Integration by reduction formulae

Integration by reduction formulae 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,i.e. I_n, in terms of an integral that involves a lower power of that function, for example I_n-2.

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 our initial function to be integrated is to the power 0 and can be computed. Then we substitude the result backwards until we have computed I_n.

Examples

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

Solution

In = $\int \cos ^{n}(x)\,dx\!$ 
= $\int \cos ^{n-1}(x)\cos(x)\,dx\!$ 
= $\int \cos ^{n-1}(x)\,d(sin(x))\!$
= $\cos ^{n-1}(x)sin(x)\!$ - $\int \sin(x)\,d(cos^{n-1}(x))\!$
= $\cos ^{n-1}(x)sin(x)\!$ + $(n-1)\int \sin(x)\cos ^{n-2}(x)\sin(x)\,dx\!$
= $\cos ^{n-1}(x)sin(x)\!$ + $(n-1)\int \cos ^{n-2}(x)\sin ^{2}(x)\,dx\!$
= $\cos ^{n-1}(x)sin(x)\!$ + $(n-1)\int \cos ^{n-2}(x)(1-cos^{2}(x))\,dx\!$
= $\cos ^{n-1}(x)sin(x)\!$ + $(n-1)\int \cos ^{n-2}(x)\,dx\!$ -$(n-1)\int \cos ^{n}(x)\,dx\!$
= $\cos ^{n-1}(x)sin(x)\!$ + $(n-1)$ In-2 $-(n-1)$ In
→ In + (n-1) In = $\cos ^{n-1}(x)sin(x)\!$  + (n-1)In → nIn= $\cos ^{n-1}(x)sin(x)\!$ + (n-1)In-2
→ In = $(1/n)\cos ^{n-1}(x)sin(x)\!$ + ${\tfrac {n-1}{n}}$ In-2

So, the reduction formula is:

$(1/n)\cos ^{n-1}(x)sin(x)\!$ + ${\tfrac {n-1}{n}}$ In-2

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

When n=5: I5 = $(1/5)\cos ^{4}(x)sin(x)\!$ + ${\tfrac {4}{5}}$ I3
When n=3: I3 = $(1/3)\cos ^{2}(x)sin(x)\!$ + ${\tfrac {2}{3}}$ I1
But I1 = $\int \cos(x)\,dx\!$ = $\sin(x)\!$ + k1, where k1 is the constant of integration
Hence, I3 = $(1/3)\cos ^{2}(x)sin(x)\!$ + ${\tfrac {2}{3}}\sin(x)\!$ + k2, where k2 = ${\tfrac {2}{3}}$k1
and I5 = $(1/5)\cos ^{4}(x)sin(x)\!$ + ${\tfrac {4}{15}}\ cos^{2}(x)sin(x)$ +${\tfrac {8}{15}}\ sin(x)$ + c, where c is a constant

References

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