Integrasyon sa pagpapaliit

Ang Integrasyon sa pagpapaliit(Integration by reduction formula) ay isang paraan para mahanap ang integral ng isang punsiyon. Ang paraang ito ang isa sa pinakaunang paraan ng integrasyon sa buong mundo.

Ang pagpapaliit na pormula ay maaaring matamo gamit ang anumang karaniwang paraan ng integrasyon gaya ng Integrasyon sa substitusyon, Integrasyon ng mga bahagi, Trigonometrikong substitusyon, Integrasyon gamit ang parsiyal na praksiyon at iba pa. Ang ideya ay ihayag ang integral na kinasasangkutan ng isang kapangyarihan ng isang punsiyon na kumakatawan sa I_n sa mga termino ng integral na kinasasangkutan ng mas mababang kapangyarihan ng punsiyong ito, halimbawa ang example I_n-2. Sa paraaang ito, ang pagpapaliit na pormula ay nagiging ugnayang rekursibo. Sa ibang salita, ang pagpapaliit na pormula ay inihahayag ang ang integral na ${\displaystyle I_{n}=\int f(x,n)\,dx}$ sa mga termino ng ${\displaystyle I_{k}=\int f(x,k)\,dx}$, kung saan ang ${\displaystyle k<n}$. Upang kwentahin ang integral, ang n ay papalitan ng halaga nito at gagamitin ang pagpapaliit na pormula ng paulit ulit hanggang maabot ang punto kung saan ang punsiyon na iintegraduhin(integrate) ay maaaring kwentahin na karawniwan kung ito ay kapangyarihan ng 0 o 1. Tapos ihahalili ang resulta ng paurong hanggang sa makwenta ang I_n.

Magtakda ng pagpapaliit na pormula na maaaring gamitin upang mahanap ang ${\displaystyle \int \cos ^{n}(x)\,dx\!}$. Samakatuwid, hanapin ang ${\displaystyle \int \cos ^{5}(x)\,dx\!}$.

${\displaystyle I_{n}\,=\int \cos ^{n}(x)\,dx\!}$

${\displaystyle =\int \cos ^{n-1}(x)\cos(x)\,dx\!}$

${\displaystyle =\int \cos ^{n-1}(x)\,d(\sin(x))\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)\ -\int \sin(x)\,d(\cos ^{n-1}(x))\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \sin(x)\cos ^{n-2}(x)\sin(x)\,dx\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)\sin ^{2}(x)\,dx\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)(1-\cos ^{2}(x))\,dx\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)\,dx\ -(n-1)\int \cos ^{n}(x)\,dx\!}$

${\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)I_{n-2}\ -(n-1)I_{n}\,}$

${\displaystyle I_{n}\ +(n-1)I_{n}\ =\cos ^{n-1}(x)\sin(x)\ +\ (n-1)I_{n-2}\,}$

${\displaystyle nI_{n}\ =\cos ^{n-1}(x)\sin(x)\ +(n-1)I_{n-2}\,}$

${\displaystyle I_{n}\ ={\frac {1}{n}}\cos ^{n-1}(x)\sin(x)\ +{\frac {n-1}{n}}I_{n-2}\,}$

Ang pagpapaliit na pormula:

${\displaystyle \int \cos ^{n}(x)\,dx\ ={\frac {1}{n}}\cos ^{n-1}(x)\sin(x)\ +{\frac {n-1}{n}}\int \cos ^{n-2}(x)\,dx\!}$

Upang mahanap ang ${\displaystyle \int \cos ^{5}(x)\,dx\!}$:

${\displaystyle n=5\,}$: ${\displaystyle I_{5}\ ={\tfrac {1}{5}}\cos ^{4}(x)\sin(x)\ +{\tfrac {4}{5}}I_{3}\,}$

${\displaystyle n=3\,}$: ${\displaystyle I_{3}\ ={\tfrac {1}{3}}\cos ^{2}(x)\sin(x)\ +{\tfrac {2}{3}}I_{1}\,}$

${\displaystyle \because I_{1}\ =\int \cos(x)\,dx\ =\sin(x)\ +C_{1}\,}$

${\displaystyle \therefore I_{3}\ ={\tfrac {1}{3}}\cos ^{2}(x)\sin(x)\ +{\tfrac {2}{3}}\sin(x)\ +C_{2}\,}$，${\displaystyle C_{2}\ ={\tfrac {2}{3}}C_{1}\,}$

${\displaystyle I_{5}\ ={\frac {1}{5}}\cos ^{4}(x)\sin(x)\ +{\frac {4}{5}}\left[{\frac {1}{3}}\cos ^{2}(x)\sin(x)+{\frac {2}{3}}\sin(x)\right]+C\,}$, where C is a constant.

Ang mga sumusunod na integral ^[1] ay naglalaman ng:

• Mga paktor ng linyar ng radikal na ${\displaystyle {\sqrt {ax+b}}\,\!}$
• Mga linyar na paktor ${\displaystyle {px+q}\,\!}$ at ang linyar na radikal na${\displaystyle {\sqrt {ax+b}}\,\!}$
• Mga kwadratikong paktor na ${\displaystyle x^{2}+a^{2}\,\!}$
• Mga kwadratikong paktor na ${\displaystyle x^{2}-a^{2}\,\!}$, para sa ${\displaystyle x>a\,\!}$
• Mga kwadratikong paktor ${\displaystyle a^{2}-x^{2}\,\!}$, para sa ${\displaystyle x<a\,\!}$
• Hindi mapapaliit na mga kwadratikong paktor(Irreducible quadratic factors) na ${\displaystyle ax^{2}+bx+c\,\!}$
• Mga radikal ng hindi mapapaliit na mga kwadratikong paktor na ${\displaystyle {\sqrt {ax^{2}+bx+c}}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int {\frac {x^{n}}{\sqrt {ax+b}}}\mathrm {d} x\,\!}$	${\displaystyle I_{n}={\frac {2x^{n}{\sqrt {ax+b}}}{a(2n+1)}}-{\frac {2nb}{a(2n+1)}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{x^{n}{\sqrt {ax+b}}}}\,\!}$	${\displaystyle I_{n}=-{\frac {\sqrt {ax+b}}{(n-1)bx^{n-1}}}-{\frac {a(2n-3)}{2b(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int x^{n}{\sqrt {ax+b}}\mathrm {d} x\,\!}$	${\displaystyle I_{n}={\frac {2x^{n}{\sqrt {(ax+b)^{3}}}}{a(2n+3)}}-{\frac {2nb}{a(2n+3)}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int {\frac {\sqrt {ax+b}}{x^{n}}}\mathrm {d} x\,\!}$	${\displaystyle I_{n}=-{\frac {\sqrt {ax+b}}{(n-1)x^{n-1}}}+{\frac {a}{2(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {\mathrm {d} x}{(ax+b)^{n}(px+q)^{m}}}\,\!}$	${\displaystyle I_{n,m}=-{\frac {1}{(n-1)(bp-aq)}}\left[{\frac {1}{(ax+b)^{m-1}(px+q)^{n-1}}}+a(n+m-2)I_{m,n-1}\right]\,\!}$
${\displaystyle I_{n,m}=\int {\frac {(ax+b)^{m}}{(px+q)^{n}}}\mathrm {d} x\,\!}$	${\displaystyle I_{n,m}={\begin{cases}-{\frac {1}{(n-1)(bp-aq)}}\left[{\frac {(ax+b)^{m+1}}{(px+q)^{n-1}}}+a(n+m-2)I_{m-1,n-1}\right]\\-{\frac {1}{(n-m-1)p}}\left[{\frac {(ax+b)^{m}}{(px+q)^{n-1}}}+m(bp-aq)I_{m-1,n}\right]\\-{\frac {1}{(n-1)p}}\left[{\frac {(ax+b)^{m}}{(px+q)^{n-1}}}-amI_{m-1,n-1}\right]\end{cases}}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int {\frac {(px+q)^{n}}{\sqrt {ax+b}}}\mathrm {d} x\,\!}$	${\displaystyle \int (px+q)^{n}{\sqrt {ax+b}}\mathrm {d} x={\frac {2(px+q)^{n+1}{\sqrt {ax+b}}}{p(2n+3)}}+{\frac {bp-aq}{p(2n+3)}}I_{n}\,\!}$ ${\displaystyle I_{n}={\frac {2(px+q)^{n}{\sqrt {ax+b}}}{a(2n+1)}}+{\frac {2n(aq-bp)}{a(2n+1)}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{(px+q)^{n}{\sqrt {ax+b}}}}\,\!}$	${\displaystyle \int {\frac {\sqrt {ax+b}}{(px+q)^{n}}}\mathrm {d} x=-{\frac {\sqrt {ax+b}}{p(n-1)(px+q)^{n-1}}}+{\frac {a}{2p(n-1)}}I_{n}\,\!}$ ${\displaystyle I_{n}=-{\frac {\sqrt {ax+b}}{(n-1)(aq-bp)(px+q)^{n-1}}}+{\frac {a(2n-3)}{2(n-1)(aq-bp)}}I_{n-1}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{(x^{2}+a^{2})^{n}}}\,\!}$	${\displaystyle I_{n}={\frac {x}{2a^{2}(n-1)(x^{2}+a^{2})^{n-1}}}+{\frac {2n-3}{2a^{2}(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {\mathrm {d} x}{x^{m}(x^{2}+a^{2})^{n}}}\,\!}$	${\displaystyle a^{2}I_{n,m}=I_{m,n-1}-I_{m-2,n}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {x^{m}}{(x^{2}+a^{2})^{n}}}\mathrm {d} x\,\!}$	${\displaystyle I_{n,m}=I_{m-2,n-1}-a^{2}I_{m-2,n}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{(x^{2}-a^{2})^{n}}}\,\!}$	${\displaystyle I_{n}=-{\frac {x}{2a^{2}(n-1)(x^{2}-a^{2})^{n-1}}}-{\frac {2n-3}{2a^{2}(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {\mathrm {d} x}{x^{m}(x^{2}-a^{2})^{n}}}\,\!}$	${\displaystyle {a^{2}}I_{n,m}=I_{m-2,n}-I_{m,n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {x^{m}}{(x^{2}-a^{2})^{n}}}\mathrm {d} x\,\!}$	${\displaystyle I_{n,m}=I_{m-2,n-1}+a^{2}I_{m-2,n}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{(a^{2}-x^{2})^{n}}}\,\!}$	${\displaystyle I_{n}={\frac {x}{2a^{2}(n-1)(a^{2}-x^{2})^{n-1}}}+{\frac {2n-3}{2a^{2}(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {\mathrm {d} x}{x^{m}(a^{2}-x^{2})^{n}}}\,\!}$	${\displaystyle {a^{2}}I_{n,m}=I_{m,n-1}+I_{m-2,n}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {x^{m}}{(a^{2}-x^{2})^{n}}}\mathrm {d} x\,\!}$	${\displaystyle I_{n,m}=a^{2}I_{m-2,n}-I_{m-2,n-1}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{{x^{n}}(ax^{2}+bx+c)}}\,\!}$	${\displaystyle -cI_{n}={\frac {1}{x^{n-1}(n-1)}}+bI_{n-1}+aI_{n-2}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {x^{m}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}\,\!}$	${\displaystyle I_{m,n}=-{\frac {x^{m-1}}{a(2n-m-1)(ax^{2}+bx+c)^{n-1}}}+{\frac {c(m-1)}{a(2n-m-1)}}I_{m-2,n}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {\mathrm {d} x}{x^{m}(ax^{2}+bx+c)^{n}}}\,\!}$	${\displaystyle -c(m-1)I_{m,n}={\frac {1}{x^{m-1}(ax^{2}+bx+c)^{n-1}}}+{a(m+2n-3)}I_{m-2,n}+{b(m+n-2)}I_{m-1,n}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int (ax^{2}+bx+c)^{n}\mathrm {d} x\,\!}$	${\displaystyle 4a(n+1)I_{n+{\frac {1}{2}}}=2(2ax+b)(ax^{2}+bx+c)^{n+{\frac {1}{2}}}+(2n+1)(4ac-b^{2})I_{n-{\frac {1}{2}}}\,\!}$
${\displaystyle I_{n}=\int {\frac {1}{(ax^{2}+bx+c)^{n}}}\mathrm {d} x\,\!}$	${\displaystyle (2n-1)(4ac-b^{2})I_{n+{\frac {1}{2}}}={\frac {2(2ax+b)}{(ax^{2}+bx+c)^{n-{\frac {1}{2}}}}}+{8a(n-1)}I_{n-{\frac {1}{2}}}\,\!}$

tandaan na sa batas ng mga indises(law of indices) ng eksponente:

${\displaystyle I_{n+{\frac {1}{2}}}=I_{\frac {2n+1}{2}}=\int {\frac {1}{(ax^{2}+bx+c)^{\frac {2n+1}{2}}}}\mathrm {d} x=\int {\frac {1}{\sqrt {(ax^{2}+bx+c)^{2n+1}}}}\mathrm {d} x\,\!}$

Ang mga sumusunod na integral ^[2] ay naglalaman ng:

• Mga paktor ng sine
• Mga paktor ng cosine
• Mga paktor ng produkto at kosiyente ng sine at cosine products
• Produkto/kosiyente ng eksponesiyal na mga paktor at mga kapangyarihan ng x
• Mga produkto ng ekponensiyal at sine/cosine na mga paktor

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int x^{n}\sin {ax}\mathrm {d} x\,\!}$	${\displaystyle a^{2}I_{n}=-ax^{n}\cos {ax}+nx^{n-1}\sin {ax}-n(n-1)I_{n-2}\,\!}$
${\displaystyle J_{n}=\int x^{n}\cos {ax}\mathrm {d} x\,\!}$	${\displaystyle a^{2}J_{n}=ax^{n}\sin {ax}+nx^{n-1}\cos {ax}-n(n-1)J_{n-2}\,\!}$
${\displaystyle I_{n}=\int {\frac {\sin {ax}}{x^{n}}}\mathrm {d} x\,\!}$ ${\displaystyle J_{n}=\int {\frac {\cos {ax}}{x^{n}}}\mathrm {d} x\,\!}$	${\displaystyle I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}J_{n-1}\,\!}$ ${\displaystyle J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}I_{n-1}\,\!}$ Ang mga pormula ay maaaring pagsamahin upang makamit ang magkahiwalay na ekwasyon sa In: ${\displaystyle I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}J_{n-1}\,\!}$ ${\displaystyle J_{n-1}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}I_{n-2}\,\!}$ ${\displaystyle I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\left[{\frac {\cos {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}I_{n-2}\right]\,\!}$ ${\displaystyle \therefore I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}-{\frac {a}{(n-1)^{2}}}\left({\frac {\cos {ax}}{x^{n-1}}}+aI_{n-2}\right)\,\!}$ and Jn: ${\displaystyle I_{n-1}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}J_{n-2}\,\!}$ ${\displaystyle J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}I_{n-1}\,\!}$ ${\displaystyle J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\left[-{\frac {\sin {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}J_{n-2}\right]\,\!}$ ${\displaystyle \therefore J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{(n-1)^{2}}}\left(-{\frac {\sin {ax}}{x^{n-1}}}+aJ_{n-2}\right)\,\!}$
${\displaystyle I_{n}=\int \sin ^{n}{ax}\mathrm {d} x\,\!}$	${\displaystyle anI_{n}=-\sin ^{n-1}{ax}\cos {ax}+a(n-1)I_{n-2}\,\!}$
${\displaystyle J_{n}=\int \cos ^{n}{ax}\mathrm {d} x\,\!}$	${\displaystyle anJ_{n}=\sin {ax}\cos ^{n-1}{ax}+a(n-1)J_{n-2}\,\!}$
${\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{\sin ^{n}{ax}}}\,\!}$	${\displaystyle (n-1)I_{n}=-{\frac {\cos {ax}}{a\sin ^{n-1}{ax}}}+(n-2)I_{n-2}\,\!}$
${\displaystyle J_{n}=\int {\frac {\mathrm {d} x}{\cos ^{n}{ax}}}\,\!}$	${\displaystyle (n-1)J_{n}={\frac {\sin {ax}}{a\cos ^{n-1}{ax}}}+(n-2)I_{n-2}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{m,n}=\int \sin ^{m}{ax}\cos ^{n}{ax}\mathrm {d} x\,\!}$	${\displaystyle I_{m,n}={\begin{cases}-{\frac {\sin ^{m-1}{ax}\cos ^{n+1}{ax}}{a(m+n)}}+{\frac {m-1}{m+n}}I_{m-2,n}\\{\frac {\sin ^{m+1}{ax}\cos ^{n-1}{ax}}{a(m+n)}}+{\frac {n-1}{m+n}}I_{m,n-2}\\\end{cases}}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {\mathrm {d} x}{\sin ^{m}{ax}\cos ^{n}{ax}}}\,\!}$	${\displaystyle I_{m,n}={\begin{cases}{\frac {1}{a(n-1)\sin ^{m-1}{ax}\cos ^{n-1}{ax}}}+{\frac {m+n-2}{n-1}}I_{m,n-2}\\{\frac {1}{a(m-1)\sin ^{m-1}{ax}\cos ^{n-1}{ax}}}+{\frac {m+n-2}{m-1}}I_{m-2,n}\\\end{cases}}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {\sin ^{m}{ax}}{\cos ^{n}{ax}}}\mathrm {d} x\,\!}$	${\displaystyle I_{m,n}={\begin{cases}{\frac {\sin ^{m-1}{ax}}{a(n-1)\cos ^{n-1}{ax}}}-{\frac {m-1}{n-1}}I_{m-2,n-2}\\{\frac {\sin ^{m+1}{ax}}{a(n-1)\cos ^{n-1}{ax}}}-{\frac {m-n+2}{n-1}}I_{m,n-2}\\{\frac {\sin ^{m-1}{ax}}{a(m-n)\cos ^{n-1}{ax}}}+{\frac {m-1}{m-n}}I_{m-2,n}\\\end{cases}}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {\cos ^{m}{ax}}{\sin ^{n}{ax}}}\mathrm {d} x\,\!}$	${\displaystyle I_{m,n}={\begin{cases}-{\frac {\cos ^{m-1}{ax}}{a(n-1)\sin ^{n-1}{ax}}}-{\frac {m-1}{n-1}}I_{m-2,n-2}\\-{\frac {\cos ^{m+1}{ax}}{a(n-1)\sin ^{n-1}{ax}}}-{\frac {m-n+2}{n-1}}I_{m,n-2}\\{\frac {\cos ^{m-1}{ax}}{a(m-n)\sin ^{n-1}{ax}}}+{\frac {m-1}{m-n}}I_{m-2,n}\\\end{cases}}\,\!}$

Integral	Pagpapaliit na pormula
${\displaystyle I_{n}=\int x^{n}e^{ax}\mathrm {d} x\,\!}$ ${\displaystyle n>0\,\!}$	${\displaystyle I_{n}=-{\frac {x^{n}e^{ax}}{a}}-{\frac {n}{a}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int x^{-n}e^{ax}\mathrm {d} x\,\!}$ ${\displaystyle n>0\,\!}$ ${\displaystyle n\neq 1\,\!}$	${\displaystyle I_{n}=-{\frac {-e^{ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int x^{ax}\sin ^{n}{bx}\mathrm {d} x\,\!}$	${\displaystyle I_{n}=-{\frac {x^{ax}\sin ^{n-1}{bx}}{a^{2}+(bn)^{2}}}\left(a\sin bx-bn\cos bx\right)+{\frac {n(n-1)b^{2}}{a^{2}+(bn)^{2}}}I_{n-2}\,\!}$
${\displaystyle I_{n}=\int x^{ax}\cos ^{n}{bx}\mathrm {d} x\,\!}$	${\displaystyle I_{n}=-{\frac {x^{ax}\cos ^{n-1}{bx}}{a^{2}+(bn)^{2}}}\left(a\cos bx+bn\sin bx\right)+{\frac {n(n-1)b^{2}}{a^{2}+(bn)^{2}}}I_{n-2}\,\!}$

Sa sumusunod, ang mga indises(indices) ay mga positibong mga intedyer na ${\displaystyle n,p,q\in \mathbf {N} \,\!}$.

Kung papansin, maraming mga pagpapaliit na pormula ng isang indeks na I_n (o hindi naglalaman ng dalawang indises kung saan ang I_m,n) ay maaaring isulat sa pangkalahatang anyo: ${\displaystyle I_{n}=A+BI_{n-p}\,\!}$

kung saan ang ${\displaystyle I_{n-p}\,\!}$ ay kumakatawan sa paglipat(shift) ng mga integral na p ng paurong relatibo sa ${\displaystyle I_{n}\,\!}$, at ang A at B ang mga tinipong mga dagdag na mga termino(hindi mga konstante), ang pangkalahatang pormula para sa mga integral na ito ay matatagpuan gamit ang paulit ulit na paghalali ng pagurong(backward subtitution): ${\displaystyle I_{n}=A+BI_{n-p}\,\!}$

${\displaystyle I_{n-p}=A+BI_{n-2p}\,\!}$

${\displaystyle I_{n-2p}=A+BI_{n-3p}\,\!}$

etc., so the general recurrence shift is:

${\displaystyle I_{n-qp}=A+BI_{n-(q-1)p}\,\!}$

kung saan ang q ay isang multiple ng p. Isagawa ang paurong na paghalili sa indeks na n - pq, at maingat na palawgin ang mga braket:

${\displaystyle I_{n}=A+B\left(A+B\left(A+B\left(\cdots A+qI_{n-pq}\cdots \right)\right)\right)\,\!}$

${\displaystyle I_{n}=A+\left(AB+B^{2}\left(A+B\left(\cdots A+qI_{n-pq}\cdots \right)\right)\right)\,\!}$

${\displaystyle I_{n}=A+\left(AB+\left(AB^{2}+B^{3}\left(\cdots A+qI_{n-pq}\cdots \right)\right)\right)\,\!}$

${\displaystyle I_{n}=A+\left(AB+\left(AB^{2}+\left(\cdots AB^{q-1}+B^{q}I_{n-pq}\cdots \right)\right)\right)\,\!}$

${\displaystyle I_{n}=A+AB+AB^{2}+\cdots +AB^{q-1}+B^{q}I_{n-pq}\,\!}$

na nagreresulta ng:

${\displaystyle I_{n}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{n-pq}\,\!}$

Ito ay maaaring gamitin upang makamit ang integral na kasing layo pabalik sa mga indises na pq relatibo sa n, ngunit makakatulong kung babaguhin ang mga indises sa q, depende sa piniling p (hindi mahalaga kung ang n o q ang indeks). Ang pagbabago ay isang simpleng paghalili:

Kung ang I₀ ay alam, maaaring itakda ang ${\displaystyle I_{n-pq}=I_{0}\,\!}$, tapos ang ${\displaystyle n-pq=0\,\!}$ upang ang ${\displaystyle n=pq\,\!}$

${\displaystyle I_{pq}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}$

Kung ang I₀ ay alam, maaaring itakda ang ${\displaystyle I_{n-pq}=I_{1}\,\!}$, tapos ${\displaystyle n-pq=1\,\!}$ upang ang ${\displaystyle n=pq+1\,\!}$

${\displaystyle I_{pq+1}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{1}\,\!}$

Malimit, gaya ng makikita sa mga tabla sa taas ang p = 1 o p = 2, upang ang pagpapaliit na pormula ay maaring itakda gaya ng:

	For ${\displaystyle p=1\,\!}$	For ${\displaystyle p=2\,\!}$
${\displaystyle n=pq\,\!}$	${\displaystyle I_{q}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}$	${\displaystyle I_{2q}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}$
${\displaystyle n=pq+1\,\!}$	${\displaystyle I_{q+1}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{1}\,\!}$ (Isa lamang paglipat ng indeks).	${\displaystyle I_{2q+1}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{1}\,\!}$

Ang isang indaktibong pagpapatunay ay magpapakita na ang pangkahalatang pormula na ${\displaystyle I_{n}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{n-pq}\,\!}$ upang maging totoo ang ${\displaystyle \forall n\in \mathbf {N} \,\!}$. Ang pagpapatunay ay maaaring hatiin sa dalawa sa katulad na paraan sa itaas:

Kasong 1: one for ${\displaystyle I_{n-pq}=I_{0}\,\!}$ so ${\displaystyle I_{pq}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}$

Kasong 2: one for ${\displaystyle I_{n-pq}=I_{1}\,\!}$ so ${\displaystyle I_{pq+1}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{1}\,\!}$

Sa kasong 2, ang paglipat ng indeks ay +1, upang ang pagpapatunay maging katulad sa kasong 1. Ang kasong 1 lamang ang pinatunayan sa ibaba:

Kasong 1':

Maaaring mahanap na ang:

${\displaystyle I_{pq}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}$

ay totoo para sa q = 1;

${\displaystyle I_{p}=A\sum _{i=0}^{0}B^{i}+BI_{0}\,\!}$

${\displaystyle I_{p}=AB^{0}+BI_{0}\,\!}$

${\displaystyle I_{p}=A+BI_{0}\,\!}$

Ngayon ipagpapalagay na:

${\displaystyle I_{pq}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}$

ay totoo para sa isang q > 1, tapos ipakita ang pormula ay totoo para sa q + 1:

${\displaystyle I_{p(q+1)}=A\sum _{i=0}^{q}B^{i}+B^{q+1}I_{0}\,\!}$

${\displaystyle I_{p(q+1)}=A\left(\sum _{i=0}^{q-1}B^{i}+B^{q}\right)+B^{q}BI_{0}\,\!}$

${\displaystyle I_{p(q+1)}=A\sum _{i=0}^{q-1}B^{i}+AB^{q}+B^{q}BI_{0}\,\!}$

${\displaystyle I_{p(q+1)}=A\sum _{i=0}^{q-1}B^{i}+B^{q}\left(A+BI_{0}\right)\,\!}$

${\displaystyle I_{p(q+1)}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{p}\,\!}$

${\displaystyle I_{pq+p}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{p}\,\!}$

na umaayon sa paglipat ng indeks mula I₀ patungo sa I_p at I_pq patungo I_pq+p sa +p. Kaya ang pormula ay tototoo ${\displaystyle \forall q\geqslant 1\,\!}$.