# Integration by reduction formulae

Integration by reduction formula in integral calculus is a technique 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] 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, 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 In-1 or In-2. This makes the reduction formula a type of recurrence relation. In other words, the reduction formula expresses the integral

${\displaystyle I_{n}=\int f(x,n)\,{\text{d}}x,}$

in terms of

${\displaystyle I_{k}=\int f(x,k)\,{\text{d}}x,}$

where

${\displaystyle k

## 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

${\displaystyle \int \cos ^{n}x\,{\text{d}}x,\,\!}$

can be evaluated by a reduction formula.

${\displaystyle \int \cos ^{n}(x)\,{\text{d}}x\!}$, for n = 1, 2 ... 30

Start by setting:

${\displaystyle I_{n}=\int \cos ^{n}x\,{\text{d}}x.\,\!}$

Now re-write as:

${\displaystyle I_{n}=\int \cos ^{n-1}x\cos x\,{\text{d}}x,\,\!}$

Integrating by this substitution:

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

Now integrating by parts:

{\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:

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

so the reduction formula is:

${\displaystyle \int \cos ^{n}x\,{\text{d}}x\ ={\frac {1}{n}}\cos ^{n-1}x\sin x+{\frac {n-1}{n}}\int \cos ^{n-2}x\,{\text{d}}x.\!}$

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

${\displaystyle I_{5}=\int \cos ^{5}x\,{\text{d}}x.\,\!}$

Calculating lower indices:

${\displaystyle n=5,\quad I_{5}={\tfrac {1}{5}}\cos ^{4}x\sin x+{\tfrac {4}{5}}I_{3},\,}$
${\displaystyle n=3,\quad I_{3}={\tfrac {1}{3}}\cos ^{2}x\sin x+{\tfrac {2}{3}}I_{1},\,}$

back-substituting:

${\displaystyle \because I_{1}\ =\int \cos x\,{\text{d}}x=\sin x+C_{1},\,}$
${\displaystyle \therefore I_{3}\ ={\tfrac {1}{3}}\cos ^{2}x\sin x+{\tfrac {2}{3}}\sin x+C_{2},\quad C_{2}\ ={\tfrac {2}{3}}C_{1},\,}$
${\displaystyle I_{5}\ ={\frac {1}{5}}\cos ^{4}x\sin x+{\frac {4}{5}}\left[{\frac {1}{3}}\cos ^{2}x\sin x+{\frac {2}{3}}\sin x\right]+C,\,}$

where C is a constant.

Exponential integral

Another typical example is:

${\displaystyle \int x^{n}e^{ax}\,{\text{d}}x.\,\!}$.

Start by setting:

${\displaystyle I_{n}=\int x^{n}e^{ax}\,{\text{d}}x.\,\!}$

Integrating by substitution:

${\displaystyle x^{n}\,{\text{d}}x={\frac {{\text{d}}(x^{n+1})}{n+1}},\,\!}$
${\displaystyle I_{n}={\frac {1}{n+1}}\int e^{ax}\,{\text{d}}(x^{n+1}),\!}$

Now integrating by parts:

{\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}}\!}
${\displaystyle (n+1)I_{n}=x^{n+1}e^{ax}-aI_{n+1},\!}$

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

${\displaystyle nI_{n-1}=x^{n}e^{ax}-aI_{n},\!}$

solving for In:

${\displaystyle I_{n}={\frac {1}{a}}\left(x^{n}e^{ax}-nI_{n-1}\right),\,\!}$

so the reduction formula is:

${\displaystyle \int x^{n}e^{ax}\,{\text{d}}x={\frac {1}{a}}\left(x^{n}e^{ax}-n\int x^{n-1}e^{ax}\,{\text{d}}x\right).\!}$

## Tables of integral reduction formulas

### Rational functions

The following integrals[3] contain:

• Factors of the linear radical ${\displaystyle {\sqrt {ax+b}}\,\!}$
• Linear factors ${\displaystyle {px+q}\,\!}$ and the linear radical ${\displaystyle {\sqrt {ax+b}}\,\!}$
• Quadratic factors ${\displaystyle x^{2}+a^{2}\,\!}$
• Quadratic factors ${\displaystyle x^{2}-a^{2}\,\!}$, for ${\displaystyle x>a\,\!}$
• Quadratic factors ${\displaystyle a^{2}-x^{2}\,\!}$, for ${\displaystyle x
• (Irreducible) quadratic factors ${\displaystyle ax^{2}+bx+c\,\!}$
• Radicals of irreducible quadratic factors ${\displaystyle {\sqrt {ax^{2}+bx+c}}\,\!}$
Integral Reduction formula
${\displaystyle I_{n}=\int {\frac {x^{n}}{\sqrt {ax+b}}}\,{\text{d}}x\,\!}$ ${\displaystyle I_{n}={\frac {2x^{n}{\sqrt {ax+b}}}{a(2n+1)}}-{\frac {2nb}{a(2n+1)}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int {\frac {{\text{d}}x}{x^{n}{\sqrt {ax+b}}}}\,\!}$ ${\displaystyle I_{n}=-{\frac {\sqrt {ax+b}}{(n-1)bx^{n-1}}}-{\frac {a(2n-3)}{2b(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int x^{n}{\sqrt {ax+b}}\,{\text{d}}x\,\!}$ ${\displaystyle I_{n}={\frac {2x^{n}{\sqrt {(ax+b)^{3}}}}{a(2n+3)}}-{\frac {2nb}{a(2n+3)}}I_{n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {{\text{d}}x}{(ax+b)^{n}(px+q)^{m}}}\,\!}$ ${\displaystyle I_{n,m}=-{\frac {1}{(n-1)(bp-aq)}}\left[{\frac {1}{(ax+b)^{m-1}(px+q)^{n-1}}}+a(n+m-2)I_{m,n-1}\right]\,\!}$
${\displaystyle I_{n,m}=\int {\frac {(ax+b)^{m}}{(px+q)^{n}}}\,{\text{d}}x\,\!}$ ${\displaystyle I_{n,m}={\begin{cases}-{\frac {1}{(n-1)(bp-aq)}}\left[{\frac {(ax+b)^{m+1}}{(px+q)^{n-1}}}+a(n+m-2)I_{m-1,n-1}\right]\\-{\frac {1}{(n-m-1)p}}\left[{\frac {(ax+b)^{m}}{(px+q)^{n-1}}}+m(bp-aq)I_{m-1,n}\right]\\-{\frac {1}{(n-1)p}}\left[{\frac {(ax+b)^{m}}{(px+q)^{n-1}}}-amI_{m-1,n-1}\right]\end{cases}}\,\!}$
Integral Reduction formula
${\displaystyle I_{n}=\int {\frac {(px+q)^{n}}{\sqrt {ax+b}}}\,{\text{d}}x\,\!}$ ${\displaystyle \int (px+q)^{n}{\sqrt {ax+b}}\,{\text{d}}x={\frac {2(px+q)^{n+1}{\sqrt {ax+b}}}{p(2n+3)}}+{\frac {bp-aq}{p(2n+3)}}I_{n}\,\!}$

${\displaystyle I_{n}={\frac {2(px+q)^{n}{\sqrt {ax+b}}}{a(2n+1)}}+{\frac {2n(aq-bp)}{a(2n+1)}}I_{n-1}\,\!}$

${\displaystyle I_{n}=\int {\frac {{\text{d}}x}{(px+q)^{n}{\sqrt {ax+b}}}}\,\!}$ ${\displaystyle \int {\frac {\sqrt {ax+b}}{(px+q)^{n}}}\,{\text{d}}x=-{\frac {\sqrt {ax+b}}{p(n-1)(px+q)^{n-1}}}+{\frac {a}{2p(n-1)}}I_{n}\,\!}$

${\displaystyle I_{n}=-{\frac {\sqrt {ax+b}}{(n-1)(aq-bp)(px+q)^{n-1}}}+{\frac {a(2n-3)}{2(n-1)(aq-bp)}}I_{n-1}\,\!}$

Integral Reduction formula
${\displaystyle I_{n}=\int {\frac {{\text{d}}x}{(x^{2}+a^{2})^{n}}}\,\!}$ ${\displaystyle I_{n}={\frac {x}{2a^{2}(n-1)(x^{2}+a^{2})^{n-1}}}+{\frac {2n-3}{2a^{2}(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {{\text{d}}x}{x^{m}(x^{2}+a^{2})^{n}}}\,\!}$ ${\displaystyle a^{2}I_{n,m}=I_{m,n-1}-I_{m-2,n}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {x^{m}}{(x^{2}+a^{2})^{n}}}\,{\text{d}}x\,\!}$ ${\displaystyle I_{n,m}=I_{m-2,n-1}-a^{2}I_{m-2,n}\,\!}$
Integral Reduction formula
${\displaystyle I_{n}=\int {\frac {{\text{d}}x}{(x^{2}-a^{2})^{n}}}\,\!}$ ${\displaystyle I_{n}=-{\frac {x}{2a^{2}(n-1)(x^{2}-a^{2})^{n-1}}}-{\frac {2n-3}{2a^{2}(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {{\text{d}}x}{x^{m}(x^{2}-a^{2})^{n}}}\,\!}$ ${\displaystyle {a^{2}}I_{n,m}=I_{m-2,n}-I_{m,n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {x^{m}}{(x^{2}-a^{2})^{n}}}\,{\text{d}}x\,\!}$ ${\displaystyle I_{n,m}=I_{m-2,n-1}+a^{2}I_{m-2,n}\,\!}$
Integral Reduction formula
${\displaystyle I_{n}=\int {\frac {{\text{d}}x}{(a^{2}-x^{2})^{n}}}\,\!}$ ${\displaystyle I_{n}={\frac {x}{2a^{2}(n-1)(a^{2}-x^{2})^{n-1}}}+{\frac {2n-3}{2a^{2}(n-1)}}I_{n-1}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {{\text{d}}x}{x^{m}(a^{2}-x^{2})^{n}}}\,\!}$ ${\displaystyle {a^{2}}I_{n,m}=I_{m,n-1}+I_{m-2,n}\,\!}$
${\displaystyle I_{n,m}=\int {\frac {x^{m}}{(a^{2}-x^{2})^{n}}}\,{\text{d}}x\,\!}$ ${\displaystyle I_{n,m}=a^{2}I_{m-2,n}-I_{m-2,n-1}\,\!}$
Integral Reduction formula
${\displaystyle I_{n}=\int {\frac {{\text{d}}x}{{x^{n}}(ax^{2}+bx+c)}}\,\!}$ ${\displaystyle -cI_{n}={\frac {1}{x^{n-1}(n-1)}}+bI_{n-1}+aI_{n-2}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {x^{m}\,{\text{d}}x}{(ax^{2}+bx+c)^{n}}}\,\!}$ ${\displaystyle I_{m,n}=-{\frac {x^{m-1}}{a(2n-m-1)(ax^{2}+bx+c)^{n-1}}}-{\frac {b(n-m)}{a(2n-m-1)}}I_{m-1,n}+{\frac {c(m-1)}{a(2n-m-1)}}I_{m-2,n}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {{\text{d}}x}{x^{m}(ax^{2}+bx+c)^{n}}}\,\!}$ ${\displaystyle -c(m-1)I_{m,n}={\frac {1}{x^{m-1}(ax^{2}+bx+c)^{n-1}}}+{a(m+2n-3)}I_{m-2,n}+{b(m+n-2)}I_{m-1,n}\,\!}$
Integral Reduction formula
${\displaystyle I_{n}=\int (ax^{2}+bx+c)^{n}\,{\text{d}}x\,\!}$ ${\displaystyle 8a(n+1)I_{n+{\frac {1}{2}}}=2(2ax+b)(ax^{2}+bx+c)^{n+{\frac {1}{2}}}+(2n+1)(4ac-b^{2})I_{n-{\frac {1}{2}}}\,\!}$
${\displaystyle I_{n}=\int {\frac {1}{(ax^{2}+bx+c)^{n}}}\,{\text{d}}x\,\!}$ ${\displaystyle (2n-1)(4ac-b^{2})I_{n+{\frac {1}{2}}}={\frac {2(2ax+b)}{(ax^{2}+bx+c)^{n-{\frac {1}{2}}}}}+{8a(n-1)}I_{n-{\frac {1}{2}}}\,\!}$

note that by the laws of indices:

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

### Transcendental functions

See main article: Transcendental function

The following integrals[4] 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
Integral Reduction formula
${\displaystyle I_{n}=\int x^{n}\sin {ax}\,{\text{d}}x\,\!}$ ${\displaystyle a^{2}I_{n}=-ax^{n}\cos {ax}+nx^{n-1}\sin {ax}-n(n-1)I_{n-2}\,\!}$
${\displaystyle J_{n}=\int x^{n}\cos {ax}\,{\text{d}}x\,\!}$ ${\displaystyle a^{2}J_{n}=ax^{n}\sin {ax}+nx^{n-1}\cos {ax}-n(n-1)J_{n-2}\,\!}$
${\displaystyle I_{n}=\int {\frac {\sin {ax}}{x^{n}}}\,{\text{d}}x\,\!}$

${\displaystyle J_{n}=\int {\frac {\cos {ax}}{x^{n}}}\,{\text{d}}x\,\!}$

${\displaystyle I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}J_{n-1}\,\!}$

${\displaystyle J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}I_{n-1}\,\!}$

the formulae can be combined to obtain separate equations in In:

${\displaystyle J_{n-1}=-{\frac {\cos {ax}}{(n-2)x^{n-2}}}-{\frac {a}{n-2}}I_{n-2}\,\!}$

${\displaystyle I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\left[{\frac {\cos {ax}}{(n-2)x^{n-2}}}+{\frac {a}{n-2}}I_{n-2}\right]\,\!}$

${\displaystyle \therefore I_{n}=-{\frac {\sin {ax}}{(n-1)x^{n-1}}}-{\frac {a}{(n-1)(n-2)}}\left({\frac {\cos {ax}}{x^{n-2}}}+aI_{n-2}\right)\,\!}$

and Jn:

${\displaystyle I_{n-1}=-{\frac {\sin {ax}}{(n-2)x^{n-2}}}+{\frac {a}{n-2}}J_{n-2}\,\!}$

${\displaystyle J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\left[-{\frac {\sin {ax}}{(n-2)x^{n-2}}}+{\frac {a}{n-2}}J_{n-2}\right]\,\!}$

${\displaystyle \therefore J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{(n-1)(n-2)}}\left(-{\frac {\sin {ax}}{x^{n-2}}}+aJ_{n-2}\right)\,\!}$

${\displaystyle I_{n}=\int \sin ^{n}{ax}\,{\text{d}}x\,\!}$ ${\displaystyle anI_{n}=-\sin ^{n-1}{ax}\cos {ax}+a(n-1)I_{n-2}\,\!}$
${\displaystyle J_{n}=\int \cos ^{n}{ax}\,{\text{d}}x\,\!}$ ${\displaystyle anJ_{n}=\sin {ax}\cos ^{n-1}{ax}+a(n-1)J_{n-2}\,\!}$
${\displaystyle I_{n}=\int {\frac {{\text{d}}x}{\sin ^{n}{ax}}}\,\!}$ ${\displaystyle (n-1)I_{n}=-{\frac {\cos {ax}}{a\sin ^{n-1}{ax}}}+(n-2)I_{n-2}\,\!}$
${\displaystyle J_{n}=\int {\frac {{\text{d}}x}{\cos ^{n}{ax}}}\,\!}$ ${\displaystyle (n-1)J_{n}={\frac {\sin {ax}}{a\cos ^{n-1}{ax}}}+(n-2)I_{n-2}\,\!}$
Integral Reduction formula
${\displaystyle I_{m,n}=\int \sin ^{m}{ax}\cos ^{n}{ax}\,{\text{d}}x\,\!}$ ${\displaystyle I_{m,n}={\begin{cases}-{\frac {\sin ^{m-1}{ax}\cos ^{n+1}{ax}}{a(m+n)}}+{\frac {m-1}{m+n}}I_{m-2,n}\\{\frac {\sin ^{m+1}{ax}\cos ^{n-1}{ax}}{a(m+n)}}+{\frac {n-1}{m+n}}I_{m,n-2}\\\end{cases}}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {{\text{d}}x}{\sin ^{m}{ax}\cos ^{n}{ax}}}\,\!}$ ${\displaystyle I_{m,n}={\begin{cases}{\frac {1}{a(n-1)\sin ^{m-1}{ax}\cos ^{n-1}{ax}}}+{\frac {m+n-2}{n-1}}I_{m,n-2}\\-{\frac {1}{a(m-1)\sin ^{m-1}{ax}\cos ^{n-1}{ax}}}+{\frac {m+n-2}{m-1}}I_{m-2,n}\\\end{cases}}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {\sin ^{m}{ax}}{\cos ^{n}{ax}}}\,{\text{d}}x\,\!}$ ${\displaystyle I_{m,n}={\begin{cases}{\frac {\sin ^{m-1}{ax}}{a(n-1)\cos ^{n-1}{ax}}}-{\frac {m-1}{n-1}}I_{m-2,n-2}\\{\frac {\sin ^{m+1}{ax}}{a(n-1)\cos ^{n-1}{ax}}}-{\frac {m-n+2}{n-1}}I_{m,n-2}\\{\frac {\sin ^{m-1}{ax}}{a(m-n)\cos ^{n-1}{ax}}}+{\frac {m-1}{m-n}}I_{m-2,n}\\\end{cases}}\,\!}$
${\displaystyle I_{m,n}=\int {\frac {\cos ^{m}{ax}}{\sin ^{n}{ax}}}\,{\text{d}}x\,\!}$ ${\displaystyle I_{m,n}={\begin{cases}-{\frac {\cos ^{m-1}{ax}}{a(n-1)\sin ^{n-1}{ax}}}-{\frac {m-1}{n-1}}I_{m-2,n-2}\\-{\frac {\cos ^{m+1}{ax}}{a(n-1)\sin ^{n-1}{ax}}}-{\frac {m-n+2}{n-1}}I_{m,n-2}\\{\frac {\cos ^{m-1}{ax}}{a(m-n)\sin ^{n-1}{ax}}}+{\frac {m-1}{m-n}}I_{m-2,n}\\\end{cases}}\,\!}$
Integral Reduction formula
${\displaystyle I_{n}=\int x^{n}e^{ax}\,{\text{d}}x\,\!}$

${\displaystyle n>0\,\!}$

${\displaystyle I_{n}={\frac {x^{n}e^{ax}}{a}}-{\frac {n}{a}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int x^{-n}e^{ax}\,{\text{d}}x\,\!}$

${\displaystyle n>0\,\!}$

${\displaystyle n\neq 1\,\!}$

${\displaystyle I_{n}={\frac {-e^{ax}}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}I_{n-1}\,\!}$
${\displaystyle I_{n}=\int e^{ax}\sin ^{n}{bx}\,{\text{d}}x\,\!}$ ${\displaystyle I_{n}={\frac {e^{ax}\sin ^{n-1}{bx}}{a^{2}+(bn)^{2}}}\left(a\sin bx-bn\cos bx\right)+{\frac {n(n-1)b^{2}}{a^{2}+(bn)^{2}}}I_{n-2}\,\!}$
${\displaystyle I_{n}=\int e^{ax}\cos ^{n}{bx}\,{\text{d}}x\,\!}$ ${\displaystyle I_{n}={\frac {e^{ax}\cos ^{n-1}{bx}}{a^{2}+(bn)^{2}}}\left(a\cos bx+bn\sin bx\right)+{\frac {n(n-1)b^{2}}{a^{2}+(bn)^{2}}}I_{n-2}\,\!}$

## References

1. ^ 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
2. ^ Further Elementary Analysis, R.I. Porter, G. Bell & Sons Ltd, 1978, ISBN 0-7135-1594-5
3. ^ http://www.sosmath.com/tables/tables.html -> Indefinite integrals list
4. ^ http://www.sosmath.com/tables/tables.html -> Indefinite integrals list

## Bibliography

• Anton, Bivens, Davis, Calculus, 7th edition.