Integration by Successive Reduction

Objective

To study integration by successive reduction.

To derive reduction formulas.

Modules

Module I- Integration of sinn x, cosnx , n being any positive integer by successive reduction

Module II- Integration of, n being a positive integer by successive reduction

Module III- Integration of by successive reduction

Module IV- Integration of by successive reduction

Module V- Reduction formula for , and

Module VI- Integration of and where a and b are constants

Introduction

We know the process of integration and standard methods of evaluation of various type indefinite integrals. An integral may be evaluated by the method of substitution, method of partial fraction or integration by parts.The time required for repeated integrations by parts can sometimes be shortened by applying reduction formulas.

A reduction formula is a formula which connects a given integral with another integral which is of the same type but of a lower degree or of a lower order, or is otherwise easier to evaluate.By applying such a formula repeatedly, we can eventually express the original integral in terms of a power low enough to be evaluated directly. In this session we shall derive some reduction formulas which will be useful in the process of integration.

Module I- Integration of sinn x, cosnx , n being any positive integer by successive reduction

When n is any positive integer, the function sinn x may be integrated with the help of a reduction formula. But, if n is an odd positive integer the function can be more easily integrated by means of the substitution cos x= t. Hence the reduction formula need be applied only when n is an even positive integer.

Reduction Formula for

We can write

Integrating by parts, we get

Transposing to the L.H.S and dividing throughout by n,we get

Example 1. Find

We have

Put cos x = t so that -sin x dx = dt

So

Example 2. Evaluate

Applying the Reduction Formula successively,

.

When n is any positive integer, the function cosn x may be integrated with the help of a reduction formula. But, if n is an odd positive integer, the function can be more easily integrated by means of the substitution sin x= t. Hence here also the reduction formula need be applied only when n is an even positive integer.

Reduction Formula for

We have

Integrating by parts, we get

Transposing to the L.H.S and dividing throughout by n, we get

.

Example 3. Evaluate

Put sin x = t so that cos x dx = dt

Example 4. Evaluate

Applying the Reduction Formula successively,

Module II- Integration of , n being a positive integer by successive reduction

We have

Thus.

We have

Thus

Example 5. Evaluate

Example 6. Evaluate

We have

Module III- Integration of ,n being positive integerby successive reduction

We have

Integrating by parts, we get

Transposing to the L.H.S and dividing throughout by (n-1), we get

We have

Integrating by parts, we get

Transposing to the L.H.S and dividing throughout by (n-1), we get

Remark: When n is a positive integer or may be integrated with the help of the reduction formulae. But if n is an even positive integer, the functions can also be integrated by the substitution tan x = t or cot x = t, as the case may be.

Example 7. Evaluate

Using the reduction formula, we have

Example 8. Evaluate

We have

Module IV- Integration of by successive reduction

We can write

Integrating by parts, we get

Transposing to the L.H.S and dividing throughout by, we get

In a similar manner by splitting cosm x instead of sinnx, we can show that

Remark. The substitution cos x = t proves effective when m is odd and sin x = t when n is odd.

Example 9. Evaluate

We have

.

Module V- Reduction formula for , and

Reduction formula for

Integrating by parts

Thus if we have

Reduction formula for

On integration by parts, we have

Above is the reduction formula which reduces the power of x by 2 and ultimately we shall be left with if n is even, or with if n is odd, which we shall integrate by parts.

Similarly, we can prove

Reduction formula for when m and n are positive integers

Let denote the given integral,

Then

.

Module VI- Integration of and where a and b are constants

Let and

Hence

and

Thus

Example 10. Evaluate

We have

Here a = 2 and b = 3

So

Example 11. Evaluate

Here we have

Example 12. Evaluate

We have

Summary

We have proved the reduction formulas

Assignment

  1. Obtain the reduction formula for .
  2. Evaluate .
  3. Prove that .
  4. Evaluate .
  5. If , n being positive, show that .
  6. Prove that .

Books

  1. Calculus, 9th Edition by George B. Thomas, Ross L. Finney, Pearson Education, New Delhi (2002).
  2. Theory and Problems of Advanced Calculus, Schaums outline series, Schaum publishing company, New York.
  3. Integral Calculus, Gorakh Prasad, Pothishala pvt. Limited, Allahabad.

Glossary

Function: A relation from a set X to another set Y which associates every element of X with a unique element of Y is called a function.

Indefinite Integral: Let f(x) and g(x) be two functions so that , then g(x) is called an indefinite integral of f(x) w.r.to x.

Partial fraction: Suppose that is a proper rational function and is a product of polynomials. Then can be expressed as sum of simpler rational functions, each of which is called a partial fraction. This process is called decomposition of into partial fractions.

The decomposition depends on the nature of the factors of.

If ie., a product of non-repeating linear functions, then

, where are constants.

If ie., some factors repeating, then

, where are constants.

If some factors are quadratic, but non-repeating, then corresponding to these factors, the partial fraction is in the form

FAQs

  1. Define reduction formula for integration.

Answer:

A reduction formula is a formula which connects a given integral with another integral which is of the same type but of a lower degree or of a lower order, or is otherwise easier to evaluate.

  1. Explain the method of integrating.

Answer:

We have

Integrating by parts, we get

Transposing to the L.H.S and dividing throughout by n,we get

.

  1. Obtain the reduction formula for

Answer:

We can write

Integrating by parts, we get

Transposing to the L.H.S and dividing throughout by, we get

.

  1. If , prove that .

Answer:

We have .

Integrating on RHS by parts, we get

.

Transposing the last term on the RHS, we get

.

Therefore.