Reduction Formulae

Discipline Courses-I
Semester-I
Paper: Calculus-I
Lesson: Reduction Formulae
Lesson Developer: Sada Nand Prasad
College/Department: Department of Mathematics, Acharya Narendra Dev
College (D.U.)

Institute of Lifelong Learning, University of Delhi pg. 1

 Reduction Formulae

Institute of Lifelong Learning, University of Delhi pg. 2

 Reduction Formulae

Table of Contents:

Chapter : Reduction Formulae

 1: Learning Outcomes
 2: Introduction
 3: Reduction Formulae
 4: Reduction Formulae for Trigonometric Functions
 5: Determination of the value of definite integral
 Summary
 Exercises
 References for further readings

1. Learning Outcomes

After reading this chapter, you should be able to understand how to derive
reduction formulae for the following functions
  sin x dx ,
n

  cos x dx ,
n

  tan x dx ,
n

  sec x dx ,
n

  sin x cos x dx ,
n m

  log x  dx ,
n


2
  sin
n
x cos m x dx etc.
0

and how to apply them to find the value of the integrals.

Institute of Lifelong Learning, University of Delhi pg. 3

 Reduction Formulae

2: Introduction:

In this lesson we will study a different method of integration to evaluate the

integrals. Integration by parts is one of the effective and most useful
methods to evaluate the integrals having an expression involving products or
logarithm or inverse trigonometric functions, which cannot be solved directly
using standard integral methods. We need to integrate powers of some
trigonometric functions and their products in many problems in engineering
and physics. In this chapter we will discuss the techniques to derive these
kinds of integrals and derive some standard forms using this technique.
Using the method of integration by parts, we will try to reduce a given
integral in terms of another simple integral.

3: Reduction Formulae:

A standard technique to solve the complex problems is to reduce the given

problem into similar problem but with lesser complexity. Sometimes
integrand depends not only on the independent variables but also on the
value n, called the parameter.
For example in order to evaluate
 cos x dx
n

we may convert it into a relation which requires the solution of

 cos x cos x dx
n 1

and will get

cos n 1 x sin x  n 1
 cos x dx    cos n  2 x dx ,
n

n n
Thus if we apply this formula repeatedly, the exponent can be reduced to 0
or 1, accordingly n is even or odd.
We can also use this technique to evaluate the integrals of the type
 sin x cos x dx
m n

by reducing the power of m and n to small values like 0, 1, 2… etc. Since

this process connects the given integral with another of the same type with
reduced power, the relation obtained is called reduction formula.

Value Addition: Definition

Reduction Formulae: A formula of the form
 g  x, n  dx  h  x    f  x, k  dx , where k < n,
is called the reduction formulae.

Institute of Lifelong Learning, University of Delhi pg. 4

 Reduction Formulae

Example 1: Show that the integration of x ea x dx depends upon x and also

n

on n, which is exponent of x.

Solution: Let I n   x n ea x dx , on integrating by parts, we have

ea x n n 1 a x ea x n
I n   x n ea x dx  x n
  x e dx  I n  x n  I n 1
a a a a
Thus, the integration of  x n ea x dx depends upon x and also on n.

4: Reduction Formulae for Trigonometric Functions:

4.1: Reduction formula for  sin x dx , n being a positive integer

n

greater than unity

Let
I n   sin n x dx   sin n 1 x sin x dx (n > 1)

n n -1
(we can rewrite sin x as sin x sin x)

Using method of integration by parts, we get


  d sin n 1 x
x  sin x dx    
  
 sin x dx  dx 
n 1
I n  sin
  dx
   
  sin n 1 x cos x   n 1  sin n  2 x cos x cos x dx


  sin n 1 x cos x   n 1  sin n  2 x 1  sin 2 x dx 
  sin n 1 x cos x   n 1  sin n  2 x dx   n 1  sin n x dx
  sin n 1 x cos x   n 1 I n  2   n 1 I n

I n   n 1 I n   sin n 1 x cos x   n  1 I n  2

 sin n 1 x cos x  n 1
 In   In2 ... (1)
n n

which is valid for n ≥ 2.

Institute of Lifelong Learning, University of Delhi pg. 5

 Reduction Formulae

Example 1: Integrate (i)  sin 6 x dx (ii)  sin 5 x dx

Solution: (i) Let I 6   sin 6 x dx ,
 sin n 1 x cos x  n 1
by equation (1), we have I n   In2
n n
Putting n = 6, we get
 sin 5 x cos x 5
I6   I4
6 6
Repeating the same for n = 4 and 2, we have
 sin 5 x cos x 5   sin 3 x cos x 3 
I6     I2  ,
6 6 4 4 
 sin 5 x cos x 5   sin 3 x cos x 3   sin x cos x 1 
I6        sin 0 x dx  
6 6 4 4 2 2 
 sin 5 x cos x 5 5 5
I6   sin 3 x cos x  sin x cos x  x
6 24 16 16
(ii) Let
I5   sin 5 x dx
proceeding the same as in part (i), we have
 sin 4 x cos x 4 8
I5   sin 2 x cos x  cos x
5 15 15

Value Addition: Note

Derivation of reduction formula is valid for all values of n except n = 1. For
negative integer n = -m, I  m   sin  m x dx   cosecm x dx, which will be derived in
next section.

4.2: Reduction formula for  cosec x dx n being a positive integer

n

greater than unity

Let
I n   cosec n x dx   cosec n  2 x cosec2 x dx
By using method of integration by parts, we get

Institute of Lifelong Learning, University of Delhi pg. 6

 Reduction Formulae

x  cosec x dx    

  d cosecn  2 x   
 dx 

n2
I n  cosec 2
cosec 2
x dx
  dx  
   
  cosec n  2 x cot x   n  2   cosec n  3 x cosec x cot x cot x dx

  cosec n  2 x cot x   n  2   cosec n  2 x  cosec 2 x 1 dx

  cosec n  2 x cot x   n  2   cosec n x dx   n  2   cosec n  2 x dx
  cosec n  2 x cot x   n  2  I n   n  2  I n  2

I n   n  2  I n   cosec n  2 x cot x   n  2  I n  2
 cosec n  2 x cot x  n  2 
 In   In2 ... (2)
n 1 n 1

which is valid for n > 2.

Example 2: Integrate  cosec x dx .

4

Solution: On putting n = 4 in equation (2), we get

 cot x cosec 2 x 2
    cosec 2 x dx
4
cosec x dx
3 3
 cot x cosec x 2
2
  cot x
3 3
4.3: Reduction formula for  cos n x dx , n being a positive integer
greater than unity

Let
I n   cos n x dx   cos n 1 x cos x dx
using method of integration by parts, we get

x  cos x dx    

  d cos n 1 x   
 dx 

n 1
I n  cos cos x dx
  dx  
   
 cos n 1 x sin x   n 1  cos n  2 x sin x sin x dx


 cos n 1 x sin x   n 1  cos n  2 x 1  cos 2 x dx 
 cos n 1 x sin x   n 1  cos n  2 x dx   n 1  cos n x dx
 cos n 1 x sin x   n 1 I n  2   n 1 I n

Institute of Lifelong Learning, University of Delhi pg. 7

 Reduction Formulae

I n   n 1 I n  cos n 1 x sin x   n 1 I n  2

cos n 1 x sin x  n 1
 In   In2 ... (3)
n n

which is valid for n ≥ 2.

Example 3: Evaluate (i)  cos 4 x dx (ii)  cos 5 x dx

Solution: (i) Putting n = 4 and 2 in equation (3), we get

cos 3 x sin x 3
 cos x dx  4  4  cos x dx
4 2

cos3 x sin x 3  cos x sin x 1 

    x
4 4 2 2 
cos3 x sin x 3cos x sin x 3
   x
4 8 8

(ii) putting n = 5 and 3 in equation (3), we get

cos 4 x sin x 4
    cos 3 x dx
5
cos x dx
5 5
cos x sin x 4  cos 2 x sin x 2
4

     cos x dx 
5 5 3 3 
cos 4 x sin x 4 cos 2 x sin x 8

  sin x
5 15 15
4.4: Reduction formula for  sec x dx , n being a positive integer
n

greater than unity

Let
I n   sec n x dx   sec n  2 x sec 2 x dx

using method of integration by parts, we get


  d sec n  2 x
x  sec x dx    
  
 sec x dx  dx 
n2
I n  sec 2 2

  dx
   
 sec n  2 x tan x   n  2   sec n  3 x sec x tan x tan x dx


 sec n  2 x tan x   n  2   sec n  2 x sec 2 x 1 dx 
 sec n  2 x tan x   n  2   sec n x dx   n  2   sec n  2 x dx
 sec n  2 x tan x   n  2  I n   n  2  I n  2

Institute of Lifelong Learning, University of Delhi pg. 8

 Reduction Formulae

I n   n  2  I n  sec n  2 x tan x   n  2  I n  2
sec n  2 x tan x  n  2 
 In   In2 ... (4)
n 1 n 1

which is valid for n > 2.

Example 4: Integrate  sec

6
x dx
Solution: Putting n = 6, 4 and 2 in equation (4), we get
sec 4 x tan x 4
    sec 4 x dx
6
sec x dx
5 5
sec 4 x tan x 4  sec 2 x tan x 2 
     sec 2 x dx 
5 5 3 3 
sec 4 x tan x 4sec 2 x tan x 8
   tan x
5 15 15

4.5: Reduction formula for  tan x dx , n being a positive integer

n

greater than unity

Let

I n   tan n x dx   tan n  2 x tan 2 x dx

 
  tan n  2 x sec2 x 1 dx   tan n  2 x sec2 x dx   tan n  2 x dx ...(5)

Value Addition: Do you know ?

 f  x m1 
 
  f  x   f   x  dx  m  1  k
m

tan n 1 x
Similarly,  tan n  2 x sec2 x dx  k
n 1

Therefore, from equation (5), we have,

tan n 1 x
I n   tan n x dx   In2 ...(6) ,
n 1

which is the required reduction formula for  tan x dx .

n

4.6: Reduction formula for  cot x dx , n being a positive integer

n

greater than unity

Institute of Lifelong Learning, University of Delhi pg. 9

 Reduction Formulae

Let

I n   cot n x dx   cot n  2 x cot 2 x dx

 
  cot n  2 x csc2 x 1 dx
  cot n  2 x csc2 x dx   cot n  2 x dx ...(7)
As from the previous example, from equation (7), we get
 cot n 1 x
I n   cot n x dx   In2 ...(8) ,
n 1

which is the required reduction formula for  cot x dx .

n

Example 5: Evaluate (i)  tan 8 x dx , (ii)  cot 5 x dx

Solution: (i) Putting n = 8, 6 and 4 in the reduction formulae obtained in

equation (6), we get

tan 7 x tan 7 x tan 5 x

 tan x dx    tan 6 x dx     tan 4 x dx
8

7 7 5
7 5 3
tan x tan x tan x
     tan 2 x dx
7 5 3
tan x tan x tan 3 x
7 5

7

5

3

  sec 2 x  1 dx 
tan x tan x tan 3 x
7 5
    tan x  x
7 5 3

(ii) Putting n = 5 and 3 in the reduction formulae obtained in equation (8),

we get

 cot 4 x  cot 4 x cot 2 x

 cot x dx    cot 3 x dx     cot x dx
5

4 4 2
 cot 4 x cot 2 x
   log sin x
4 2

4.7: Reduction formula for   log x 

n
dx

Let
I n    log x  dx
n

By using method of integration by parts, we get,

Institute of Lifelong Learning, University of Delhi pg. 10

 Reduction Formulae

 d  log x  n 
I n    log x  dx   log x     dx   dx

n n
dx  dx
 

1
 x  log x   n   log x  x dx  x  log x   n I n 1
n n 1 n

x
 I n    log x  dx  x  log x   n I n 1 ,
n n
...(9)
Which is the required reduction formula for   log x  dx .
n

Example 6: Write reduction formula for  x  log x 

m n
dx
Solution: On integrating by parts, we have

x m 1 n n 1 1
 x  log x  dx   log x    x m 1  log x 
m n n
dx
m 1 m 1 x
x m 1 n
 x m  log x  dx   log x   x m  log x 
n 1

n n
dx
m 1 m 1
is the required reduction formula for  x  log x  dx .
m n

4.8: Reduction formula for  sin

n
x cos m x dx

Let
I n, m   sin n x cos m x dx
If m = 1, then
 sin n 1 x
  k , where n   1
I n ,1   sin n x cos x dx   n  1
1n sin x  k , where n   1

So, we take m > 1, and we get,
I n, m   sin n x cos m x dx   cos m 1 x  sin n x cos x  dx
By using method of integration by parts, we get,

Institute of Lifelong Learning, University of Delhi pg. 11

 Reduction Formulae

cos m 1 x sin n 1 x sin n 1 x

I n, m    m  1  cos m  2 x   sin x  dx, when n   1
n 1 n 1
cos m 1 x sin n 1 x m 1

n 1

n 1   
sin n x cos m  2 x 1  cos 2 x dx,

cos m 1 x sin n 1 x m 1
   I n , m  2  I n , m  ,
n 1 n 1 
m 1 nm cos m 1 x sin n 1 x m 1
 I n, m  I n, m  I n, m   I n, m  2
n 1 n 1 n 1 nm
cos m 1 x sin n 1 x m 1
or , I n , m   I n, m  2 ...(10)
nm nm

Value Addition: Cautions

This formula is not valid when m+n = 0. So, when m+n=0, we take n = -m
and get
I n, n   sin m x cos m x dx   cot m x dx,
which we have already derived in section (4.6).
Similarly if we consider n > 1 instead of m >1, and proceeding in the same
way, we can derive
 cos m 1 x sin n 1 x n 1
I n, m   I n  2, m if n  m  0. ...(11)
nm nm

Example 7: Evaluate

 sin
4
x cos 2 x dx

Solution: Putting n = 4 and m = 2 in equation (11), we get

Institute of Lifelong Learning, University of Delhi pg. 12

 Reduction Formulae

 cos3 x sin 3 x 3
 sin x cos x dx    sin 2 x cos 2 x dx
4 2

6 6
 cos3 x sin 3 x 1   cos3 x sin x 1 
     cos 2 x dx 
6 2 4 4 
 cos3 x sin 3 x cos3 x sin x 1 1
    1  cos 2 x  dx
6 8 8 2
 cos3 x sin 3 x cos3 x sin x 1 1
   x  sin 2 x
6 8 16 32
1 1  6 6 
 x  cos x  cos 2 x sin 3 x  cos 2 x sin x  sin x 
16 6  8 16 
 

1
16
1
  6
  6
x  cos x  1  sin 2 x sin 3 x  1  sin 2 x sin x  sin x 
6  8 16 
1 1   3  3 3 
 x  cos x sin 5 x  sin 3 x  1    sin x     
16 6   4  4 8 
1 1  1 3 
 x  cos x sin 5 x  sin 3 x  sin x 
16 6  4 8 

Example 8: Write reduction formula for x

m
cos n x dx
Solution: On integrating by parts, we have
m sin n x m m 1
 x cos n x dx  x n  n  x sin n x dx
m

sin n x m  m 1 cos n x m 1 m  2 
 xm
n
  x
n n

n  x cos n x dx 

sin n x m x m 1 cos n x m  m 1 m  2
x    x cos n x dx
m

n n2 n2
n x m sin n x  m x m 1 cos n x m  m 1 m  2
    x cos n x dx
m
x cos n x dx
n2 n2
is the required reduction formula for  x m cos n x dx .

5: Determination of the value of definite integrals


2
5.1: Reduction formula for  sin
n
x dx,
0

we have

Institute of Lifelong Learning, University of Delhi pg. 13

 Reduction Formulae

  /2  
2
 sin n 1 x cos x n 1 2 n  2 n 1 2 n  2
0  
n 0

n 0
sin x dx, n  2.
n
sin x dx sin x dx
n 0

by repeating this formula we have

 n 1 n  3 n  5 3 1  2
       dx, when n is even and n  2.
2 
 n n  2 n  4 4 2 0
0 sin x dx  
n

 n 1 n  3 n  5 4 2 2
      sin x dx, when n is odd and n  3.
 n n2 n4 5 3 0

 n 1 n  3 n  5 3 1 

 n  n  2  n  4  4  2  2 , when n is even and n  2.
2

or ,  sin x dx  
n

0  n 1  n  3  n  5  4  2 , when n is odd and n  3.


 n n2 n4 5 3
 n 1
 
2
Example 9: If I n   x sin x dx and n  1, then provethat I n  n  n 1 I n  2  n  
n

0 2

Solution: On integrating In by parts, xn as first function, we get


 2
I n   x   cos x  
n 2
 n x   cos x  dx
n 1
0
0
 
  
n 1

 

 
2 2
 0  n  x sin x

   n 1 x sin x dx   n    n  n 1  x n  2 sin x dx

n 1 2 n 2

 0  2
 0
 0

n 1
 
 I n  n    n  n  1 I n  2
2
n 1
 
 I n  n  n 1 I n  2  n  
2


2
5.2: Reduction formula for  cos
n
x dx,
0

In the same way we can derive and get

Institute of Lifelong Learning, University of Delhi pg. 14

 Reduction Formulae

 n 1 n  3 n  5 3 1 

 n  n  2  n  4  4  2  2 , when n is even and n  2.
2

 cos n x dx  
0  n 1  n  3  n  5  4  2 , when n is odd and n  3.

 n n2 n4 5 3
   
2 2 2 2
Example 10: Evaluate (i)  sin x dx, (ii)  sin x dx, (iii)  cos x dx, (iv)  cos 6 x dx
9 10 5

0 0 0 0

Solution: (i) We have


2
8 6 4 2 128
 sin x dx     
9

0
9 7 5 3 315
(ii) we have

2
9 7 5 3 1  63
 sin x dx        
10

0
10 8 6 4 2 2 512
(iii) we have

2
4 2 8
 cos x dx   
5

0
5 3 15
(iv) we have

2
5 3 1  5
 cos x dx      
6

0
6 4 2 2 32


2
5.3: Reduction formula for  sin
n
x cos m x dx, where n and m are positive integers
0

By equation (11), we have,

 cos m 1 x sin n 1 x n 1
 sin x cos x dx  
nm 
sin n  2 x cos m x dx
n m

nm


Taking the limits 0 to on both sides, we get,
2
 /2  /2  /2
 cos m 1 x sin n 1 x n 1
  sin
n2
In, m  sin x cos x dx 
n m
 x cos m x dx ...(12)
0
nm 0
nm 0

n 1
 In, m  I n  2, m ...(13)
nm
interchanging n to n-2, n-4, n-6 and so on, we obtain

Institute of Lifelong Learning, University of Delhi pg. 15

 Reduction Formulae

n 3 n 5 n7
I n  2, m  I n  2, m , I n  4, m  I n  6, m , I n  6, m  I n  8, m ...
mn2 mn4 nm6
2 1
I3, m  I1, m , when n is odd , I 2, m  I0, m , when n is even,
3 m 2m
 /2  /2  /2
1
0
Also, I1, m  sin x cos m x dx 
, and I0, m   sin 0 x cos m x dx   cos
m
x dx
m 1 0 0
Substituting for I 0, m , I 2, m , and I1, m , I 3, m , we get ,
 /2
n 1 n3 n5 1
I n, m      cos
m
x dx, when n is even
nm nm 2 nm4 2m 0

n 1 n3 n5 2 1
and I n , m      , when n is odd
n  m n  m  2 n  m  4 3  m m 1
n 1 n3 n5 1 m 1 m  3 1 
 I n, m         , when n is even, m is even
nm nm 2 nm4 2m m m2 2 2
n 1 n3 n5 1 m 1 m  3 2
and I n , m        , when n is even, m is odd
nm nm 2 nm4 2m m m2 3
 
2 2
Example 11: Write down the value of (i)  sin x cos x dx, (ii)  sin 6 x cos 8 x dx 4 5

0 0

2
3 1 4 2 8
Solution: (i)  sin x cos 5 x dx     
4
,
0
9 7 5 3 315

2
5 3 1 7 5 3 1  5
(ii)  sin x cos 8 x dx         
6
,
0
14 12 10 8 6 4 2 2 4096

Summary:

Reduction formula is a formula which expresses a given integral in terms of

another simple integral. In this chapter we have derived reduction formulae
for the followings.
 sin n 1 x cos x  n 1
  sin n x dx    sin n  2 x dx ,
n n
cos x sin x  1
n 1
n 
  cos n x dx    cos n  2 x dx ,
n n
n 1
tan x
  tan n x dx    tan n  2 x dx ,
n 1

Institute of Lifelong Learning, University of Delhi pg. 16

 Reduction Formulae

csc n  2 x cot x  n  2 
  csc x dx    csc n  2 x dx
n

n 1 n 1
sec n  2 x tan x  n  2 
  sec n x dx    sec n  2 x dx ,
n 1 n 1
 cot n 1 x
  cot x dx    cot n  2 x dx
n

n 1
cos m 1 x sin n 1 x m 1
 sin x cos x dx    sin n x cos m  2 x dx , m  1
n m

nm nm

 cos m 1 x sin n 1 x n 1

nm

nm  sin n  2 x cos m x dx , n  1

  log x  dx    log x  dx  x  log x   n   log x 

n n n n 1
 dx ,
 n 1 n  3 n  5 3 1 

 n  n  2  n  4  4  2  2 , when n is even and n  2.
2

  sin x dx  
n

0  n 1  n  3  n  5  4  2 , when n is odd and n  3.

 n n  2 n  4 5 3
 n 1 n  3 n  5 3 1 

 n  n  2  n  4  4  2  2 , when n is even and n  2.
2

  cos x dx  
n

0  n 1  n  3  n  5  4  2 , when n is odd and n  3.


 n n2 n4 5 3
 n 1 n 3 n 5 1 m 1 m  3 1 
 n  m  n  m  2  n  m  4  2  m  m  m  2  2  2 ,

 /2
 when n is even, m is even
  sin n x cos m x dx  
0  n 1  n  3  n  5  1  m 1  m  3  2 ,
n m n m 2 n m4 2 m m m2 3

 when n is even, m is odd

We have used method of integration by parts to derive reduction formulae

for integrals. These are the formulas that express an integral involving a
power of a function in terms of an integral that involves a lower power of
that function. It has been also observed that by using this method we can
derive many integrations related to algebraic and trigonometric functions.

Exercises:

(1) Evaluate :

Institute of Lifelong Learning, University of Delhi pg. 17

 Reduction Formulae

(a)  sin 5 x dx (b)  cos 10 x dx (c)  tan 11 x dx

(d)  csc 8 x dx (e)  sec 5 x dx (f)  cot 9 x dx
(2) Show that
5 1  1 5 5 
 sin x cos 6 x dx  x  sin x  cos 7 x  cos 5 x  cos 3 x  cos x 
2

128 8  6 24 16 
(3) Show that
1 1
 tan x sec 4 x dx  tan 5 x  tan 3 x
2

5 3

(4) Evaluate :
 /2  /2

 sin 15 x dx  cos
10
(a) (b) x dx
0 0
 /2  /2

 sin 4 x cos 6 x dx  sin

6
(c) (d) x cos 7 x dx
0 0

(5) Evaluate
 /4 

 tan  cos
5 2
(a) x dx (b) 5 x dx
0 
 /2  /4

 csc 4 x dx  sec
6
(c) (d) x dx
/4 0

(6) Prove that

3
1
 a

  a 
2
(a)  x 1  x 2 2
dx  (b)  x 4 2
 x 2 dx  a2
0
32 0
32
x m cos nx m x m 1 sin nx m  m 1 m  2
(7) Prove that  x sin n x dx      x sin n x dx
m

n n2 n2
(8) Obtain a reduction formula for  cos
m
x sin n x dx
(9) Obtain a reduction formula for  cos
m
x sin n x dx

2
n 1 1
(10) If I n   x sin x dx and n  1, then provethat I n  In2  2
n

0
n n

Institute of Lifelong Learning, University of Delhi pg. 18

 Reduction Formulae

Institute of Lifelong Learning, University of Delhi pg. 19