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 In sa mga termino ng integral na kinasasangkutan ng mas mababang kapangyarihan ng punsiyong ito, halimbawa ang example In-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
I
n
=
∫
f
(
x
,
n
)
d
x
{\displaystyle I_{n}=\int f(x,n)\,dx}
I
k
=
∫
f
(
x
,
k
)
d
x
{\displaystyle I_{k}=\int f(x,k)\,dx}
k
<
n
{\displaystyle k<n}
n .
∫
cos
n
(
x
)
d
x
{\displaystyle \int \cos ^{n}(x)\,dx\!}
Magtakda ng pagpapaliit na pormula na maaaring gamitin upang mahanap ang
∫
cos
n
(
x
)
d
x
{\displaystyle \int \cos ^{n}(x)\,dx\!}
∫
cos
5
(
x
)
d
x
{\displaystyle \int \cos ^{5}(x)\,dx\!}
I
n
=
∫
cos
n
(
x
)
d
x
{\displaystyle I_{n}\,=\int \cos ^{n}(x)\,dx\!}
=
∫
cos
n
−
1
(
x
)
cos
(
x
)
d
x
{\displaystyle =\int \cos ^{n-1}(x)\cos(x)\,dx\!}
=
∫
cos
n
−
1
(
x
)
d
(
sin
(
x
)
)
{\displaystyle =\int \cos ^{n-1}(x)\,d(\sin(x))\!}
=
cos
n
−
1
(
x
)
sin
(
x
)
−
∫
sin
(
x
)
d
(
cos
n
−
1
(
x
)
)
{\displaystyle =\cos ^{n-1}(x)\sin(x)\ -\int \sin(x)\,d(\cos ^{n-1}(x))\!}
=
cos
n
−
1
(
x
)
sin
(
x
)
+
(
n
−
1
)
∫
sin
(
x
)
cos
n
−
2
(
x
)
sin
(
x
)
d
x
{\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \sin(x)\cos ^{n-2}(x)\sin(x)\,dx\!}
=
cos
n
−
1
(
x
)
sin
(
x
)
+
(
n
−
1
)
∫
cos
n
−
2
(
x
)
sin
2
(
x
)
d
x
{\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)\sin ^{2}(x)\,dx\!}
=
cos
n
−
1
(
x
)
sin
(
x
)
+
(
n
−
1
)
∫
cos
n
−
2
(
x
)
(
1
−
cos
2
(
x
)
)
d
x
{\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)(1-\cos ^{2}(x))\,dx\!}
=
cos
n
−
1
(
x
)
sin
(
x
)
+
(
n
−
1
)
∫
cos
n
−
2
(
x
)
d
x
−
(
n
−
1
)
∫
cos
n
(
x
)
d
x
{\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)\,dx\ -(n-1)\int \cos ^{n}(x)\,dx\!}
=
cos
n
−
1
(
x
)
sin
(
x
)
+
(
n
−
1
)
I
n
−
2
−
(
n
−
1
)
I
n
{\displaystyle =\cos ^{n-1}(x)\sin(x)\ +(n-1)I_{n-2}\ -(n-1)I_{n}\,}
I
n
+
(
n
−
1
)
I
n
=
cos
n
−
1
(
x
)
sin
(
x
)
+
(
n
−
1
)
I
n
−
2
{\displaystyle I_{n}\ +(n-1)I_{n}\ =\cos ^{n-1}(x)\sin(x)\ +\ (n-1)I_{n-2}\,}
n
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}\,}
I
n
=
1
n
cos
n
−
1
(
x
)
sin
(
x
)
+
n
−
1
n
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:
∫
cos
n
(
x
)
d
x
=
1
n
cos
n
−
1
(
x
)
sin
(
x
)
+
n
−
1
n
∫
cos
n
−
2
(
x
)
d
x
{\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
∫
cos
5
(
x
)
d
x
{\displaystyle \int \cos ^{5}(x)\,dx\!}
n
=
5
{\displaystyle n=5\,}
I
5
=
1
5
cos
4
(
x
)
sin
(
x
)
+
4
5
I
3
{\displaystyle I_{5}\ ={\tfrac {1}{5}}\cos ^{4}(x)\sin(x)\ +{\tfrac {4}{5}}I_{3}\,}
n
=
3
{\displaystyle n=3\,}
I
3
=
1
3
cos
2
(
x
)
sin
(
x
)
+
2
3
I
1
{\displaystyle I_{3}\ ={\tfrac {1}{3}}\cos ^{2}(x)\sin(x)\ +{\tfrac {2}{3}}I_{1}\,}
∵
I
1
=
∫
cos
(
x
)
d
x
=
sin
(
x
)
+
C
1
{\displaystyle \because I_{1}\ =\int \cos(x)\,dx\ =\sin(x)\ +C_{1}\,}
∴
I
3
=
1
3
cos
2
(
x
)
sin
(
x
)
+
2
3
sin
(
x
)
+
C
2
{\displaystyle \therefore I_{3}\ ={\tfrac {1}{3}}\cos ^{2}(x)\sin(x)\ +{\tfrac {2}{3}}\sin(x)\ +C_{2}\,}
C
2
=
2
3
C
1
{\displaystyle C_{2}\ ={\tfrac {2}{3}}C_{1}\,}
I
5
=
1
5
cos
4
(
x
)
sin
(
x
)
+
4
5
[
1
3
cos
2
(
x
)
sin
(
x
)
+
2
3
sin
(
x
)
]
+
C
{\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\,}
Ang mga sumusunod na integral
[1]
Mga paktor ng linyar ng radikal na
a
x
+
b
{\displaystyle {\sqrt {ax+b}}\,\!}
Mga linyar na paktor
p
x
+
q
{\displaystyle {px+q}\,\!}
a
x
+
b
{\displaystyle {\sqrt {ax+b}}\,\!}
Mga kwadratikong paktor na
x
2
+
a
2
{\displaystyle x^{2}+a^{2}\,\!}
Mga kwadratikong paktor na
x
2
−
a
2
{\displaystyle x^{2}-a^{2}\,\!}
x
>
a
{\displaystyle x>a\,\!}
Mga kwadratikong paktor
a
2
−
x
2
{\displaystyle a^{2}-x^{2}\,\!}
x
<
a
{\displaystyle x<a\,\!}
Hindi mapapaliit na mga kwadratikong paktor(Irreducible quadratic factors) na
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c\,\!}
Mga radikal ng hindi mapapaliit na mga kwadratikong paktor na
a
x
2
+
b
x
+
c
{\displaystyle {\sqrt {ax^{2}+bx+c}}\,\!}
Integral
Pagpapaliit na pormula
I
n
=
∫
x
n
a
x
+
b
d
x
{\displaystyle I_{n}=\int {\frac {x^{n}}{\sqrt {ax+b}}}\mathrm {d} x\,\!}
I
n
=
2
x
n
a
x
+
b
a
(
2
n
+
1
)
−
2
n
b
a
(
2
n
+
1
)
I
n
−
1
{\displaystyle I_{n}={\frac {2x^{n}{\sqrt {ax+b}}}{a(2n+1)}}-{\frac {2nb}{a(2n+1)}}I_{n-1}\,\!}
I
n
=
∫
d
x
x
n
a
x
+
b
{\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{x^{n}{\sqrt {ax+b}}}}\,\!}
I
n
=
−
a
x
+
b
(
n
−
1
)
b
x
n
−
1
−
a
(
2
n
−
3
)
2
b
(
n
−
1
)
I
n
−
1
{\displaystyle I_{n}=-{\frac {\sqrt {ax+b}}{(n-1)bx^{n-1}}}-{\frac {a(2n-3)}{2b(n-1)}}I_{n-1}\,\!}
I
n
=
∫
x
n
a
x
+
b
d
x
{\displaystyle I_{n}=\int x^{n}{\sqrt {ax+b}}\mathrm {d} x\,\!}
I
n
=
2
x
n
(
a
x
+
b
)
3
a
(
2
n
+
3
)
−
2
n
b
a
(
2
n
+
3
)
I
n
−
1
{\displaystyle I_{n}={\frac {2x^{n}{\sqrt {(ax+b)^{3}}}}{a(2n+3)}}-{\frac {2nb}{a(2n+3)}}I_{n-1}\,\!}
I
n
=
∫
a
x
+
b
x
n
d
x
{\displaystyle I_{n}=\int {\frac {\sqrt {ax+b}}{x^{n}}}\mathrm {d} x\,\!}
I
n
=
−
a
x
+
b
(
n
−
1
)
x
n
−
1
+
a
2
(
n
−
1
)
I
n
−
1
{\displaystyle I_{n}=-{\frac {\sqrt {ax+b}}{(n-1)x^{n-1}}}+{\frac {a}{2(n-1)}}I_{n-1}\,\!}
I
n
,
m
=
∫
d
x
(
a
x
+
b
)
n
(
p
x
+
q
)
m
{\displaystyle I_{n,m}=\int {\frac {\mathrm {d} x}{(ax+b)^{n}(px+q)^{m}}}\,\!}
I
n
,
m
=
−
1
(
n
−
1
)
(
b
p
−
a
q
)
[
1
(
a
x
+
b
)
m
−
1
(
p
x
+
q
)
n
−
1
+
a
(
n
+
m
−
2
)
I
m
,
n
−
1
]
{\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]\,\!}
I
n
,
m
=
∫
(
a
x
+
b
)
m
(
p
x
+
q
)
n
d
x
{\displaystyle I_{n,m}=\int {\frac {(ax+b)^{m}}{(px+q)^{n}}}\mathrm {d} x\,\!}
I
n
,
m
=
{
−
1
(
n
−
1
)
(
b
p
−
a
q
)
[
(
a
x
+
b
)
m
+
1
(
p
x
+
q
)
n
−
1
+
a
(
n
+
m
−
2
)
I
m
−
1
,
n
−
1
]
−
1
(
n
−
m
−
1
)
p
[
(
a
x
+
b
)
m
(
p
x
+
q
)
n
−
1
+
m
(
b
p
−
a
q
)
I
m
−
1
,
n
]
−
1
(
n
−
1
)
p
[
(
a
x
+
b
)
m
(
p
x
+
q
)
n
−
1
−
a
m
I
m
−
1
,
n
−
1
]
{\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
I
n
=
∫
d
x
(
x
2
+
a
2
)
n
{\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{(x^{2}+a^{2})^{n}}}\,\!}
I
n
=
x
2
a
2
(
n
−
1
)
(
x
2
+
a
2
)
n
−
1
+
2
n
−
3
2
a
2
(
n
−
1
)
I
n
−
1
{\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}\,\!}
I
n
,
m
=
∫
d
x
x
m
(
x
2
+
a
2
)
n
{\displaystyle I_{n,m}=\int {\frac {\mathrm {d} x}{x^{m}(x^{2}+a^{2})^{n}}}\,\!}
a
2
I
n
,
m
=
I
m
,
n
−
1
−
I
m
−
2
,
n
{\displaystyle a^{2}I_{n,m}=I_{m,n-1}-I_{m-2,n}\,\!}
I
n
,
m
=
∫
x
m
(
x
2
+
a
2
)
n
d
x
{\displaystyle I_{n,m}=\int {\frac {x^{m}}{(x^{2}+a^{2})^{n}}}\mathrm {d} x\,\!}
I
n
,
m
=
I
m
−
2
,
n
−
1
−
a
2
I
m
−
2
,
n
{\displaystyle I_{n,m}=I_{m-2,n-1}-a^{2}I_{m-2,n}\,\!}
Integral
Pagpapaliit na pormula
I
n
=
∫
d
x
(
x
2
−
a
2
)
n
{\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{(x^{2}-a^{2})^{n}}}\,\!}
I
n
=
−
x
2
a
2
(
n
−
1
)
(
x
2
−
a
2
)
n
−
1
−
2
n
−
3
2
a
2
(
n
−
1
)
I
n
−
1
{\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}\,\!}
I
n
,
m
=
∫
d
x
x
m
(
x
2
−
a
2
)
n
{\displaystyle I_{n,m}=\int {\frac {\mathrm {d} x}{x^{m}(x^{2}-a^{2})^{n}}}\,\!}
a
2
I
n
,
m
=
I
m
−
2
,
n
−
I
m
,
n
−
1
{\displaystyle {a^{2}}I_{n,m}=I_{m-2,n}-I_{m,n-1}\,\!}
I
n
,
m
=
∫
x
m
(
x
2
−
a
2
)
n
d
x
{\displaystyle I_{n,m}=\int {\frac {x^{m}}{(x^{2}-a^{2})^{n}}}\mathrm {d} x\,\!}
I
n
,
m
=
I
m
−
2
,
n
−
1
+
a
2
I
m
−
2
,
n
{\displaystyle I_{n,m}=I_{m-2,n-1}+a^{2}I_{m-2,n}\,\!}
Integral
Pagpapaliit na pormula
I
n
=
∫
d
x
(
a
2
−
x
2
)
n
{\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{(a^{2}-x^{2})^{n}}}\,\!}
I
n
=
x
2
a
2
(
n
−
1
)
(
a
2
−
x
2
)
n
−
1
+
2
n
−
3
2
a
2
(
n
−
1
)
I
n
−
1
{\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}\,\!}
I
n
,
m
=
∫
d
x
x
m
(
a
2
−
x
2
)
n
{\displaystyle I_{n,m}=\int {\frac {\mathrm {d} x}{x^{m}(a^{2}-x^{2})^{n}}}\,\!}
a
2
I
n
,
m
=
I
m
,
n
−
1
+
I
m
−
2
,
n
{\displaystyle {a^{2}}I_{n,m}=I_{m,n-1}+I_{m-2,n}\,\!}
I
n
,
m
=
∫
x
m
(
a
2
−
x
2
)
n
d
x
{\displaystyle I_{n,m}=\int {\frac {x^{m}}{(a^{2}-x^{2})^{n}}}\mathrm {d} x\,\!}
I
n
,
m
=
a
2
I
m
−
2
,
n
−
I
m
−
2
,
n
−
1
{\displaystyle I_{n,m}=a^{2}I_{m-2,n}-I_{m-2,n-1}\,\!}
Integral
Pagpapaliit na pormula
I
n
=
∫
d
x
x
n
(
a
x
2
+
b
x
+
c
)
{\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{{x^{n}}(ax^{2}+bx+c)}}\,\!}
−
c
I
n
=
1
x
n
−
1
(
n
−
1
)
+
b
I
n
−
1
+
a
I
n
−
2
{\displaystyle -cI_{n}={\frac {1}{x^{n-1}(n-1)}}+bI_{n-1}+aI_{n-2}\,\!}
I
m
,
n
=
∫
x
m
d
x
(
a
x
2
+
b
x
+
c
)
n
{\displaystyle I_{m,n}=\int {\frac {x^{m}\mathrm {d} x}{(ax^{2}+bx+c)^{n}}}\,\!}
I
m
,
n
=
−
x
m
−
1
a
(
2
n
−
m
−
1
)
(
a
x
2
+
b
x
+
c
)
n
−
1
+
c
(
m
−
1
)
a
(
2
n
−
m
−
1
)
I
m
−
2
,
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}\,\!}
I
m
,
n
=
∫
d
x
x
m
(
a
x
2
+
b
x
+
c
)
n
{\displaystyle I_{m,n}=\int {\frac {\mathrm {d} x}{x^{m}(ax^{2}+bx+c)^{n}}}\,\!}
−
c
(
m
−
1
)
I
m
,
n
=
1
x
m
−
1
(
a
x
2
+
b
x
+
c
)
n
−
1
+
a
(
m
+
2
n
−
3
)
I
m
−
2
,
n
+
b
(
m
+
n
−
2
)
I
m
−
1
,
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
I
n
=
∫
(
a
x
2
+
b
x
+
c
)
n
d
x
{\displaystyle I_{n}=\int (ax^{2}+bx+c)^{n}\mathrm {d} x\,\!}
4
a
(
n
+
1
)
I
n
+
1
2
=
2
(
2
a
x
+
b
)
(
a
x
2
+
b
x
+
c
)
n
+
1
2
+
(
2
n
+
1
)
(
4
a
c
−
b
2
)
I
n
−
1
2
{\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}}}\,\!}
I
n
=
∫
1
(
a
x
2
+
b
x
+
c
)
n
d
x
{\displaystyle I_{n}=\int {\frac {1}{(ax^{2}+bx+c)^{n}}}\mathrm {d} x\,\!}
(
2
n
−
1
)
(
4
a
c
−
b
2
)
I
n
+
1
2
=
2
(
2
a
x
+
b
)
(
a
x
2
+
b
x
+
c
)
n
−
1
2
+
8
a
(
n
−
1
)
I
n
−
1
2
{\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 :
I
n
+
1
2
=
I
2
n
+
1
2
=
∫
1
(
a
x
2
+
b
x
+
c
)
2
n
+
1
2
d
x
=
∫
1
(
a
x
2
+
b
x
+
c
)
2
n
+
1
d
x
{\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]
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
I
n
=
∫
x
n
sin
a
x
d
x
{\displaystyle I_{n}=\int x^{n}\sin {ax}\mathrm {d} x\,\!}
a
2
I
n
=
−
a
x
n
cos
a
x
+
n
x
n
−
1
sin
a
x
−
n
(
n
−
1
)
I
n
−
2
{\displaystyle a^{2}I_{n}=-ax^{n}\cos {ax}+nx^{n-1}\sin {ax}-n(n-1)I_{n-2}\,\!}
J
n
=
∫
x
n
cos
a
x
d
x
{\displaystyle J_{n}=\int x^{n}\cos {ax}\mathrm {d} x\,\!}
a
2
J
n
=
a
x
n
sin
a
x
+
n
x
n
−
1
cos
a
x
−
n
(
n
−
1
)
J
n
−
2
{\displaystyle a^{2}J_{n}=ax^{n}\sin {ax}+nx^{n-1}\cos {ax}-n(n-1)J_{n-2}\,\!}
I
n
=
∫
sin
a
x
x
n
d
x
{\displaystyle I_{n}=\int {\frac {\sin {ax}}{x^{n}}}\mathrm {d} x\,\!}
J
n
=
∫
cos
a
x
x
n
d
x
{\displaystyle J_{n}=\int {\frac {\cos {ax}}{x^{n}}}\mathrm {d} x\,\!}
I
n
=
−
sin
a
x
(
n
−
1
)
x
n
−
1
+
a
n
−
1
J
n
−
1
{\displaystyle I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}J_{n-1}\,\!}
J
n
=
−
cos
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
I
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 :
I
n
=
−
sin
a
x
(
n
−
1
)
x
n
−
1
+
a
n
−
1
J
n
−
1
{\displaystyle I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}J_{n-1}\,\!}
J
n
−
1
=
−
cos
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
I
n
−
2
{\displaystyle J_{n-1}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}I_{n-2}\,\!}
I
n
=
−
sin
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
[
cos
a
x
(
n
−
1
)
x
n
−
1
+
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]\,\!}
∴
I
n
=
−
sin
a
x
(
n
−
1
)
x
n
−
1
−
a
(
n
−
1
)
2
(
cos
a
x
x
n
−
1
+
a
I
n
−
2
)
{\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 :
I
n
−
1
=
−
sin
a
x
(
n
−
1
)
x
n
−
1
+
a
n
−
1
J
n
−
2
{\displaystyle I_{n-1}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}J_{n-2}\,\!}
J
n
=
−
cos
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
I
n
−
1
{\displaystyle J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}I_{n-1}\,\!}
J
n
=
−
cos
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
[
−
sin
a
x
(
n
−
1
)
x
n
−
1
+
a
n
−
1
J
n
−
2
]
{\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]\,\!}
∴
J
n
=
−
cos
a
x
(
n
−
1
)
x
n
−
1
−
a
(
n
−
1
)
2
(
−
sin
a
x
x
n
−
1
+
a
J
n
−
2
)
{\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)\,\!}
I
n
=
∫
sin
n
a
x
d
x
{\displaystyle I_{n}=\int \sin ^{n}{ax}\mathrm {d} x\,\!}
a
n
I
n
=
−
sin
n
−
1
a
x
cos
a
x
+
a
(
n
−
1
)
I
n
−
2
{\displaystyle anI_{n}=-\sin ^{n-1}{ax}\cos {ax}+a(n-1)I_{n-2}\,\!}
J
n
=
∫
cos
n
a
x
d
x
{\displaystyle J_{n}=\int \cos ^{n}{ax}\mathrm {d} x\,\!}
a
n
J
n
=
sin
a
x
cos
n
−
1
a
x
+
a
(
n
−
1
)
J
n
−
2
{\displaystyle anJ_{n}=\sin {ax}\cos ^{n-1}{ax}+a(n-1)J_{n-2}\,\!}
I
n
=
∫
d
x
sin
n
a
x
{\displaystyle I_{n}=\int {\frac {\mathrm {d} x}{\sin ^{n}{ax}}}\,\!}
(
n
−
1
)
I
n
=
−
cos
a
x
a
sin
n
−
1
a
x
+
(
n
−
2
)
I
n
−
2
{\displaystyle (n-1)I_{n}=-{\frac {\cos {ax}}{a\sin ^{n-1}{ax}}}+(n-2)I_{n-2}\,\!}
J
n
=
∫
d
x
cos
n
a
x
{\displaystyle J_{n}=\int {\frac {\mathrm {d} x}{\cos ^{n}{ax}}}\,\!}
(
n
−
1
)
J
n
=
sin
a
x
a
cos
n
−
1
a
x
+
(
n
−
2
)
I
n
−
2
{\displaystyle (n-1)J_{n}={\frac {\sin {ax}}{a\cos ^{n-1}{ax}}}+(n-2)I_{n-2}\,\!}
Integral
Pagpapaliit na pormula
I
m
,
n
=
∫
sin
m
a
x
cos
n
a
x
d
x
{\displaystyle I_{m,n}=\int \sin ^{m}{ax}\cos ^{n}{ax}\mathrm {d} x\,\!}
I
m
,
n
=
{
−
sin
m
−
1
a
x
cos
n
+
1
a
x
a
(
m
+
n
)
+
m
−
1
m
+
n
I
m
−
2
,
n
sin
m
+
1
a
x
cos
n
−
1
a
x
a
(
m
+
n
)
+
n
−
1
m
+
n
I
m
,
n
−
2
{\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}}\,\!}
I
m
,
n
=
∫
d
x
sin
m
a
x
cos
n
a
x
{\displaystyle I_{m,n}=\int {\frac {\mathrm {d} x}{\sin ^{m}{ax}\cos ^{n}{ax}}}\,\!}
I
m
,
n
=
{
1
a
(
n
−
1
)
sin
m
−
1
a
x
cos
n
−
1
a
x
+
m
+
n
−
2
n
−
1
I
m
,
n
−
2
1
a
(
m
−
1
)
sin
m
−
1
a
x
cos
n
−
1
a
x
+
m
+
n
−
2
m
−
1
I
m
−
2
,
n
{\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}}\,\!}
I
m
,
n
=
∫
sin
m
a
x
cos
n
a
x
d
x
{\displaystyle I_{m,n}=\int {\frac {\sin ^{m}{ax}}{\cos ^{n}{ax}}}\mathrm {d} x\,\!}
I
m
,
n
=
{
sin
m
−
1
a
x
a
(
n
−
1
)
cos
n
−
1
a
x
−
m
−
1
n
−
1
I
m
−
2
,
n
−
2
sin
m
+
1
a
x
a
(
n
−
1
)
cos
n
−
1
a
x
−
m
−
n
+
2
n
−
1
I
m
,
n
−
2
sin
m
−
1
a
x
a
(
m
−
n
)
cos
n
−
1
a
x
+
m
−
1
m
−
n
I
m
−
2
,
n
{\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}}\,\!}
I
m
,
n
=
∫
cos
m
a
x
sin
n
a
x
d
x
{\displaystyle I_{m,n}=\int {\frac {\cos ^{m}{ax}}{\sin ^{n}{ax}}}\mathrm {d} x\,\!}
I
m
,
n
=
{
−
cos
m
−
1
a
x
a
(
n
−
1
)
sin
n
−
1
a
x
−
m
−
1
n
−
1
I
m
−
2
,
n
−
2
−
cos
m
+
1
a
x
a
(
n
−
1
)
sin
n
−
1
a
x
−
m
−
n
+
2
n
−
1
I
m
,
n
−
2
cos
m
−
1
a
x
a
(
m
−
n
)
sin
n
−
1
a
x
+
m
−
1
m
−
n
I
m
−
2
,
n
{\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
I
n
=
∫
x
n
e
a
x
d
x
{\displaystyle I_{n}=\int x^{n}e^{ax}\mathrm {d} x\,\!}
n
>
0
{\displaystyle n>0\,\!}
I
n
=
−
x
n
e
a
x
a
−
n
a
I
n
−
1
{\displaystyle I_{n}=-{\frac {x^{n}e^{ax}}{a}}-{\frac {n}{a}}I_{n-1}\,\!}
I
n
=
∫
x
−
n
e
a
x
d
x
{\displaystyle I_{n}=\int x^{-n}e^{ax}\mathrm {d} x\,\!}
n
>
0
{\displaystyle n>0\,\!}
n
≠
1
{\displaystyle n\neq 1\,\!}
I
n
=
−
−
e
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
I
n
−
1
{\displaystyle I_{n}=-{\frac {-e^{ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}I_{n-1}\,\!}
I
n
=
∫
x
a
x
sin
n
b
x
d
x
{\displaystyle I_{n}=\int x^{ax}\sin ^{n}{bx}\mathrm {d} x\,\!}
I
n
=
−
x
a
x
sin
n
−
1
b
x
a
2
+
(
b
n
)
2
(
a
sin
b
x
−
b
n
cos
b
x
)
+
n
(
n
−
1
)
b
2
a
2
+
(
b
n
)
2
I
n
−
2
{\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}\,\!}
I
n
=
∫
x
a
x
cos
n
b
x
d
x
{\displaystyle I_{n}=\int x^{ax}\cos ^{n}{bx}\mathrm {d} x\,\!}
I
n
=
−
x
a
x
cos
n
−
1
b
x
a
2
+
(
b
n
)
2
(
a
cos
b
x
+
b
n
sin
b
x
)
+
n
(
n
−
1
)
b
2
a
2
+
(
b
n
)
2
I
n
−
2
{\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
n
,
p
,
q
∈
N
{\displaystyle n,p,q\in \mathbf {N} \,\!}
Kung papansin, maraming mga pagpapaliit na pormula ng isang indeks na I n I m,n
I
n
=
A
+
B
I
n
−
p
{\displaystyle I_{n}=A+BI_{n-p}\,\!}
kung saan ang
I
n
−
p
{\displaystyle I_{n-p}\,\!}
p ng paurong relatibo sa
I
n
{\displaystyle I_{n}\,\!}
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):
I
n
=
A
+
B
I
n
−
p
{\displaystyle I_{n}=A+BI_{n-p}\,\!}
I
n
−
p
=
A
+
B
I
n
−
2
p
{\displaystyle I_{n-p}=A+BI_{n-2p}\,\!}
I
n
−
2
p
=
A
+
B
I
n
−
3
p
{\displaystyle I_{n-2p}=A+BI_{n-3p}\,\!}
etc., so the general recurrence shift is:
I
n
−
q
p
=
A
+
B
I
n
−
(
q
−
1
)
p
{\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 :
I
n
=
A
+
B
(
A
+
B
(
A
+
B
(
⋯
A
+
q
I
n
−
p
q
⋯
)
)
)
{\displaystyle I_{n}=A+B\left(A+B\left(A+B\left(\cdots A+qI_{n-pq}\cdots \right)\right)\right)\,\!}
I
n
=
A
+
(
A
B
+
B
2
(
A
+
B
(
⋯
A
+
q
I
n
−
p
q
⋯
)
)
)
{\displaystyle I_{n}=A+\left(AB+B^{2}\left(A+B\left(\cdots A+qI_{n-pq}\cdots \right)\right)\right)\,\!}
I
n
=
A
+
(
A
B
+
(
A
B
2
+
B
3
(
⋯
A
+
q
I
n
−
p
q
⋯
)
)
)
{\displaystyle I_{n}=A+\left(AB+\left(AB^{2}+B^{3}\left(\cdots A+qI_{n-pq}\cdots \right)\right)\right)\,\!}
I
n
=
A
+
(
A
B
+
(
A
B
2
+
(
⋯
A
B
q
−
1
+
B
q
I
n
−
p
q
⋯
)
)
)
{\displaystyle I_{n}=A+\left(AB+\left(AB^{2}+\left(\cdots AB^{q-1}+B^{q}I_{n-pq}\cdots \right)\right)\right)\,\!}
I
n
=
A
+
A
B
+
A
B
2
+
⋯
+
A
B
q
−
1
+
B
q
I
n
−
p
q
{\displaystyle I_{n}=A+AB+AB^{2}+\cdots +AB^{q-1}+B^{q}I_{n-pq}\,\!}
na nagreresulta ng:
I
n
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
n
−
p
q
{\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 I 0 ay alam, maaaring itakda ang
I
n
−
p
q
=
I
0
{\displaystyle I_{n-pq}=I_{0}\,\!}
n
−
p
q
=
0
{\displaystyle n-pq=0\,\!}
n
=
p
q
{\displaystyle n=pq\,\!}
I
p
q
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
0
{\displaystyle I_{pq}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}
Kung ang I 0 ay alam, maaaring itakda ang
I
n
−
p
q
=
I
1
{\displaystyle I_{n-pq}=I_{1}\,\!}
n
−
p
q
=
1
{\displaystyle n-pq=1\,\!}
n
=
p
q
+
1
{\displaystyle n=pq+1\,\!}
I
p
q
+
1
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
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:
Ang isang indaktibong pagpapatunay ay magpapakita na ang pangkahalatang pormula na
I
n
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
n
−
p
q
{\displaystyle I_{n}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{n-pq}\,\!}
∀
n
∈
N
{\displaystyle \forall n\in \mathbf {N} \,\!}
Kasong 1 : one for
I
n
−
p
q
=
I
0
{\displaystyle I_{n-pq}=I_{0}\,\!}
I
p
q
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
0
{\displaystyle I_{pq}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}
Kasong 2 : one for
I
n
−
p
q
=
I
1
{\displaystyle I_{n-pq}=I_{1}\,\!}
I
p
q
+
1
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
1
{\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:
I
p
q
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
0
{\displaystyle I_{pq}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{0}\,\!}
ay totoo para sa q = 1;
I
p
=
A
∑
i
=
0
0
B
i
+
B
I
0
{\displaystyle I_{p}=A\sum _{i=0}^{0}B^{i}+BI_{0}\,\!}
I
p
=
A
B
0
+
B
I
0
{\displaystyle I_{p}=AB^{0}+BI_{0}\,\!}
I
p
=
A
+
B
I
0
{\displaystyle I_{p}=A+BI_{0}\,\!}
Ngayon ipagpapalagay na:
I
p
q
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
0
{\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:
I
p
(
q
+
1
)
=
A
∑
i
=
0
q
B
i
+
B
q
+
1
I
0
{\displaystyle I_{p(q+1)}=A\sum _{i=0}^{q}B^{i}+B^{q+1}I_{0}\,\!}
I
p
(
q
+
1
)
=
A
(
∑
i
=
0
q
−
1
B
i
+
B
q
)
+
B
q
B
I
0
{\displaystyle I_{p(q+1)}=A\left(\sum _{i=0}^{q-1}B^{i}+B^{q}\right)+B^{q}BI_{0}\,\!}
I
p
(
q
+
1
)
=
A
∑
i
=
0
q
−
1
B
i
+
A
B
q
+
B
q
B
I
0
{\displaystyle I_{p(q+1)}=A\sum _{i=0}^{q-1}B^{i}+AB^{q}+B^{q}BI_{0}\,\!}
I
p
(
q
+
1
)
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
(
A
+
B
I
0
)
{\displaystyle I_{p(q+1)}=A\sum _{i=0}^{q-1}B^{i}+B^{q}\left(A+BI_{0}\right)\,\!}
I
p
(
q
+
1
)
=
A
∑
i
=
0
q
−
1
B
i
+
B
q
I
p
{\displaystyle I_{p(q+1)}=A\sum _{i=0}^{q-1}B^{i}+B^{q}I_{p}\,\!}
I
p
q
+
p
=
A
∑
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 I 0 patungo sa I p I pq I pq+p
∀
q
⩾
1
{\displaystyle \forall q\geqslant 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\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e78011c3e0317d0c662b7d1bffb368882709656)
![{\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]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39e0266ea24d176b6004afe85b072b79fec2d05a)
![{\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}}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6219e3f7d92afca1589d9686cab0b65932a9e8ec)
![{\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]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4b89a9c0fa31eeb55e404535cdfb3c72b934e5a)
![{\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]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51ac8e27f7631aaeb6bcb5481f7698bf7d789fea)
