Integration by reduction formulae

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 I_n:

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

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

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

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 + 1 → n, n → n – 1):

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

solving for I_n:

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

so the reduction formula is:

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

An alternative way in which the derivation could be done starts by substituting ${\displaystyle e^{ax}}$.

Integration by substitution:

${\displaystyle e^{ax}\,{\text{d}}x={\frac {{\text{d}}(e^{ax})}{a}},\,\!}$

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

Now integrating by parts:

${\displaystyle {\begin{aligned}\int x^{n}\,{\text{d}}(e^{ax})&=x^{n}e^{ax}-\int e^{ax}\,{\text{d}}(x^{n})\\&=x^{n}e^{ax}-n\int e^{ax}x^{n-1}\,{\text{d}}x,\end{aligned}}\!}$

which gives the reduction formula when substituting back:

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

which is equivalent to:

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

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 ${\displaystyle x<a\,\!}$
• (Irreducible) quadratic factors ${\displaystyle ax^{2}+bx+c\,\!}$
• Radicals of irreducible quadratic factors ${\displaystyle {\sqrt {ax^{2}+bx+c}}\,\!}$

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

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

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:

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

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: ${\displaystyle I_{n-1}=-{\frac {\sin {ax}}{(n-2)x^{n-2}}}+{\frac {a}{n-2}}J_{n-2}\,\!}$ ${\displaystyle J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\left[-{\frac {\sin {ax}}{(n-2)x^{n-2}}}+{\frac {a}{n-2}}J_{n-2}\right]\,\!}$ ${\displaystyle \therefore J_{n}=-{\frac {\cos {ax}}{(n-1)x^{n-1}}}-{\frac {a}{(n-1)(n-2)}}\left(-{\frac {\sin {ax}}{x^{n-2}}}+aJ_{n-2}\right)\,\!}$
${\displaystyle I_{n}=\int \sin ^{n}{ax}\,{\text{d}}x\,\!}$	${\displaystyle anI_{n}=-\sin ^{n-1}{ax}\cos {ax}+a(n-1)I_{n-2}\,\!}$
${\displaystyle J_{n}=\int \cos ^{n}{ax}\,{\text{d}}x\,\!}$	${\displaystyle anJ_{n}=\sin {ax}\cos ^{n-1}{ax}+a(n-1)J_{n-2}\,\!}$
${\displaystyle I_{n}=\int {\frac {{\text{d}}x}{\sin ^{n}{ax}}}\,\!}$	${\displaystyle (n-1)I_{n}=-{\frac {\cos {ax}}{a\sin ^{n-1}{ax}}}+(n-2)I_{n-2}\,\!}$
${\displaystyle J_{n}=\int {\frac {{\text{d}}x}{\cos ^{n}{ax}}}\,\!}$	${\displaystyle (n-1)J_{n}={\frac {\sin {ax}}{a\cos ^{n-1}{ax}}}+(n-2)J_{n-2}\,\!}$

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