms.
+We can rewrite the integral this way due to the substitution of the ''x'' terms with the ''u'' terms.
-Now we can solve the integral in terms of u.
+Now we can solve the integral in terms of ''u''.
-:<big><math>\int\frac{1}{u}du=ln|u|+c</math></big>
+:<math>\int\frac{1}{u}du=ln|u|+c</math>
-Now we replace u with the term :<big><math>x^2+3</math></big> to get,
+Now we replace ''u'' with the term ''x''<sup>2</sup>+3 to get,
-:<big><math>ln|x^2+3|+c</math></big>
+:<math>\ln|x^2+3|+c</math>
-We can check this by taking the derivative of :<big><math>ln|x^2+3|</math></big>,
+We can check this by taking the derivative of <math display="inline">\ln|x^2+3|</math>,
-:<big><math>\frac{d}{dx}ln|x^2+3|=\left (\frac{1}{x^2+3} \right)(2x)=\frac{2x}{x^2+3}</math></big>
+:<math>\frac{d}{dx}\ln|x^2+3|=\left (\frac{1}{x^2+3} \right)(2x)=\frac{2x}{x^2+3}</math>
 ===Reverse Chain Rule===

@@ Line 101: / Line 101: @@
 Given an integral
-:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx</math></big>
+:<math>\int\frac{1}{\sqrt{9-x^2}}dx</math>
 By looking at the radical we can determine that it represents the base of a right triangle by understanding the [[Pythagorean theorem]].
 :<math>\sqrt{9-x^2}=3+x</math>
-:<big><math>\sqrt{9-x^2}=3+x</math></big> where 3 is the hypotenuse and x is the height of the triangle.
+where 3 is the hypotenuse and ''x'' is the height of the triangle.
-This allows us to rewrite the expression to :<big><math>sin\theta=\frac{x}{3}</math></big>.  This allows us to substitute x with :<big><math>3sin\theta</math></big>.
+This allows us to rewrite the expression to <math display="inline">\sin\theta=\frac{x}{3}</math>. This allows us to substitute ''x'' with <math display="inline">3\sin\theta</math>.
 Now to do the substitution
-:<big><math>9-x^2=9-(3sin\theta )^2</math></big>
+:<math>9-x^2=9-(3\sin\theta )^2</math>
-:<big><math>9-(3sin\theta )^2=9-9sin^2\theta </math></big>
+:<math>9-(3\sin\theta )^2=9-9\sin^2\theta </math>
-:<big><math>9-9sin^2\theta=9(1- sin^2\theta) </math></big>
+:<math>9-9\sin^2\theta=9(1- \sin^2\theta) </math>
 And by use of trigonometric identities we know that
-:<big><math>1- sin^2\theta=cos^2\theta </math></big>
+:<math>1- \sin^2\theta=\cos^2\theta </math>
-:<big><math>9(1- sin^2\theta)=9cos^2\theta </math></big>
+:<math>9(1- \sin^2\theta)=9\cos^2\theta </math>
 Therefore,
-:<big><math>\sqrt{9-x^2}=3cos\theta </math></big>
+:<math>\sqrt{9-x^2}=3\cos\theta </math>
-We are not done yet, we must also take the derivative of :<big><math>3sin\theta</math></big>
+We are not done yet, we must also take the derivative of <math display="inline">3\sin\theta</math>
-:<big><math>\frac{dx}{d\theta}3sin\theta=3cos\theta </math></big>
+:<math>\frac{dx}{d\theta} 3\sin\theta=3\cos\theta </math>
-By partial derivatives we move the :<big><math>{d\theta}</math></big> over.
+By partial derivatives we move the <math display="inline">d\theta</math> over.
-:<big><math>dx=3cos\theta d\theta </math></big>
+:<math>dx=3\cos\theta d\theta </math>
 Now we are ready to rewrite our integral.
-:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx=\int\frac{3cos\theta}{3cos\theta}d\theta=\int d\theta=\theta+c</math></big>
+:<math>\int\frac{1}{\sqrt{9-x^2}}dx=\int\frac{3\cos\theta}{3\cos\theta}d\theta=\int d\theta=\theta+c</math>
-:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx=sin^{-1}\left(\frac{x}{3}\right)+c </math></big>
+From our trigonometric expression <math display="inline">x=3\sin\theta</math> we can see that
 == Other methods of integration ==

@@ Line 145: / Line 145: @@
 *[[Calculus]]
 *[[Integration]]
 ==External links==

@@ Line 82: / Line 82: @@
 </math>
-This means that if a function can be recognised in the form f'(x)g'(f(x)), then it can be integrated almost immeadiately. As an example, consider the function:
+This means that if a function can be recognised in the form f'(x)g'(f(x)), then it can be integrated almost immediately. As an example, consider the function:
 :<math>

← Older revision		Revision as of 12:18, August 28, 2017
Line 68:		Line 68:

	:<big><math>\frac{d}{dx}ln\|x^2+3\|=\left (\frac{1}{x^2+3} \right)(2x)=\frac{2x}{x^2+3}</math></big>		:<big><math>\frac{d}{dx}ln\|x^2+3\|=\left (\frac{1}{x^2+3} \right)(2x)=\frac{2x}{x^2+3}</math></big>
		+
		+	===Reverse Chain Rule===
		+	The reverse chain rule is a trick for integration by substitution without having to do all the steps involved with integration by substitution. It uses a reversed form of the [[chain rule]] in [[differentiation]] to integrate functions of the form f'(x)g'(f(x)), where f and g are two [[function]]s and f'(x) is the derivative of f(x). From the chain rule, which states:
		+
		+	:<math>
		+	\frac{d}{dx} \left( g(f(x)) \right) = f'(x) g'(f(x))
		+	</math>
		+
		+	Integrating both sides produces the reverse chain rule:
		+
		+	:<math>
		+	\int f'(x) g'(f(x)) \, dx = g(f(x))
		+	</math>
		+
		+	This means that if a function can be recognised in the form f'(x)g'(f(x)), then it can be integrated almost immeadiately. As an example, consider the function:
		+
		+	:<math>
		+	y(x) = 2x \cos{\left( x^2 \right)}
		+	</math>
		+
		+	Comparing to the above, f(x)=x<sup>2</sup>, f'(x)=2x and g'(x) is represented by a [[cosine]]. All wee need to do is know the integral of a cosine is [[sine]] so g(x) is sine. Therefore:
		+
		+	:<math>
		+	\int 2x \cos{\left( x^2 \right)} \, dx = \sin{\left( x^2 \right)}
		+	</math>
		+
		+	If the first part of the function is f'(x) multiplied by a constant, then this constant can be taken out of the integral so that it the integral is of the form f'(x)g'(f(x)). This method can also be applied to the fraction example above.

	==Trigonometric Substitution==		==Trigonometric Substitution==

← Older revision		Revision as of 15:58, July 13, 2016
Line 1:		Line 1:
−	{{~~Template:~~Math-h}}	+	{{Math-h}}

	This article details several methods of [[integral\|integration]] for advanced high school or early university student, each with an example.		This article details several methods of [[integral\|integration]] for advanced high school or early university student, each with an example.

← Older revision		Revision as of 19:16, August 20, 2010
Line 1:		Line 1:
−	[[CP is ~~raped~~ by ~~vandals!~~]]	+	{{Template:Math-h}}
		+
		+	This article details several methods of [[integral\|integration]] for advanced high school or early university student, each with an example.
		+
		+	==Estimation==
		+	There are various methods of estimating integrals. This is particularly useful if the rule for integrating is unknown or extremely complicated, as in the case of <math>e^{x^2}</math>. The primary method of estimation is the Riemann integral.
		+
		+	<br />Another major method, known as [[Simpson's rule]] uses the Riemann integral. Instead of using linear estimations, Simpson's rule allows for the use of [[parabola]]s or other higher order expressions to estimate integrals.
		+	:{{main\|Simpson's rule}}
		+
		+	==Integration by Parts==
		+	{{main\|Integration by parts}}
		+	'''Integration by parts''' is a special technique to facilitate the integration of the product of two functions that otherwise lack an obvious integral. This technique can be proven with the [[product rule]].
		+
		+	The rule for '''integration by parts''' is stated as follows:
		+	:<big><math>\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,</math></big>
		+	:or
		+	:<big><math>\int u\,dv = uv - \int v\,du\,</math></big>
		+
		+	This rule is often useful when one function is a power of ''x'' and the other function is a trigonometric function or ''e'' raised to a power of ''x''.
		+
		+	Note that it may be necessary to repeat the '''integration by parts''' several times, one for each power of ''x''.
		+
		+	==Partial Fractions==
		+	{{Main\|Partial fractions in integration}}
		+	'''Integration by partial fractions''' is a technique to facilitate the integration of a rational expression by partial fraction decomposition.
		+
		+	Given an integral
		+	<br />
		+	:<math>\int \frac {f(x)}{g(x)}dx</math>
		+	<br />
		+	where <math>f(x)</math> and <math>g(x)</math> are both polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.
		+
		+	==Algebraic Substitution==
		+	'''Integration by Algebraic Substitution''' is a technique to facilitate the integration of a rational expression by substituting a more complicated expression with a variable.
		+
		+	Given an integral
		+
		+	:<big><math>\int\frac{2x}{x^2+3}dx</math></big>
		+
		+	We can substitute the term :<big><math>x^2+3</math></big> with a u. Giving us
		+
		+	:<big><math>u=x^2+3</math></big>
		+
		+	We then take the derivative of u with respect to x,
		+
		+	:<big><math>\frac{du}{dx}=2x</math></big>
		+
		+	We then set the terms equal to du,
		+
		+	:<big><math>du=2xdx</math></big>
		+
		+	Now we are ready to rewrite the integral,
		+
		+	:<big><math>\int\frac{2x}{x^2+3}dx=\int\frac{1}{u}du</math></big>
		+
		+	We can rewrite the integral this way due to the subsitution of the x terms with the u terms.
		+
		+	Now we can solve the integral in terms of u.
		+
		+	:<big><math>\int\frac{1}{u}du=ln\|u\|+c</math></big>
		+
		+	Now we replace u with the term :<big><math>x^2+3</math></big> to get,
		+
		+	:<big><math>ln\|x^2+3\|+c</math></big>
		+
		+	We can check this by taking the derivative of :<big><math>ln\|x^2+3\|</math></big>,
		+
		+	:<big><math>\frac{d}{dx}ln\|x^2+3\|=\left (\frac{1}{x^2+3} \right)(2x)=\frac{2x}{x^2+3}</math></big>
		+
		+	==Trigonometric Substitution==
		+	'''Integration by Trigonometric Substitution''' is a technique to facilitate the integration of a rational expression by substituting a more complicated radical expression with a trigonometric expression.
		+
		+	Given an integral
		+
		+	:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx</math></big>
		+
		+	By looking at the radical we can determine that it represents the base of a right triangle by understanding the [[Pythagorean theorem]].
		+
		+	:<big><math>\sqrt{9-x^2}=3+x</math></big> where 3 is the hypotenuse and x is the height of the triangle.
		+	This allows us to rewrite the expression to :<big><math>sin\theta=\frac{x}{3}</math></big>. This allows us to substitute x with :<big><math>3sin\theta</math></big>.
		+	Now to do the substitution
		+	:<big><math>9-x^2=9-(3sin\theta )^2</math></big>
		+	:<big><math>9-(3sin\theta )^2=9-9sin^2\theta </math></big>
		+	:<big><math>9-9sin^2\theta=9(1- sin^2\theta) </math></big>
		+	And by use of trigonometric identities we know that
		+	:<big><math>1- sin^2\theta=cos^2\theta </math></big>
		+	:<big><math>9(1- sin^2\theta)=9cos^2\theta </math></big>
		+	Therefore
		+	:<big><math>\sqrt{9-x^2}=3cos\theta </math></big>
		+	We are not done yet, we must also take the derivative of :<big><math>3sin\theta</math></big>
		+	:<big><math>\frac{dx}{d\theta}3sin\theta=3cos\theta </math></big>
		+	By partial derivatives we move the :<big><math>{d\theta}</math></big> over.
		+	:<big><math>dx=3cos\theta d\theta </math></big>
		+
		+	Now we are ready to rewrite our integral.
		+
		+	:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx=\int\frac{3cos\theta}{3cos\theta}d\theta=\int d\theta=\theta+c</math></big>
		+
		+	From our trigonometric expression :<big><math>x=3sin\theta</math></big> we can see that
		+	:<big><math>\theta=sin^{-1}\left(\frac{x}{3}\right)+c</math></big> giving us the final solution.
		+
		+	:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx=sin^{-1}\left(\frac{x}{3}\right)+c </math></big>
		+
		+	== Other methods of integration ==
		+	The primary methods of integration include:
		+
		+	* [[Integration by parts]]
		+	* [[Integration using polar coordinates]]
		+	* [[Multiple integration]]
		+	* [[Residue calculus]] (for definite integrals)
		+	* [[Method of simultaneous convolutions]]
		+	* [[Mellin transform]]s
		+	* [[Inflation-restriction sequence]]s
		+
		+	==See also==
		+	*[[Fundamental Theorem of Calculus]]
		+	*[[Calculus]]
		+	*[[Integration]]
		+
		+	==External links==
		+	*[http://user.mendelu.cz/marik/maw/index.php?lang=en&form=integral Antiderivative Solver]
		+
		+	[[Category:calculus]]
		+	[[Category:integration]]