View source for Integration by reduction formulae

{{Calculus |Integral}}

'''Integration by reduction formula''' in [[integral calculus]] is a technique or procedure of integration, in the form of a [[recurrence relation]]. It is used when an [[expression (mathematics)|expression]] containing an [[integer]] [[parameter#Mathematical functions|parameter]], usually in the form of powers of elementary functions, or [[Product (mathematics)|product]]s of [[transcendental function]]s and [[polynomial]]s of arbitrary [[degree of a polynomial|degree]], can't be integrated directly. But using other [[integral#Methods|methods of integration]] a '''reduction formula''' can be set up to obtain the integral of the same or similar expression with a lower integer parameter, progressively simplifying the integral until it can be evaluated. <ref>Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, {{ISBN|978-0-521-86153-3}}</ref> This method of integration is one of the earliest used.

== How to find the reduction formula ==

The reduction formula can be derived using any of the common methods of integration, like [[integration by substitution]], [[integration by parts]], [[Trigonometric substitution|integration by trigonometric substitution]], [[integration by partial fractions]], etc. The main idea is to express an integral involving an integer parameter (e.g. power) of a function, represented by I<sub>n</sub>, in terms of an integral that involves a lower value of the parameter (lower power) of that function, for example ''I''<sub>''n''-1</sub> or ''I''<sub>''n''-2</sub>. This makes the reduction formula a type of [[recurrence relation]]. In other words, the reduction formula expresses the integral 
:<math>I_n =\int f(x,n) \,\text{d}x, </math> 
in terms of 
:<math>I_k =  \int f(x,k) \,\text{d}x, </math> 
where 
:<math>k < n.</math>

== How to compute the integral ==

To compute the integral, we set ''n'' to its value and use the reduction formula to calculate the (''n'' – 1) or (''n'' – 2) integral. The higher index integral can be used to calculate lower index ones; the process is continued repeatedly until we reach a point where the function to be integrated can be computed, usually when its index is 0 or 1. Then we back-substitute the previous results until we have computed ''I<sub>n</sub>''. <ref>Further Elementary Analysis, R.I. Porter, G. Bell & Sons Ltd, 1978, {{ISBN|0-7135-1594-5}}</ref>

===Examples=== 

Below are examples of the procedure.

'''Cosine integral'''

Typically, integrals like 

:<math>\int \cos^n x \,\text{d}x , \,\!</math>

can be evaluated by a reduction formula. 

[[Image:Cos to the n.png|thumb|<math>\int \cos^n (x) \,\text{d}x\!</math>, for ''n'' = 1, 2 ... 30]]

Start by setting:

:<math>I_n = \int \cos^n x\,\text{d}x . \,\!</math>

Now re-write as:

:<math>I_n = \int \cos^ {n-1} x \cos x \,\text{d}x , \,\!</math>

Integrating by this substitution:

:<math>\cos x \,\text{d}x = \text{d} ( \sin x) , \,\!</math>

:<math>I_n = \int \cos^{n-1} x \,\text{d}(\sin x) . \!</math>

Now integrating by parts:

:<math> \begin{align} \int \cos^n x \,\text{d}x & = \cos^{n-1} x \sin x - \int \sin x \,\text{d}(\cos^{n-1} x) \\
& = \cos^{n-1} x \sin x + (n-1) \int \sin x \cos^{n-2} x\sin x \,\text{d}x\\
& = \cos^{n-1} x \sin x + (n-1) \int \cos^{n-2} x \sin^2 x \,\text{d}x\\
& = \cos^{n-1} x \sin x + (n-1) \int \cos^{n-2} x (1-\cos^2 x )\,\text{d}x\\
& = \cos^{n-1} x \sin x + (n-1) \int \cos^{n-2} x \,\text{d}x - (n-1)\int \cos^n x \,\text{d}x\\
& = \cos^{n-1} x \sin x + (n-1) I_{n-2} - (n-1) I_n ,
\end{align} \,</math>

solving for ''I<sub>n</sub>'':

:<math>I_n \ + (n-1) I_n\ = \cos^{n-1} x \sin x\ + \ (n-1) I_{n-2} , \,</math>

:<math>n I_n\ = \cos^{n-1} (x) \sin x\ + (n-1) I_{n-2} , \,</math>

:<math>I_n \ = \frac{1}{n}\cos^{n-1} x \sin x\ + \frac{n-1}{n} I_{n-2} , \,</math>

so the reduction formula is:

:<math>\int \cos^n x \,\text{d}x\ = \frac{1}{n}\cos^{n-1} x \sin x + \frac{n-1}{n} \int \cos^{n-2} x \,\text{d}x . \!</math>

To supplement the example, the above can be used to evaluate the integral for (say) ''n'' = 5;

:<math> I_5 = \int \cos^5 x \,\text{d}x . \,\!</math>

Calculating lower indices:

:<math>n=5, \quad I_5 = \tfrac{1}{5} \cos^4 x \sin x + \tfrac{4}{5} I_3 , \,</math>

:<math>n=3, \quad I_3 = \tfrac{1}{3} \cos^2 x \sin x + \tfrac{2}{3} I_1, \,</math>

back-substituting:

:<math>\because I_1\ = \int \cos x \,\text{d}x = \sin x + C_1,\,</math>

:<math>\therefore I_3\ = \tfrac{1}{3} \cos^2 x \sin x + \tfrac{2}{3}\sin x + C_2, \quad C_2\ = \tfrac{2}{3} C_1,\,</math>

:<math>I_5\ = \frac{1}{5} \cos^4 x \sin x + \frac{4}{5}\left[\frac{1}{3} \cos^2 x \sin x + \frac{2}{3} \sin x\right] + C,\,</math>

where ''C'' is a constant.

'''Exponential integral'''

Another typical example is:

:<math>\int x^n e^{ax} \,\text{d}x . \,\!</math>

Start by setting:

:<math>I_n = \int x^n e^{ax} \,\text{d}x . \,\!</math>

Integrating by substitution:

:<math> x^n \,\text{d}x = \frac{\text{d} ( x^{n+1})}{n+1} , \,\!</math>

:<math>I_n = \frac{1}{n+1} \int e^{ax} \,\text{d}(x^{n+1}) , \!</math>

Now integrating by parts:

:<math>\begin{align} \int e^{ax} \,\text{d}(x^{n+1}) & = x^{n+1}e^{ax} - \int x^{n+1} \,\text{d}(e^{ax}) \\
& = x^{n+1}e^{ax} - a \int x^{n+1} e^{ax}\,\text{d}x ,
\end{align} \!</math>

:<math>(n+1) I_n = x^{n+1}e^{ax} - a I_{n+1} , \!</math>

shifting indices back by 1 (so ''n + 1'' → ''n'', ''n'' → ''n'' – 1):

:<math>n I_{n-1} = x^ne^{ax} - a I_n , \!</math>

solving for ''I<sub>n</sub>'':

:<math> I_n = \frac{1}{a} \left ( x^ne^{ax} - n I_{n-1} \right ) , \,\!</math>

so the reduction formula is:

:<math> \int x^n e^{ax} \,\text{d}x = \frac{1}{a} \left ( x^ne^{ax} - n \int x^{n-1} e^{ax} \,\text{d}x \right ). \!</math>
An alternative way in which the derivation could be done starts by substituting <math>e^{ax}</math>.

Integration by substitution:

<math> e^{ax} \,\text{d}x = \frac{\text{d} ( e^{ax})}{a} , \,\!</math>

<math>I_n = \frac{1}{a} \int x^{n} \,\text{d}(e^{ax}) , \!</math>

Now integrating by parts:

<math>\begin{align} \int x^{n} \,\text{d}(e^{ax}) & = x^{n}e^{ax} - \int e^{ax} \,\text{d}(x^{n}) \\
& = x^{n}e^{ax} - n \int e^{ax} x^{n-1}\,\text{d}x ,
\end{align} \!</math>

which gives the reduction formula when substituting back:

<math> I_n = \frac{1}{a} \left ( x^ne^{ax} - n I_{n-1} \right ) , \,\!</math>

which is equivalent to:

:<math> \int x^n e^{ax} \,\text{d}x = \frac{1}{a} \left ( x^ne^{ax} - n \int x^{n-1} e^{ax} \,\text{d}x \right ). \!</math>

== Tables of integral reduction formulas ==

===Rational functions===

The following integrals<ref>http://www.sosmath.com/tables/tables.html -> Indefinite integrals list</ref> contain:
*Factors of the [[Linear equation|linear]] [[Square root|radical]] <math>\sqrt{ax+b}\,\!</math>
*Linear factors <math>{px+q}\,\!</math> and the linear radical <math>\sqrt{ax+b}\,\!</math>
*[[Quadratic polynomial|Quadratic]] factors <math>x^2+a^2\,\!</math>
*Quadratic factors <math>x^2-a^2\,\!</math>, for <math>x>a\,\!</math>
*Quadratic factors <math>a^2-x^2\,\!</math>, for <math>x<a\,\!</math>
*([[Irreducible polynomial|Irreducible]]) quadratic factors <math>ax^2+bx+c\,\!</math>
*Radicals of irreducible quadratic factors <math>\sqrt{ax^2+bx+c}\,\!</math>

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_n = \int \frac{x^n}{\sqrt{ax+b}} \,\text{d}x\,\!</math> || <math>I_n = \frac{2x^n\sqrt{ax+b}}{a(2n+1)} - \frac{2nb}{a(2n+1)} I_{n-1}\,\!</math>
|-
| <math>I_n = \int \frac{\text{d}x}{x^n\sqrt{ax+b}}\,\!</math> || <math>I_n = -\frac{\sqrt{ax+b}}{(n-1)bx^{n-1}}-\frac{a(2n-3)}{2b(n-1)}I_{n-1}\,\!</math>
|-
| <math>I_n = \int x^n\sqrt{ax+b}\,\text{d}x\,\!</math> || <math>I_n = \frac{2x^n\sqrt{(ax+b)^3}}{a(2n+3)}-\frac{2nb}{a(2n+3)}I_{n-1}\,\!</math>
|-
| <math>I_{m,n} = \int \frac{\text{d}x}{(ax+b)^m(px+q)^n}\,\!</math> || <math>I_{m,n} = 
\begin{cases}
  -\frac{1}{(n-1)(bp-aq)} \left [ \frac{1}{(ax+b)^{m-1}(px+q)^{n-1}}+a(m+n-2)I_{m,n-1} \right ]   \\
  \frac{1}{(m-1)(bp-aq)} \left [ \frac{1}{(ax+b)^{m-1}(px+q)^{n-1}}+p(m+n-2)I_{m-1,n} \right ]
\end{cases}\,\!</math>
|-
| <math>I_{m,n} = \int  \frac{(ax+b)^m}{(px+q)^n} \,\text{d}x\,\!</math> || <math>I_{m,n} =
\begin{cases}
  -\frac{1}{(n-1)(bp-aq)}\left [ \frac{(ax+b)^{m+1}}{(px+q)^{n-1}}+a(n-m-2)I_{m,n-1} \right ]   \\
  -\frac{1}{(n-m-1)p}\left [ \frac{(ax+b)^m}{(px+q)^{n-1}}+m(bp-aq)I_{m-1,n} \right ]  \\
  -\frac{1}{(n-1)p}\left [ \frac{(ax+b)^m}{(px+q)^{n-1}}-amI_{m-1,n-1} \right ]
\end{cases}\,\!</math>
|}

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_n=\int \frac{(px+q)^n}{\sqrt{ax+b}} \,\text{d}x\,\!</math> || <math>\int (px+q)^n\sqrt{ax+b} \,\text{d}x  = \frac{2(px+q)^{n+1}\sqrt{ax+b}}{p(2n+3)}+\frac{bp-aq}{p(2n+3)}I_n\,\!</math>

<math>I_n=\frac{2(px+q)^n\sqrt{ax+b}}{a(2n+1)}+\frac{2n(aq-bp)}{a(2n+1)}I_{n-1}\,\!</math>
|-
| <math>I_n=\int \frac{\text{d}x}{(px+q)^n\sqrt{ax+b}}\,\!</math> || <math>\int \frac{\sqrt{ax+b}}{(px+q)^n}\,\text{d}x = -\frac{\sqrt{ax+b}}{p(n-1)(px+q)^{n-1}}+\frac{a}{2p(n-1)}I_{n}\,\!</math>

<math>I_n= -\frac{\sqrt{ax+b}}{(n-1)(aq-bp)(px+q)^{n-1}}+\frac{a(2n-3)}{2(n-1)(aq-bp)}I_{n-1}\,\!</math>
|}

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_n= \int \frac{\text{d}x}{(x^2+a^2)^n}\,\!</math> || <math>I_n= \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}I_{n-1}\,\!</math>
|-
| <math>I_{n,m}= \int \frac{\text{d}x}{x^m(x^2+a^2)^n}\,\!</math> || <math>a^2I_{n,m}= I_{m,n-1}-I_{m-2,n}\,\!</math>
|-
| <math>I_{n,m}= \int \frac{x^m}{(x^2+a^2)^n} \,\text{d}x\,\!</math> || <math>I_{n,m}= I_{m-2,n-1}-a^2I_{m-2,n}\,\!</math>
|}

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_n= \int \frac{\text{d}x}{(x^2-a^2)^n}\,\!</math> || <math>I_n= -\frac{x}{2a^2(n-1)(x^2-a^2)^{n-1}}-\frac{2n-3}{2a^2(n-1)}I_{n-1}\,\!</math>
|-
| <math>I_{n,m}= \int \frac{\text{d}x}{x^m(x^2-a^2)^n}\,\!</math> || <math>{a^2}I_{n,m}= I_{m-2,n}-I_{m,n-1}\,\!</math>
|-
| <math>I_{n,m}= \int \frac{x^m}{(x^2-a^2)^n} \,\text{d}x\,\!</math> || <math>I_{n,m}= I_{m-2,n-1}+a^2I_{m-2,n}\,\!</math>
|}

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_n= \int \frac{\text{d}x}{(a^2-x^2)^n}\,\!</math> || <math>I_n= \frac{x}{2a^2(n-1)(a^2-x^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}I_{n-1}\,\!</math>
|-
| <math>I_{n,m}= \int \frac{\text{d}x}{x^m(a^2-x^2)^n}\,\!</math>  || <math>{a^2}I_{n,m}= I_{m,n-1}+I_{m-2,n}\,\!</math>
|-
| <math>I_{n,m}= \int \frac{x^m}{(a^2-x^2)^n} \,\text{d}x\,\!</math> || <math>I_{n,m}= a^2I_{m-2,n}-I_{m-2,n-1}\,\!</math>
|}

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_n = \int \frac{\text{d}x}{{x^n}(ax^2+bx+c)}\,\!</math> || <math>-cI_n =\frac{1}{x^{n-1}(n-1)}+ bI_{n-1}+aI_{n-2}\,\!</math>
|-
| <math>I_{m,n}=\int \frac{x^m \,\text{d}x}{(ax^2+bx+c)^n}\,\!</math> || <math>I_{m,n}= -\frac{x^{m-1}}{a(2n-m-1)(ax^2+bx+c)^{n-1}} - \frac{b(n-m)}{a(2n-m-1)}I_{m-1,n} + \frac{c(m-1)}{a(2n-m-1)}I_{m-2,n}\,\!</math>
|-
| <math>I_{m,n}= \int \frac{\text{d}x}{x^m(ax^2+bx+c)^n}\,\!</math> || <math>-c(m-1)I_{m,n}= \frac{1}{x^{m-1}(ax^2+bx+c)^{n-1}}+{a(m+2n-3)}I_{m-2,n}+{b(m+n-2)}I_{m-1,n}\,\!</math>
|-
|}

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_n = \int (ax^2+bx+c)^n\,\text{d}x\,\!</math> || <math>8a(n+1)I_{n+\frac{1}{2}} = 2(2ax+b)(ax^2+bx+c)^{n+\frac{1}{2}} + (2n+1)(4ac-b^2)I_{n-\frac{1}{2}}\,\!</math>
|-
| <math>I_n = \int \frac{1}{(ax^2+bx+c)^n}\,\text{d}x\,\!</math> || <math>(2n-1)(4ac-b^2)I_{n+\frac{1}{2}} = \frac{2(2ax+b)}{(ax^2+bx+c)^{n-\frac{1}{2}}}+{8a(n-1)}I_{n-\frac{1}{2}}\,\!</math>
|-
|}

note that by the [[Exponentiation|laws of indices]]:

:<math>I_{n+\frac{1}{2}} =  I_{\frac{2n+1}{2}} =\int \frac{1}{(ax^2+bx+c)^{\frac{2n+1}{2}}}\,\text{d}x = \int \frac{1}{\sqrt{(ax^2+bx+c)^{2n+1}}}\,\text{d}x\,\!</math>

===Transcendental functions===
{{main article|Transcendental function}}

The following integrals<ref>http://www.sosmath.com/tables/tables.html -> Indefinite integrals list</ref> contain:
*Factors of sine
*Factors of cosine
*Factors of sine and cosine products and quotients
*Products/quotients of exponential factors and powers of ''x''
*Products of exponential and sine/cosine factors

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_n=\int x^n \sin{ax} \,\text{d}x\,\!</math> || <math>a^2I_n=-ax^n \cos{ax} + nx^{n-1} \sin{ax} - n(n-1) I_{n-2} \,\!</math>
|-
| <math>J_n=\int x^n \cos{ax} \,\text{d}x \,\!</math> || <math>a^2J_n=ax^n \sin{ax} + nx^{n-1} \cos{ax} - n(n-1) J_{n-2} \,\!</math>
|-
| <math> I_n = \int \frac{\sin{ax}}{x^n} \,\text{d}x\,\!</math>

<math>J_n = \int \frac{\cos{ax}}{x^n} \,\text{d}x \,\!</math>
|| <math>I_n = -\frac{\sin{ax}}{(n-1)x^{n-1}}+\frac{a}{n-1}J_{n-1}\,\!</math>

<math>J_n = -\frac{\cos{ax}}{(n-1)x^{n-1}}-\frac{a}{n-1}I_{n-1}\,\!</math>

the formulae can be combined to obtain separate equations in ''I<sub>n</sub>'':

<math>J_{n-1} = -\frac{\cos{ax}}{(n-2)x^{n-2}}-\frac{a}{n-2}I_{n-2}\,\!</math>

<math>I_n = -\frac{\sin{ax}}{(n-1)x^{n-1}}-\frac{a}{n-1}\left [\frac{\cos{ax}}{(n-2)x^{n-2}}+\frac{a}{n-2}I_{n-2}\right ] \,\!</math>

<math> \therefore I_n = -\frac{\sin{ax}}{(n-1)x^{n-1}}-\frac{a}{(n-1)(n-2)}\left (\frac{\cos{ax}}{x^{n-2}}+aI_{n-2}\right ) \,\!</math>

and ''J<sub>n</sub>'':

<math>I_{n-1} = -\frac{\sin{ax}}{(n-2)x^{n-2}}+\frac{a}{n-2}J_{n-2}\,\!</math>

<math>J_n = -\frac{\cos{ax}}{(n-1)x^{n-1}}-\frac{a}{n-1}\left [-\frac{\sin{ax}}{(n-2)x^{n-2}}+\frac{a}{n-2}J_{n-2}  \right ]\,\!</math>

<math> \therefore J_n = -\frac{\cos{ax}}{(n-1)x^{n-1}}-\frac{a}{(n-1)(n-2)}\left (-\frac{\sin{ax}}{x^{n-2}}+aJ_{n-2}  \right )\,\!</math>
|-
| <math>I_n = \int \sin^n{ax} \,\text{d}x\,\!</math> || <math>anI_n = -\sin^{n-1}{ax}\cos{ax}+a(n-1)I_{n-2}\,\!</math>
|-
| <math>J_n = \int \cos^n{ax} \,\text{d}x\,\!</math> || <math>anJ_n = \sin{ax}\cos^{n-1}{ax}+a(n-1)J_{n-2}\,\!</math>
|-
| <math>I_n = \int \frac{\text{d}x}{\sin^n{ax}}\,\!</math> || <math>(n-1)I_n = - \frac{\cos{ax}}{a\sin^{n-1}{ax}}+ (n-2)I_{n-2}\,\!</math>
|-
| <math>J_n = \int \frac{\text{d}x}{\cos^n{ax}}\,\!</math> || <math>(n-1)J_n = \frac{\sin{ax}}{a\cos^{n-1}{ax}}+ (n-2)J_{n-2}\,\!</math>
|-
|}

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_{m,n} = \int \sin^m{ax}\cos^n{ax}\,\text{d}x\,\!</math> || <math>I_{m,n} = \begin{cases}
    -\frac{\sin^{m-1}{ax}\cos^{n+1}{ax}}{a(m+n)}+\frac{m-1}{m+n}I_{m-2,n} \\
    \frac{\sin^{m+1}{ax}\cos^{n-1}{ax}}{a(m+n)}+\frac{n-1}{m+n}I_{m,n-2} \\
\end{cases}\,\!</math>
|-
| <math>I_{m,n} = \int \frac{\text{d}x}{\sin^m{ax}\cos^n{ax}}\,\!</math> || <math>I_{m,n} = \begin{cases}
    \frac{1}{a(n-1)\sin^{m-1}{ax}\cos^{n-1}{ax}}+\frac{m+n-2}{n-1}I_{m,n-2} \\
    -\frac{1}{a(m-1)\sin^{m-1}{ax}\cos^{n-1}{ax}}+\frac{m+n-2}{m-1}I_{m-2,n} \\
\end{cases}\,\!</math>
|-
| <math>I_{m,n} = \int \frac{\sin^m{ax}}{\cos^n{ax}}\,\text{d}x\,\!</math> || <math>I_{m,n} = \begin{cases}
    \frac{\sin^{m-1}{ax}}{a(n-1)\cos^{n-1}{ax}}-\frac{m-1}{n-1}I_{m-2,n-2} \\
    \frac{\sin^{m+1}{ax}}{a(n-1)\cos^{n-1}{ax}}-\frac{m-n+2}{n-1}I_{m,n-2} \\
    -\frac{\sin^{m-1}{ax}}{a(m-n)\cos^{n-1}{ax}}+\frac{m-1}{m-n}I_{m-2,n} \\
\end{cases}\,\!</math>
|-
| <math>I_{m,n} =  \int \frac{\cos^m{ax}}{\sin^n{ax}}\,\text{d}x\,\!</math> || <math>I_{m,n} = \begin{cases}
    -\frac{\cos^{m-1}{ax}}{a(n-1)\sin^{n-1}{ax}}-\frac{m-1}{n-1}I_{m-2,n-2} \\
    -\frac{\cos^{m+1}{ax}}{a(n-1)\sin^{n-1}{ax}}-\frac{m-n+2}{n-1}I_{m,n-2} \\
    \frac{\cos^{m-1}{ax}}{a(m-n)\sin^{n-1}{ax}}+\frac{m-1}{m-n}I_{m-2,n} \\
\end{cases}\,\!</math>
|-
|}

{| class="wikitable"
|-
! Integral!! Reduction formula
|-
| <math>I_{n} = \int x^n e^{ax}\,\text{d}x\,\!</math>

<math>n > 0\,\!</math>
| <math> I_{n} = \frac{x^n e^{ax}}{a} - \frac{n}{a}I_{n-1} \,\!</math>
|-
| <math>I_{n} = \int x^{-n} e^{ax} \,\text{d}x\,\!</math>

<math>n > 0\,\!</math>

<math>n \neq 1\,\!</math>
| <math> I_{n} = \frac{- e^{ax}}{(n-1)x^{n-1}} + \frac{a}{n-1}I_{n-1} \,\!</math>
|-
| <math>I_{n} = \int e^{ax} \sin^n{bx} \,\text{d}x\,\!</math>
| <math> I_{n} = \frac{e^{ax} \sin^{n-1}{bx}}{a^2+(bn)^2}\left ( a\sin bx - bn\cos bx \right ) + \frac{n(n-1)b^2}{a^2+(bn)^2}I_{n-2} \,\!</math>
|-
| <math>I_{n} = \int e^{ax} \cos^n{bx} \,\text{d}x\,\!</math>
| <math> I_{n} = \frac{e^{ax} \cos^{n-1}{bx}}{a^2+(bn)^2}\left ( a\cos bx + bn\sin bx \right ) + \frac{n(n-1)b^2}{a^2+(bn)^2}I_{n-2} \,\!</math>
|-
|}

==References==

{{Reflist}}

==Bibliography==
{{wikibooks|Calculus/Integration techniques/Reduction Formula}}
*Anton, Bivens, Davis, Calculus, 7th edition.


[[Category:Integral calculus]]