PDF Dr. Akul Rana, Assistant Prof., Dept. of Math, Narajole Raj College

C1T (Sem-I), Reduction Formulae Page 1 Reduction Formulae:

Reduction formulae are formulae that may be used repeatedly to express the integral of a complicated function in terms of simpler ones. The reduction formulae are generally obtained by the application of the integration by parts.

1. Reduction formula for



x e dx^{n ax} , n being a positive integer.

Let

In =



x e dx^{n ax} ,

= ^{1 1} ₁

ax ax ax ax

n n n n ax n

n

e e e n e

x nx dx x x e dx x I

a a a a a

− −

−



= −



= − − ^.

2. Reduction formula for



sinⁿxdx, n being a positive integer greater than 1.

Let In =



sinⁿxdx⁼



^{sinn−1xsinxdx}

=sinⁿ⁻¹x( cos )− x −



(n−1) sinⁿ⁻²xcos ( cos )x − x dx = −sinⁿ⁻¹xcosx+(n−1) sin



ⁿ⁻²x(1 sin− ² x dx) = −sinⁿ⁻¹xcosx+(n−1)I_n−2− −(n 1)I_n, Therefore,

1

2

sinⁿ cos 1

n n

x x n

I I

n n

−

−

− −

= + .

If

2

0

sinⁿ

Jn xdx



=



^{, then}

2

2 0

1 1

sin cos

n n

n n

J J

n

x x

n



−

− −

 

+

=− 

 

^,

i.e., J_n ^{n 1}J_{n 2}

n ⁻

= − .

3. Reduction formula for



cosⁿxdx, n being a positive integer greater than 1.

Similar to previous one .

C1T (Sem-I), Reduction Formulae Page 2 4. Reduction formula for



tanⁿxdx, n being a positive integer greater than 1.

Let I_n =



tanⁿxdx ⁼



^{tann−2xtan2xdx =}



^{tann−2x(sec2x−1)dx}

=



tanⁿ⁻²xsec²xdx−I_n−2

1

2

tan 1

n

n

x I n

−

= − −

−

5. Reduction formula for



cotⁿxdx, n being a positive integer greater than 1.

Similar to previous Ex.

6. Reduction formula for



secⁿxdx, n being a positive integer greater than 1.

Let

I_n =



secⁿxdx ⁼



^{secn−2xsec2xdx}

=secⁿ⁻² xtanx−



(n−2) secⁿ⁻³xsec tan . tanx x xdx =secⁿ⁻²xtanx− −(n 2) sec



ⁿ⁻²x(sec²x−1)dx =secⁿ⁻²xtanx− −(n 2)(I_n−I_n−2)

Transposing, we have

2

2

sec tan 2

1 1

n

n n

x x n

I I

n n

−

−

= − −

− − .

7. Reduction formula for



cosecⁿxdx, n being a positive integer greater than 1.

Let I_n =



cosecⁿxdx=



cosecⁿ⁻²xcosec dx²

Proceeding in the same way as in previous ex., we have

C1T (Sem-I), Reduction Formulae Page 3

2

2

co sec cot 2

1 1 .

n

n n

x x n

I I

n n

−

−

= − − −

− −

Ex. 1. Prove that

1 1

m

m,n , 2

sin cos 1

I = sin cos

m n

n

m n

x x n

x xdx I

m n m n

+ −

−

= + −

+ +



, m, n being

positive integers greater than 1.

Solution: I_m,n= sin



^mxcosⁿxdx⁼



^{(sinmxcos ) cosx n−1xdx}

[applying int. by parts, taking cos^n-1x as first function and sin^m x cos x as second function]

=

1 1

1 sin 2 sin

cos ( 1) cos ( sin )

1 1

m m

n x n x

x n x x dx

m m

+ +

− − − − −

+



+

=

1 1

2 2

sin cos 1

cos sin sin

1 1

m n

n m

x x n

x x xdx

m m

+ −

− −

+ + +



=

1 1

2 2

sin cos 1

cos sin (1 s )

1 1

m n

n m

x x n

x x co x dx

m m

+ −

− −

+ −

+ +



=

1 1

, 2 ,

sin cos 1 1

1 1 1

m n

m n m n

x x n n

I I

m m m

+ −

−

− −

+ −

+ + +

Simplifying,

1 1

, , 2

sin cos 1

.

m n

m n m n

x x n

I I

m n m n

+ −

−

= + −

+ +

Ex. 2. Prove that

1 1

m

m,n 2,

sin cos 1

I = sin cos

m n

n

m n

x x m

x xdx I

m n m n

− +

−

= + −

+ +



, m, n being

positive integers greater than 1.

C1T (Sem-I), Reduction Formulae Page 4 Solution: I_m,n= sin



^mxcosⁿxdx ^{= sin}



^{m-1x(cosnxsin )x dx}

[Hints: apply int. by parts, taking sin^m-1x as first function and sinx cosⁿ x as second function]

Ex. 3. Obtain a reduction formula for sin



^mxcosⁿxdx where either m or n or both are negative integers .

Solution: Let I_m,n =



sin^mxcosⁿ xdx^.

Then

1 1

, , 2

sin cos 1

.

m n

m n m n

x x n

I I

m n m n

+ −

−

= + −

+ + [ex. 1]

Replacing n by n+2,

1 1

, 2 ,

sin cos 1

2 2

m n

m n m n

x x n

I I

m n m n

+ +

+

= + +

+ + + +

Transposing,

1 1

, , 2

sin cos 2

( 1 0)

1 1

m n

m n m n

x x m n

I I n

n n

+ +

+

= − + + + + 

+ + which is a reduction

formula for Im, n.

Again,

1 1

m

m,n 2,

sin cos 1

I = sin cos

m n

n

m n

x x m

x xdx I

m n m n

− +

−

= + −

+ +



^{[Ex. 2]}

Replacing m by m+2 and transposing, we get

1 1

, 2,

sin cos 2

( 1 0)

1 1

m n

m n m n

x x m n

I I m

m m

+ +

+

= + + + + 

+ + which is another

reduction formula for Im, n.

C1T (Sem-I), Reduction Formulae Page 5 Ex. 4. Obtain reduction formula for sin

cos

m n

xdx



x , where m and n are both positive integers.

Solution: Let _, sin cos

m

m n n

I xdx

=



x

1 1

2

1 1

1 1

sin cos ( sin cos ) sin ( cos )( sin )

( )

cos cos

sin sin

=

cos cos

m m m m n

n n

m m

n n

d x x m x x x n x x

dx x x

x x

m n

x x

− −

− +

− +

− −

=

+

Integrating, we have

sin _{1, 1 1, 1} cos =

m

m n m n

n

x mI nI

x ^{− −} + ^{+ +}

Replacing m by m-1 and n by n-1, we get

1

2, 2 ,

1

sin =( 1) ( 1)

cos

m

m n m n

n

x m I n I

x

−

− −

− − + −

Hence

1

, 1 2, 2

1 sin 1

1 cos - 1

m

m n n m n

I x m I

n x n

−

− −

−

= −

− −

Ex. 5. If

2

0 nsin In x xdx



=



and n>1, show that ( 1) ₂ ( ) ¹

2

n

n n

I n n I n  ₋

+ − − =

Solution:

2

0 nsin In x xdx



=



^{= 2 2 1}

0 0

( cos )

( cos )

ⁿ

n nx x dx

x x

 

− − −

 −  

 

=

2 1 0

n cos

n x xdx





−

2 2 2

0 0

1

(sin )

( 1) ⁿ sin

n

x

n

x

n n x xdx

 

− −

=

   

−



−

C1T (Sem-I), Reduction Formulae Page 6

( ) ¹ ( 1) ₂ 2

n

n  ₋ n n In

= − − −

Hence ( 1) ₂ ( ) ¹ 2

n

n n

I n n I n  −

+ − − = .

Ex. 6. If I_{m n,} =



x^m(1−x dx)ⁿ , show that (m+n I) _{m n, −1} =x^m+1(1−x)ⁿ⁻¹+ −(n 1)I_{m n, −2}.

Solution: I_{m n,} =



x^m(1−x dx)^{n = −(1 x)n} _m^xm₊⁺¹₁⁻



^{n(1−x)n−1( 1)−} _m^xm₊⁺¹₁^dx

1

(1 ) (1 ) 1(1 1)

1 1

m

n x n n m

x x x x dx

m m

+ −

= − − − − −

+ +



1

, , 1

(1 )

1 1 1

m n

m n m n

x n n

x I I

m m m

+

= − − + −

+ + +

or,

1

, , 1

(1 ) (1 )

1 1 1

m n

m n m n

n x n

I x I

m m m

+

+ = − + −

+ + +

or, (m+ +n 1)I_{m n,} = −(1 x x)^{n m+1}+nI_{m n, −1}

Replacing n by n-1, we get

(m+n I) _{m n, −1} =x^m+1(1−x)ⁿ⁻¹+ −(n 1)I_{m n, −2}.

Ex. 7. Show that

2 2

0 0

( 1)( 3)( 5) 1

sin cos

( 2)( 4) 2 2

n n n n n

xdx xdx

n n n

 



− − −

= =

− −

 

, if n be any even

positive integer and n>1.

C1T (Sem-I), Reduction Formulae Page 7 Solution: Let

2

0

sinⁿ

In xdx



=



^{. Then In 2 2}

0 0

sin ( ) cos

2

n n

x dx xdx

 

=



 − =



^.

2 2

0 0

sinⁿ xdx cosⁿ xdx.

 





=



Now,

2

0

sinⁿ

In xdx



=



^{=2 1}

0

sinⁿ x(sin )x dx





−

=

2 2 2

0 0

1 ( 1) sin cos ( cos )

sin

ⁿ

x ( cos ) x

^{n n x x x dx}

 

− − − − −

 −  

 

=

2

2 2

0

(n 1) sinⁿ x(sin x 1)dx



− −



− − ^{=(n−1)In−2 −(n−1)In}

Transposing, we have

I_n ^{(n 1)}I_{n 2}

n ⁻

= −

=( 1) ( 3) ₄

( 2) ⁿ

n n

n n I ⁻

− −

−

=( 1) ( 3) 1 ₀

( 2) 2

n n

n n I

− −

− , since n is even.

=( 1) ( 3) 1

( 2) 2 2

n n

n n



− −

− .

Ex. 8. Show that

2 2

0 0

( 1) ( 3) 4 2

sin cos

( 2) 3 1

n n n n

xdx xdx

n n

 

− −

= =

 

− , if n be any odd positive

integer and n>1.

Try Yourself

C1T (Sem-I), Reduction Formulae Page 8 Ex. 9. Prove that

2

0

( 1)( 3) 1( 1)( 3) 1

sin cos

( )( 2) 1 2

m n m m n n

x xdx

m n m n





− − − −

= + + −



, if m and n are

positive integers and both even. (m>1, n>1).

Solution: Let

2 ,

0

sin^m cosⁿ

Im n x xdx



=



Then

2

, , 2

0

1 1

1 .

sin cos

m n m n

m n

I n I

m n

x x

m n



−

+ −

= + −

+

 

 + 

 

[by Ex 1]

= n ¹ _{m n, 2} m nI ⁻

− +

( 1) ( 3) _{, 4}

=( ) ( 2) ^{m n}

n n

m n m n I ⁻

− −

+ + −

( 1) ( 3) 1 _,0

=( ) ( 2) 2 ^m

n n

m n m n m I

− −

+ + − + …(i)

Now,

2 ,0

0

( 1) ( 3) 1

sin ( 2) 2 2

m m

m m

I xdx

m m





− −

= =



− [by Ex. 6]

Therefore from (i),

2

0

( 1)( 3) 1( 1)( 3) 1

sin cos

( )( 2) 1 2

m n m m n n

x xdx

m n m n





− − − −

= + + −



^.

C1T (Sem-I), Reduction Formulae Page 9 Ex. 10: If I_{m n,} =



cos^mxcosnxdx, show that

_, cos^m sin _{1, 1}

m n m n

x nx m

I I

m n m n ^{− −}

= +

+ +

n^sinnx^cosx₂ m₂^cosmx^sinxcos^m1 m m₂^{( 1)}_{2 m 2,n}

x I

n m n m

− −

− −

= −

− − .

Solution:

I_{m n,} =



cos^mxcosnxdx

=cos sin ¹ sin

cos ( sin )

m

x nx m nx

m x x dx

n n

−



− −

=cos sin ¹

cos sin sin

m

x nx m m

x x nxdx

n n

+



−

=cos sin ¹

cos { cos cos cos( 1) }

m

x nx m m

x x nx n x dx

n n

+



− − + −

cos^m sin _{, 1, 1}

m n m n

x nx m m

I I

n n n ^{− −}

= − +

Therefore, _, cos^m sin _{1, 1}

m n m n

x nx m

I I

m n m n ^{− −}

= +

+ + .

Again I_{m n,} =



cos^mxcosnxdx

=cos sin ¹ sin

cos ( sin )

m

x nx m nx

m x x dx

n n

−



− −

=cos sin ¹

(cos sin ) sin

m

x nx m m

x x nxdx

n n

+



−

C1T (Sem-I), Reduction Formulae Page 10

1 2 1

cos sin cos cos

[cos sin ( ) {( 1) cos ( sin ) sin cos .cos }( ) ]

m

m m m

x nx m nx nx

x x m x x x x x dx

n n n n

− − −

= + − −



− − + −

1

, 2, ,

2 2 2 2

cos sin ( 1) ( 1)

cos sin cos

m

m

m n m n m n

x nx m m m m m m

x x nx I I I

n n n n n

− −

− −

= − + − +

Transposing, we have

Im, n sin cos ₂ cos₂ sin ¹ ₂( 1)_{2 2,}

cos^m _{m n}

n nx x m mx x m m

x I

n m n m

− −

− −

= −

− − .

Ex.11: If I_{m n,} =



x^mcosec xdxⁿ , show that

2 1 1

, , 2 2, 2

(n−1)(n−2)I_{m n} =(n−2) I_{m n−} +m m( −1)I_{m− n−} −x^m− cosecⁿ⁻ x m{ sinx+(n−2) cos }x x

Solution: I_{m n,} =



x^mcosec xdx^{n =cosec xn} _m^xm₊⁺¹₁⁻



^{ncosecn−1x( cos− ecxcot )x} _m^xm₊⁺¹₁^dx

(m+1)I_{m n,} =x^m+1cosec xⁿ +n x



^m+1cosec xⁿ cotxdx

2 2

1 1 2

cos [cos cot { cos ( cos cot ) cot cos ( cos )} ]

2 2

m m

m n n x n n x

x ec x n ec x x n ec x ecx x x ec x ec x dx

m m

+ +

+ −

= + − − + −

+



+

1 2 1 2 2 2 2 2

(m 1)(m 2)I_{m n}, (m 2)x^m+ cosec xⁿ nx^m+ cosecⁿ⁺xcosx n x^m+ cosec xⁿ (cosec x 1)dx n x^m+ cosecⁿ⁺ xdx

 + + = + + +  − + 

=(m+2)x^m+1cosec xⁿ +nx^m+2cosecⁿ⁺¹xcosx+n n( +1)I_m+2,n+2−n²I_{m 2,+ n} Replacing m by m-2, n by n-2 and transposing, we have

2 1 1

, , 2 2, 2

(n−1)(n−2)I_{m n} =(n−2) I_{m n−} +m m( −1)I_{m− n−} −x^m− cosecⁿ⁻ x m{ sinx+(n−2) cos }x x

Ex.12: If I_n =



(sinx+cos )x dxⁿ , show that

nI_n =(sinx+cos )x ⁿ⁻¹(sinx−cos )x +2(n−1)I_n−2.

Solution: I_n =



(sinx+cos )x dx^{n =}



^{(sinx+cos )x n−1(sinx+cos )x dx}

C1T (Sem-I), Reduction Formulae Page 11

1 2

(sinx cos )x ⁿ⁻ (sinx cos )x (n 1)(sinx cos )x ⁿ⁻ (cosx sin )(sinx x cos )x dx

= + − −



− + − −

1 2

(sinx cos )x ⁿ⁻ (sinx cos )x (n 1) (sinx cos )x ⁿ⁻ (1 2sin cos )x x dx

= + − + −



+ −

1 2 2

(sinx cos )x ⁿ⁻ (sinx cos )x (n 1) (sinx cos )x ⁿ⁻ {2 (sinx cos ) }x dx

= + − + −



+ − +

1

(sinx cos )x ⁿ⁻ (sinx cos )x 2(n 1)I_n−2 (n 1)I_n

= + − + − − −

Transposing we get,

nI_n =(sinx+cos )x ⁿ⁻¹(sinx−cos )x +2(n−1)I_n−2 Ex.13: If 𝑰_𝒏 = ∫(𝑎+𝑏 sin 𝑥)^{𝑑𝑥 𝑛} , show that

(𝒏 − 𝟏)(𝒂^𝟐− 𝑏²)𝐼_𝑛 = 𝑏 cos 𝑥

(𝑎 + 𝑏 sin 𝑥)^𝑛−1+ (2𝑛 − 3)𝑎𝐼_𝑛−1− (𝑛 − 2)𝐼_𝑛−2 Solution: Let 𝑢 = ^{cos 𝑥}

(𝑎+𝑏 sin 𝑥)^𝑛−1. Then

𝑑𝑢

𝑑𝑥 = (𝑎 + 𝑏 sin 𝑥)^𝑛−1(− sin 𝑥) − (𝑛 − 1)(𝑎 + 𝑏 sin 𝑥)^𝑛−2. 𝑏 cos 𝑥. cos 𝑥 (𝑎 + 𝑏 sin 𝑥)^2𝑛−2

= −(𝑎 + 𝑏 sin 𝑥) sin 𝑥 − (𝑛 − 1). 𝑏(1 − sin²𝑥) (𝑎 + 𝑏 sin 𝑥)^𝑛

= 𝐴 + 𝐵(𝑎 + 𝑏 sin 𝑥) + 𝐶(𝑎 + 𝑏 sin 𝑥)²

(𝑎 + 𝑏 sin 𝑥)^𝑛 … … … . (𝑖)

Therefore, 𝐴 + 𝐵𝑎 + 𝐶𝑎² = (1 − 𝑛)𝑏 𝐵𝑏 + 2𝑎𝑏𝐶 = −𝑎 𝐶𝑏² = −𝑏 + (𝑛 − 1)𝑏

From the last equation,

C1T (Sem-I), Reduction Formulae Page 12 𝐶 =^𝑛−2

𝑏

From second equation , 𝐵 = −^{(2𝑛−3)𝑎}

𝑏

Then, from the first equation 𝐴 =^{(𝑛−1)(𝑎2−𝑏2)}

𝑏

Therefore, (i) becomes 𝑑𝑢 =(𝑛 − 1)(𝑎²− 𝑏²)

𝑏

𝑑𝑥

(𝑎 + 𝑏 sin 𝑥)^𝑛−(2𝑛 − 3)𝑎 𝑏

𝑑𝑥

(𝑎 + 𝑏 sin 𝑥)^𝑛−1 𝑛 − 2

𝑏 +𝑛 − 2

𝑏

𝑑𝑥 (𝑎 + 𝑏 sin 𝑥)^𝑛−2

Integrating and transposing, we get

(𝒏 − 𝟏)(𝒂^𝟐− 𝑏²)𝐼_𝑛 = 𝑏 cos 𝑥

(𝑎 + 𝑏 sin 𝑥)^𝑛−1+ (2𝑛 − 3)𝑎𝐼_𝑛−1− (𝑛 − 2)𝐼_𝑛−2

Exercise:

1.

If 𝑰_𝒎,𝒏= ∫ 𝒙_𝟎^{𝟏 𝒎}(𝟏 − 𝒙)^𝒏𝒅𝒙, where m, n are positive integers, then prove that (𝒎 + 𝒏)𝑰_𝒎,𝒏 = 𝑛𝐼_{𝑚,𝑛−1} . Hence deduce that 𝐼_𝑚,𝑛 = ^𝑚!𝑛!

(𝑚+𝑛+1)!.

2.

Find a reduction formula for

2 ,

0

cos^m sin

Im n x nxdxdx



=



; m, n being positive

integers and hence deduce that

2 3

, 1

1 2 2 2

[2 ]

2 3

2

m

m n m

I = ⁺ + + + + m .

3.

Obtain a reduction formula for

( cos )ⁿ dx a+b x



; n being a positive integer greater than 1 and a² ≠ b².

C1T (Sem-I), Reduction Formulae Page 13

4.

Find a reduction formula for _{2 2}

( )ⁿ

dx x +a



where n is a positive integer.

5.

If ₂

( 1)

n n

I dx

= x



+ , show that (2 2) (2 3) _{1 2 1}

( 1)

n n n

n I n I x

− x −

− − − =

+ .

6.

If

2 n n

I x dx

ax c

=



+ then show that anI_n =xⁿ⁻¹ ax² + −c (n−1)cI_n−2.

7.

If I_n =

1 2 0

(1−x )ⁿdx



, prove that (2n+1)I_n =2nI_n−1. Hence find In.

8.

If

1

1 0

tan ,

n

un =



x ⁻ xdx prove that for n>2, ( 1) ( 1) ₂ ¹

n n 2

n u n u

n

 + + − − = − .