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 I_n, in terms of an integral that involves a lower value of the parameter (lower power) of that function, for example I_n-1 or I_n-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

${\displaystyle k<n.}$

Reference works contain the general forms for recursive integration (see, for example Gradshteyn and Ryzhik).

To compute the integral, we set n to its value and use the reduction formula to express it in terms of the (n – 1) or (n – 2) integral. The lower index integral can be used to calculate the higher 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 I_n. ^[2]

Below are examples of the procedure.

Typically, integrals like

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

can be evaluated by a reduction formula.

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:

${\displaystyle {\begin{aligned}\int \cos ^{n}x\,{\text{d}}x&=\int \cos ^{n-1}x\,{\text{d}}(\sin x)\!=\cos ^{n-1}x\sin x-\int \sin x\,{\text{d}}(\cos ^{n-1}x)\\&=\cos ^{n-1}x\sin x+(n-1)\int \sin x\cos ^{n-2}x\sin x\,{\text{d}}x\\&=\cos ^{n-1}x\sin x+(n-1)\int \cos ^{n-2}x\sin ^{2}x\,{\text{d}}x\\&=\cos ^{n-1}x\sin x+(n-1)\int \cos ^{n-2}x(1-\cos ^{2}x)\,{\text{d}}x\\&=\cos ^{n-1}x\sin x+(n-1)\int \cos ^{n-2}x\,{\text{d}}x-(n-1)\int \cos ^{n}x\,{\text{d}}x\\&=\cos ^{n-1}x\sin x+(n-1)I_{n-2}-(n-1)I_{n},\end{aligned}}\,}$

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.

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:

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

Another alternative way in which the derivation could be done by integrating by parts:

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

${\displaystyle u=x^{n}{\text{ , }}\ dv=e^{ax},}$

${\displaystyle {\frac {du}{dx}}\ =nx^{n-1}{\text{ , }}\ v={\frac {e^{ax}}{a}}\ }$

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

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

Remember:

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

${\displaystyle \therefore \ I_{n}={\frac {x^{n}e^{ax}}{a}}\ -{\frac {n}{a}}\ I_{n-1}}$

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