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 containing an integer parameter , usually in the form of powers of elementary functions, or products of transcendental functions and polynomials of arbitrary degree , can't be integrated directly. But using other 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. [1]
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 , 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 In , in terms of an integral that involves a lower value of the parameter (lower power) of that function, for example I n -1I n -2recurrence relation . In other words, the reduction formula expresses the integral
I
n
=
∫
f
(
x
,
n
)
d
x
,
{\displaystyle I_{n}=\int f(x,n)\,{\text{d}}x,}
in terms of
I
k
=
∫
f
(
x
,
k
)
d
x
,
{\displaystyle I_{k}=\int f(x,k)\,{\text{d}}x,}
where
k
<
n
.
{\displaystyle k<n.}
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 In . [2]
Examples Below are examples of the procedure.
Cosine integral
Typically, integrals like
∫
cos
n
x
d
x
,
{\displaystyle \int \cos ^{n}x\,{\text{d}}x,\,\!}
can be evaluated by a reduction formula.
∫
cos
n
(
x
)
d
x
{\displaystyle \int \cos ^{n}(x)\,{\text{d}}x\!}
n = 1, 2 ... 30Start by setting:
I
n
=
∫
cos
n
x
d
x
.
{\displaystyle I_{n}=\int \cos ^{n}x\,{\text{d}}x.\,\!}
Now re-write as:
I
n
=
∫
cos
n
−
1
x
cos
x
d
x
,
{\displaystyle I_{n}=\int \cos ^{n-1}x\cos x\,{\text{d}}x,\,\!}
Integrating by this substitution:
cos
x
d
x
=
d
(
sin
x
)
,
{\displaystyle \cos x\,{\text{d}}x={\text{d}}(\sin x),\,\!}
I
n
=
∫
cos
n
−
1
x
d
(
sin
x
)
.
{\displaystyle I_{n}=\int \cos ^{n-1}x\,{\text{d}}(\sin x).\!}
Now integrating by parts:
∫
cos
n
x
d
x
=
cos
n
−
1
x
sin
x
−
∫
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
=
cos
n
−
1
x
sin
x
+
(
n
−
1
)
∫
cos
n
−
2
x
sin
2
x
d
x
=
cos
n
−
1
x
sin
x
+
(
n
−
1
)
∫
cos
n
−
2
x
(
1
−
cos
2
x
)
d
x
=
cos
n
−
1
x
sin
x
+
(
n
−
1
)
∫
cos
n
−
2
x
d
x
−
(
n
−
1
)
∫
cos
n
x
d
x
=
cos
n
−
1
x
sin
x
+
(
n
−
1
)
I
n
−
2
−
(
n
−
1
)
I
n
,
{\displaystyle {\begin{aligned}\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{aligned}}\,}
solving for In :
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},\,}
so the reduction formula is:
∫
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\,{\text{d}}x\ ={\frac {1}{n}}\cos ^{n-1}x\sin x+{\frac {n-1}{n}}\int \cos ^{n-2}x\,{\text{d}}x.\!}
To supplement the example, the above can be used to evaluate the integral for (say) n = 5;
I
5
=
∫
cos
5
x
d
x
.
{\displaystyle I_{5}=\int \cos ^{5}x\,{\text{d}}x.\,\!}
Calculating lower indices:
n
=
5
,
I
5
=
1
5
cos
4
x
sin
x
+
4
5
I
3
,
{\displaystyle n=5,\quad I_{5}={\tfrac {1}{5}}\cos ^{4}x\sin x+{\tfrac {4}{5}}I_{3},\,}
n
=
3
,
I
3
=
1
3
cos
2
x
sin
x
+
2
3
I
1
,
{\displaystyle n=3,\quad I_{3}={\tfrac {1}{3}}\cos ^{2}x\sin x+{\tfrac {2}{3}}I_{1},\,}
back-substituting:
∵
I
1
=
∫
cos
x
d
x
=
sin
x
+
C
1
,
{\displaystyle \because I_{1}\ =\int \cos x\,{\text{d}}x=\sin x+C_{1},\,}
∴
I
3
=
1
3
cos
2
x
sin
x
+
2
3
sin
x
+
C
2
,
C
2
=
2
3
C
1
,
{\displaystyle \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},\,}
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,\,}
where C is a constant.
Exponential integral
Another typical example is:
∫
x
n
e
a
x
d
x
.
{\displaystyle \int x^{n}e^{ax}\,{\text{d}}x.\,\!}
Start by setting:
I
n
=
∫
x
n
e
a
x
d
x
.
{\displaystyle I_{n}=\int x^{n}e^{ax}\,{\text{d}}x.\,\!}
Integrating by substitution:
x
n
d
x
=
d
(
x
n
+
1
)
n
+
1
,
{\displaystyle x^{n}\,{\text{d}}x={\frac {{\text{d}}(x^{n+1})}{n+1}},\,\!}
I
n
=
1
n
+
1
∫
e
a
x
d
(
x
n
+
1
)
,
{\displaystyle I_{n}={\frac {1}{n+1}}\int e^{ax}\,{\text{d}}(x^{n+1}),\!}
Now integrating by parts:
∫
e
a
x
d
(
x
n
+
1
)
=
x
n
+
1
e
a
x
−
∫
x
n
+
1
d
(
e
a
x
)
=
x
n
+
1
e
a
x
−
a
∫
x
n
+
1
e
a
x
d
x
,
{\displaystyle {\begin{aligned}\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{aligned}}\!}
(
n
+
1
)
I
n
=
x
n
+
1
e
a
x
−
a
I
n
+
1
,
{\displaystyle (n+1)I_{n}=x^{n+1}e^{ax}-aI_{n+1},\!}
shifting indices back by 1 (so n + 1 → n , n → n – 1):
n
I
n
−
1
=
x
n
e
a
x
−
a
I
n
,
{\displaystyle nI_{n-1}=x^{n}e^{ax}-aI_{n},\!}
solving for In :
I
n
=
1
a
(
x
n
e
a
x
−
n
I
n
−
1
)
,
{\displaystyle I_{n}={\frac {1}{a}}\left(x^{n}e^{ax}-nI_{n-1}\right),\,\!}
so the reduction formula is:
∫
x
n
e
a
x
d
x
=
1
a
(
x
n
e
a
x
−
n
∫
x
n
−
1
e
a
x
d
x
)
.
{\displaystyle \int x^{n}e^{ax}\,{\text{d}}x={\frac {1}{a}}\left(x^{n}e^{ax}-n\int x^{n-1}e^{ax}\,{\text{d}}x\right).\!}
An alternative way in which the derivation could be done starts by substituting
e
a
x
{\displaystyle e^{ax}}
Integration by substitution:
e
a
x
d
x
=
d
(
e
a
x
)
a
,
{\displaystyle e^{ax}\,{\text{d}}x={\frac {{\text{d}}(e^{ax})}{a}},\,\!}
I
n
=
1
a
∫
x
n
d
(
e
a
x
)
,
{\displaystyle I_{n}={\frac {1}{a}}\int x^{n}\,{\text{d}}(e^{ax}),\!}
Now integrating by parts:
∫
x
n
d
(
e
a
x
)
=
x
n
e
a
x
−
∫
e
a
x
d
(
x
n
)
=
x
n
e
a
x
−
n
∫
e
a
x
x
n
−
1
d
x
,
{\displaystyle {\begin{aligned}\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{aligned}}\!}
which gives the reduction formula when substituting back:
I
n
=
1
a
(
x
n
e
a
x
−
n
I
n
−
1
)
,
{\displaystyle I_{n}={\frac {1}{a}}\left(x^{n}e^{ax}-nI_{n-1}\right),\,\!}
which is equivalent to:
∫
x
n
e
a
x
d
x
=
1
a
(
x
n
e
a
x
−
n
∫
x
n
−
1
e
a
x
d
x
)
.
{\displaystyle \int x^{n}e^{ax}\,{\text{d}}x={\frac {1}{a}}\left(x^{n}e^{ax}-n\int x^{n-1}e^{ax}\,{\text{d}}x\right).\!}
Tables of integral reduction formulas Rational functions The following integrals[3]
Factors of the linear radical
a
x
+
b
{\displaystyle {\sqrt {ax+b}}\,\!}
Linear factors
p
x
+
q
{\displaystyle {px+q}\,\!}
a
x
+
b
{\displaystyle {\sqrt {ax+b}}\,\!}
Quadratic factors
x
2
+
a
2
{\displaystyle x^{2}+a^{2}\,\!}
Quadratic factors
x
2
−
a
2
{\displaystyle x^{2}-a^{2}\,\!}
x
>
a
{\displaystyle x>a\,\!}
Quadratic factors
a
2
−
x
2
{\displaystyle a^{2}-x^{2}\,\!}
x
<
a
{\displaystyle x<a\,\!}
(Irreducible ) quadratic factors
a
x
2
+
b
x
+
c
{\displaystyle ax^{2}+bx+c\,\!}
Radicals of irreducible quadratic factors
a
x
2
+
b
x
+
c
{\displaystyle {\sqrt {ax^{2}+bx+c}}\,\!}
Integral
Reduction formula
I
n
=
∫
x
n
a
x
+
b
d
x
{\displaystyle I_{n}=\int {\frac {x^{n}}{\sqrt {ax+b}}}\,{\text{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 {{\text{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}}\,{\text{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
m
,
n
=
∫
d
x
(
a
x
+
b
)
m
(
p
x
+
q
)
n
{\displaystyle I_{m,n}=\int {\frac {{\text{d}}x}{(ax+b)^{m}(px+q)^{n}}}\,\!}
I
m
,
n
=
{
−
1
(
n
−
1
)
(
b
p
−
a
q
)
[
1
(
a
x
+
b
)
m
−
1
(
p
x
+
q
)
n
−
1
+
a
(
m
+
n
−
2
)
I
m
,
n
−
1
]
1
(
m
−
1
)
(
b
p
−
a
q
)
[
1
(
a
x
+
b
)
m
−
1
(
p
x
+
q
)
n
−
1
+
p
(
m
+
n
−
2
)
I
m
−
1
,
n
]
{\displaystyle 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}}\,\!}
I
m
,
n
=
∫
(
a
x
+
b
)
m
(
p
x
+
q
)
n
d
x
{\displaystyle I_{m,n}=\int {\frac {(ax+b)^{m}}{(px+q)^{n}}}\,{\text{d}}x\,\!}
I
m
,
n
=
{
−
1
(
n
−
1
)
(
b
p
−
a
q
)
[
(
a
x
+
b
)
m
+
1
(
p
x
+
q
)
n
−
1
+
a
(
n
−
m
−
2
)
I
m
,
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_{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}}\,\!}
Integral
Reduction formula
I
n
=
∫
d
x
(
x
2
+
a
2
)
n
{\displaystyle I_{n}=\int {\frac {{\text{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 {{\text{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}}}\,{\text{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
Reduction formula
I
n
=
∫
d
x
(
x
2
−
a
2
)
n
{\displaystyle I_{n}=\int {\frac {{\text{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 {{\text{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}}}\,{\text{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
Reduction formula
I
n
=
∫
d
x
(
a
2
−
x
2
)
n
{\displaystyle I_{n}=\int {\frac {{\text{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 {{\text{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}}}\,{\text{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
Reduction formula
I
n
=
∫
d
x
x
n
(
a
x
2
+
b
x
+
c
)
{\displaystyle I_{n}=\int {\frac {{\text{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}\,{\text{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
−
b
(
n
−
m
)
a
(
2
n
−
m
−
1
)
I
m
−
1
,
n
+
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 {b(n-m)}{a(2n-m-1)}}I_{m-1,n}+{\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 {{\text{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
Reduction formula
I
n
=
∫
(
a
x
2
+
b
x
+
c
)
n
d
x
{\displaystyle I_{n}=\int (ax^{2}+bx+c)^{n}\,{\text{d}}x\,\!}
8
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 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}}}\,\!}
I
n
=
∫
1
(
a
x
2
+
b
x
+
c
)
n
d
x
{\displaystyle I_{n}=\int {\frac {1}{(ax^{2}+bx+c)^{n}}}\,{\text{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}}}\,\!}
note that by the laws of indices :
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}}}}\,{\text{d}}x=\int {\frac {1}{\sqrt {(ax^{2}+bx+c)^{2n+1}}}}\,{\text{d}}x\,\!}
Transcendental functions The following integrals[4]
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
Integral
Reduction formula
I
n
=
∫
x
n
sin
a
x
d
x
{\displaystyle I_{n}=\int x^{n}\sin {ax}\,{\text{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}\,{\text{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}}}\,{\text{d}}x\,\!}
J
n
=
∫
cos
a
x
x
n
d
x
{\displaystyle J_{n}=\int {\frac {\cos {ax}}{x^{n}}}\,{\text{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}\,\!}
the formulae can be combined to obtain separate equations in In :
J
n
−
1
=
−
cos
a
x
(
n
−
2
)
x
n
−
2
−
a
n
−
2
I
n
−
2
{\displaystyle J_{n-1}=-{\frac {\cos {ax}}{(n-2)x^{n-2}}}-{\frac {a}{n-2}}I_{n-2}\,\!}
I
n
=
−
sin
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
[
cos
a
x
(
n
−
2
)
x
n
−
2
+
a
n
−
2
I
n
−
2
]
{\displaystyle 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]\,\!}
∴
I
n
=
−
sin
a
x
(
n
−
1
)
x
n
−
1
−
a
(
n
−
1
)
(
n
−
2
)
(
cos
a
x
x
n
−
2
+
a
I
n
−
2
)
{\displaystyle \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)\,\!}
and Jn :
I
n
−
1
=
−
sin
a
x
(
n
−
2
)
x
n
−
2
+
a
n
−
2
J
n
−
2
{\displaystyle I_{n-1}=-{\frac {\sin {ax}}{(n-2)x^{n-2}}}+{\frac {a}{n-2}}J_{n-2}\,\!}
J
n
=
−
cos
a
x
(
n
−
1
)
x
n
−
1
−
a
n
−
1
[
−
sin
a
x
(
n
−
2
)
x
n
−
2
+
a
n
−
2
J
n
−
2
]
{\displaystyle 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]\,\!}
∴
J
n
=
−
cos
a
x
(
n
−
1
)
x
n
−
1
−
a
(
n
−
1
)
(
n
−
2
)
(
−
sin
a
x
x
n
−
2
+
a
J
n
−
2
)
{\displaystyle \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)\,\!}
I
n
=
∫
sin
n
a
x
d
x
{\displaystyle I_{n}=\int \sin ^{n}{ax}\,{\text{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}\,{\text{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 {{\text{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 {{\text{d}}x}{\cos ^{n}{ax}}}\,\!}
(
n
−
1
)
J
n
=
sin
a
x
a
cos
n
−
1
a
x
+
(
n
−
2
)
J
n
−
2
{\displaystyle (n-1)J_{n}={\frac {\sin {ax}}{a\cos ^{n-1}{ax}}}+(n-2)J_{n-2}\,\!}
Integral
Reduction formula
I
m
,
n
=
∫
sin
m
a
x
cos
n
a
x
d
x
{\displaystyle I_{m,n}=\int \sin ^{m}{ax}\cos ^{n}{ax}\,{\text{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 {{\text{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}}}\,{\text{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}}}\,{\text{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
Reduction formula
I
n
=
∫
x
n
e
a
x
d
x
{\displaystyle I_{n}=\int x^{n}e^{ax}\,{\text{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}\,{\text{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
=
∫
e
a
x
sin
n
b
x
d
x
{\displaystyle I_{n}=\int e^{ax}\sin ^{n}{bx}\,{\text{d}}x\,\!}
I
n
=
e
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 {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}\,\!}
I
n
=
∫
e
a
x
cos
n
b
x
d
x
{\displaystyle I_{n}=\int e^{ax}\cos ^{n}{bx}\,{\text{d}}x\,\!}
I
n
=
e
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 {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}\,\!}
References
^ 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
^ Further Elementary Analysis, R.I. Porter, G. Bell & Sons Ltd, 1978, ISBN 0-7135-1594-5
^ http://www.sosmath.com/tables/tables.html -> Indefinite integrals list^ http://www.sosmath.com/tables/tables.html -> Indefinite integrals list
Bibliography Anton, Bivens, Davis, Calculus, 7th edition. 
![{\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/7803bf2d86071f40537d4590e76ab498354b7c7f)
![{\displaystyle 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}}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22e62922721540ee6636beadc5183d305fa0a98b)
![{\displaystyle 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}}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18cf63aea753eb3517e589623b4728e1545c012b)
![{\displaystyle 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]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c59ab79c10f4eed2620c496fac8b476174a84477)
![{\displaystyle 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]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf2ed60b99d5d7b4dfa736c3627c68021c847b44)
