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Undo revision 808256 by CPdefeated (Talk)
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{{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]]