Integration by reduction formulae

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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.

Contents

[edit] 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 In, in terms of an integral that involves a lower power of that function, for example In-2. This makes reduction formulae a type of recurrence relation.

[edit] 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 In.

[edit] 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}{5}[(1/3)\ cos^2 (x) sin (x) +\tfrac{2}{3}\ sin (x)] + c, where c is a constant

[edit] References

Anton, Bivens, Davis, Calculus, 7th edition


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