Integration by reduction formulae

Integration by Reduction Formula can be used when we want to integrate a function raised to the power n. If we have such an integral we can establish a reduction formula which can be used to calculate the integral for any value of n.

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 a power of a function, represented by I_n, in terms of an integral that involves a lower power of that function, for example I_n-2. This makes reduction formulae a type of recurrence relation. In other words, the reduction formulae express the integral $I_{n}=\int f(x,n)\,dx$ in terms of $I_{k}=\int f(x,k)\,dx$, where $k<n$. This method of integration is one of the earliest used methods of integration in the world.

How to compute the integral

To compute the integral, we replace n by its value and use the reduction formula repeatedly until we reach a point where the function to be integrated can be computed, usually when it is to the power 0 or 1. Then we substitute the result backwards until we have computed I_n.

Examples

$\int \cos ^{n}(x)\,dx\!$
n = 1..30

Establish a reduction formula that could be used to find $\int \cos ^{n}(x)\,dx\!$. Hence, find $\int \cos ^{5}(x)\,dx\!$.

Solution

$I_{n}\,=\int \cos ^{n}(x)\,dx\!$

$=\int \cos ^{n-1}(x)\cos(x)\,dx\!$

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

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

$=\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \sin(x)\cos ^{n-2}(x)\sin(x)\,dx\!$

$=\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)\sin ^{2}(x)\,dx\!$

$=\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)(1-\cos ^{2}(x))\,dx\!$

$=\cos ^{n-1}(x)\sin(x)\ +(n-1)\int \cos ^{n-2}(x)\,dx\ -(n-1)\int \cos ^{n}(x)\,dx\!$

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

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

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

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

So, the reduction formula is:

$\int \cos ^{n}(x)\,dx\ ={\frac {1}{n}}\cos ^{n-1}(x)\sin(x)\ +{\frac {n-1}{n}}\int \cos ^{n-2}(x)\,dx\!$

Hence, to find $\int \cos ^{5}(x)\,dx\!$:

$n=5\,$: $I_{5}\ ={\tfrac {1}{5}}\cos ^{4}(x)\sin(x)\ +{\tfrac {4}{5}}I_{3}\,$

$n=3\,$: $I_{3}\ ={\tfrac {1}{3}}\cos ^{2}(x)\sin(x)\ +{\tfrac {2}{3}}I_{1}\,$

$\because I_{1}\ =\int \cos(x)\,dx\ =\sin(x)\ +C_{1}\,$

$\therefore I_{3}\ ={\tfrac {1}{3}}\cos ^{2}(x)\sin(x)\ +{\tfrac {2}{3}}\sin(x)\ +C_{2}\,$，$C_{2}\ ={\tfrac {2}{3}}C_{1}\,$

$I_{5}\ ={\frac {1}{5}}\cos ^{4}(x)\sin(x)\ +{\frac {4}{5}}\left[{\frac {1}{3}}\cos ^{2}(x)\sin(x)+{\frac {2}{3}}\sin(x)\right]+C\,$, where C is a constant.

References

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