pertemuan 8

Chapter 44

Reduction formulae

44.1 Introduction

When using integration by parts in Chapter 43, anintegral such as

∫x2ex dx requires integration by

parts twice. Similarly,∫

x3ex dx requires integra-tion by parts three times. Thus, integrals such as∫

x5ex dx,∫

x6 cos x dx and∫

x8 sin2x dx for example,would take a long time to determine using integra-tion by parts. Reduction formulae provide a quickermethod for determining such integrals and the methodis demonstrated in the following sections.

44.2 Using reduction formulae forintegrals of the form

∫xnexdx

To determine∫

xnex dx using integration by parts,

let u = xn from which,

du

dx= nxn−1 and du =nxn−1 dx

and dv= ex dx from which,

v=∫

ex dx =ex

Thus,∫

xnex dx = xnex −∫

ex nxn−1 dx

using the integration by parts formula,

= xnex −n∫

xn−1ex dx

The integral on the far right is seen to be of the sameform as the integral on the left-hand side, except that nhas been replaced by n −1.Thus, if we let,∫

xnex dx = In ,

then∫

xn−1ex dx = In−1

Hence∫

xnex dx = xnex −n∫

xn−1ex dx

can be written as:

In = xnex − nIn−1 (1)

Equation (1) is an example of a reduction formula sinceit expresses an integral in n in terms of the same integralin n −1.

Problem 1. Determine∫

x2ex dx using areduction formula.

Using equation (1) with n =2 gives:∫x2ex dx = I2 = x2ex −2I1

and I1 = x1ex −1I0

I0 =∫

x0ex dx =∫

ex dx =ex +c1

Hence I2 = x2ex −2[xex −1I0]

= x2ex −2[xex −1(ex +c1)]

i.e.∫

x2ex dx= x2ex −2xex+2ex +2c1

= ex(x2 −2x+2)+c

(where c =2c1)

As with integration by parts, in the following examplesthe constant of integration will be added at the last stepwith indefinite integrals.

Problem 2. Use a reduction formula to determine∫x3ex dx .

Reduction formulae 427

From equation (1), In = xnex −n In−1

Hence∫

x3ex dx = I3 = x3ex −3I2

I2 = x2ex −2I1

I1 = x1ex −1I0

and I0 =∫

x0ex dx =∫

ex dx =ex

Thus∫

x3ex dx = x3ex −3[x2ex −2I1]

= x3ex −3[x2ex −2(xex − I0)]

= x3ex −3[x2ex −2(xex −ex )]

= x3ex −3x2ex +6(xex −ex )

= x3ex −3x2ex +6xex −6ex

i.e.∫

x3ex dx= ex(x3−3x2+6x−6)+c

Now try the following exercise

Exercise 169 Further problems on usingreduction formulae for integrals of the form∫

xnex dx

1. Use a reduction formula to determine∫x4ex dx .

[ex (x4 −4x3 +12x2 −24x +24)+c]

2. Determine∫

t3e2t dt using a reduction for-mula. [

e2t( 1

2 t3 − 34 t2 + 3

4 t − 38

) +c]

3. Use the result of Problem 2 to evaluate∫ 10 5t3e2tdt, correct to 3 decimal places.

[6.493]


∫xn cosx dx

and∫

xn sin x dx

(a)∫

xncosxdx

Let In = ∫xn cos x dx then, using integration by parts:

if u = xn thendu

dx=nxn−1

and if dv= cos x dx then

v=∫

cos x dx = sin x

Hence In = xn sin x −∫

(sin x)nxn−1 dx

= xn sin x − n∫

xn−1 sin x dx

Using integration by parts again, this time withu = xn−1:

du

dx= (n − 1)xn−2 , and dv= sin x dx,

from which,

v =∫

sin x dx =−cos x

Hence In = xn sin x − n

[xn−1(−cos x)

−∫

(−cos x)(n − 1)xn−2 dx

]= xn sin x +nxn−1 cos x

−n(n − 1)

∫xn−2 cos x dx

i.e.In =xnsinx + nxn−1cosx

−n(n−1)In−2(2)

Problem 3. Use a reduction formula to determine∫x2 cos x dx .

Using the reduction formula of equation (2):∫x2 cos x dx = I2

= x2 sin x +2x1 cos x − 2(1)I0

and I0 =∫

x0 cos x dx

=∫

cos x dx = sin x

Hence ∫x2cosx dx= x2sinx+2x cosx −2 sinx+c

Problem 4. Evaluate∫ 2

1 4t3 cos t dt , correct to 4significant figures.

Let us firstly find a reduction formula for∫t3 cos t dt .

428 Higher Engineering Mathematics

From equation (2),∫t3 cos t dt = I3 = t3 sin t + 3t2 cos t − 3(2)I1

and

I1 = t1 sin t + 1t0 cos t − 1(0)In−2

= t sin t + cos t

Hence∫t3 cos t dt = t3 sin t + 3t2 cos t

− 3(2)[t sin t + cos t ]

= t3sin t + 3t2cos t − 6t sin t − 6 cos t

Thus∫ 2

14t3 cos t dt

= [4(t3 sin t +3t2 cos t −6t sin t −6 cos t)]21

= [4(8 sin2+12 cos2−12 sin2 − 6 cos2)]

− [4(sin 1+3 cos1 − 6 sin1−6 cos1)]

= (−24.53628)−(−23.31305)

= −1.223

Problem 5. Determine a reduction formulafor

∫ π0 xn cos x dx and hence evaluate∫ π

0 x4 cos x dx , correct to 2 decimal places.

From equation (2),

In = xn sin x + nxn−1 cos x − n(n − 1)In−2.

hence∫ π

0xn cos x dx = [xn sin x + nxn−1 cos x]π0

−n(n − 1)In−2

= [(πn sinπ + nπn−1 cosπ)

−(0 + 0)] − n(n − 1)In−2

= −nπn−1 − n(n − 1)In−2

Hence∫ π

0x4 cosx dx = I4

=−4π3 − 4(3)I2 since n = 4

When n =2,∫ π

0x2 cos x dx = I2 = −2π1 − 2(1)I0

and I0 =∫ π

0x0 cos x dx

=∫ π

0cos x dx

= [sin x]π0 = 0

Hence∫ π

0x4 cos x dx = −4π3 − 4(3)[−2π − 2(1)(0)]

= −4π 3 + 24π or −48.63,

correct to 2 decimal places.

(b)∫

xnsinx dx

Let In = ∫xn sin x dx

Using integration by parts, if u =xn thendu

dx=nxn−1 and if dv= sin x dx then

v= ∫sin x dx =−cos x . Hence∫xn sin x dx

= In = xn(−cos x)−∫

(−cos x)nxn−1 dx

= −xn cos x + n∫

xn−1 cos x dx

Using integration by parts again, with u = xn−1, from

which,du

dx=(n − 1)xn−2 and dv= cos x , from which,

v= ∫cos x dx = sin x . Hence

In = −xn cos x + n

[xn−1(sin x)

−∫

(sin x)(n − 1)xn−2 dx

]

= −xn cos x + nxn−1(sin x)

− n(n − 1)

∫xn−2 sin x dx

i.e. In =−xncosx + nxn−1 sinx − n(n − 1)In−2 (3)


Problem 6. Use a reduction formula to determine∫x3 sin x dx .

Using equation (3),∫x3 sin x dx = I3

= −x3 cos x + 3x2 sin x − 3(2)I1

and I1 = −x1 cos x + 1x0 sin x

= −x cosx + sin x

Hence∫x3 sin x dx = −x3 cos x + 3x2 sin x

−6[−x cos x + sin x]

= −x3cosx + 3x2sinx

+6x cosx −6 sinx + c

Problem 7. Evaluate∫ π

2

03θ4 sinθ dθ , correct to 2

decimal places.

From equation (3),

In = [−xn cos x + nxn−1(sin x)]π2

0 − n(n − 1)In−2

=[(

−(π

2

)ncos

π

2+ n

(π

2

)n−1sin

π

2

)− (0)

]− n(n − 1)In−2

= n(π

2

)n−1 −n(n − 1)In−2

Hence

∫ π2

03θ4 sinθ dθ = 3

∫ π2

0θ4 sinθ dθ

= 3I4

= 3

[4(π

2

)3 − 4(3)I2

]

I2 = 2(π

2

)1 − 2(1)I0 and

I0 =∫ π

2

0θ0 sinθ dθ = [−cos x]

π20

= [−0 − (−1)] = 1

Hence

3∫ π

2

0θ4 sinθ dθ

= 3I4

= 3

[4(π

2

)3 − 4(3)

{2(π

2

)1 − 2(1)I0

}]

= 3

[4(π

2

)3 − 4(3)

{2(π

2

)1 − 2(1)(1)

}]

= 3

[4(π

2

)3 − 24(π

2

)1 + 24

]= 3(15.503 − 37.699 + 24)

= 3(1.8039) = 5.41


Exercise 170 Further problems onreduction formulae for integrals of the form∫

xncosx dx and∫

xnsinx dx

1. Use a reduction formula to determine∫x5 cos x dx .⎡

⎢⎣ x5 sin x + 5x4 cos x − 20x3 sin x

−60x2 cos x + 120x sin x

+120 cosx + c

⎤⎥⎦

2. Evaluate∫ π

0 x5 cos x dx , correct to 2 decimalplaces. [−134.87]

3. Use a reduction formula to determine∫x5 sin x dx .⎡

⎢⎣−x5 cos x + 5x4 sin x + 20x3 cos x

−60x2 sin x − 120x cos x

+120 sin x + c

⎤⎥⎦

4. Evaluate∫ π

0 x5 sin x dx , correct to 2 decimalplaces. [62.89]


∫sinn xdx

and∫

cosn xdx

(a)∫

sinnx dx

Let In = ∫sinn x dx ≡ ∫

sinn−1 x sin x dx from laws ofindices.Using integration by parts, let u = sinn−1 x , from which,


du

dx= (n − 1)sinn−2 x cos x and

du = (n − 1)sinn−2 x cos x dx

and let dv= sin x dx , from which,v=∫

sin x dx =−cos x . Hence,

In =∫

sinn−1 x sin x dx

= (sinn−1 x)(−cos x)

−∫

(−cos x)(n − 1)sinn−2 x cos x dx

= −sinn−1 x cos x

+ (n − 1)

∫cos2 x sinn−2 x dx


+ (n − 1)

∫(1 − sin2 x)sinn−2 x dx


+ (n − 1)

{∫sinn−2 x dx −

∫sinn x dx

}

i.e. In = −sinn−1 x cos x

+ (n − 1)In−2 −(n − 1)In

i.e. In + (n − 1)In

= −sinn−1 x cos x + (n − 1)In−2

and n In = −sinn−1 x cos x + (n − 1)In−2

from which,∫sinn x dx =

In = −1n

sinn−1 xcosx + n − 1n

In−2 (4)

Problem 8. Use a reduction formula to determine∫sin4 x dx .

Using equation (4),∫sin4 x dx = I4 = −1

4sin3 x cos x + 3

4I2

I2 = −1

2sin1 x cos x + 1

2I0

and I0 =∫

sin0 x dx =∫

1 dx = x

Hence∫sin4 x dx = I4 = −1

4sin3 x cos x

+ 3

4

[−1

2sin x cos x + 1

2(x)

]

= −14

sin3x cosx − 38

sinx cosx

+ 38

x + c

Problem 9. Evaluate∫ 1

0 4 sin5 t dt , correct to 3significant figures.

Using equation (4),∫sin5 t dt = I5 = −1

5sin4 t cos t + 4

5I3

I3 = −1

3sin2 t cos t + 2

3I1

and I1 = −1

1sin0 t cos t + 0 = −cos t

Hence∫sin5 t dt = −1

5sin4 t cos t

+ 4

5

[−1

3sin2 t cos t + 2

3(−cos t)

]

= −1

5sin4 t cos t − 4

15sin2 t cos t

− 8

15cos t + c

and∫ t

04 sin5 t dt

= 4

[−1

5sin4 t cos t

− 4

15sin2 t cos t − 8

15cos t

]1

0

= 4

[(−1

5sin4 1 cos1 − 4

15sin2 1 cos1

− 8

15cos1

)−

(−0 − 0 − 8

15

)]


= 4[(−0.054178 − 0.1020196

− 0.2881612)− (−0.533333)]

= 4(0.0889745) = 0.356

Problem 10. Determine a reduction formula for∫ π2

0sinn x dx and hence evaluate

∫ π2

0sin6 x dx

From equation (4),∫sinn x dx

= In =− 1

nsinn−1 x cos x + n − 1

nIn−2

hence∫ π2

0sinn x dx =

[−1

nsinn−1 x cos x

] π2

0+ n −1

nIn−2

= [0−0]+ n −1

nIn−2

i.e. In = n−1n

In−2

Hence∫ π2

0sin6 x dx = I6 = 5

6I4

I4 = 3

4I2, I2 = 1

2I0

and I0 =∫ π

2

0sin0 x dx =

∫ π2

01 dx = π

2

Thus

∫ π2

0sin6 x dx = I6 = 5

6I4 = 5

6

[3

4I2

]

= 5

6

[3

4

{1

2I0

}]

= 5

6

[3

4

{1

2

[π

2

]}]= 15

96π

(b)∫

cosnx dx

Let In =∫cosn x dx ≡∫

cosn−1 x cos x dx from lawsof indices.

Using integration by parts, let u =cosn−1 x fromwhich,

du

dx=(n −1)cosn−2 x(−sin x)

and du =(n −1)cosn−2 x(−sin x)dx

and let dv =cos x dx

from which, v=∫

cos x dx = sin x

Then

In = (cosn−1 x)(sin x)

−∫

(sin x)(n −1)cosn−2 x(−sin x)dx

= (cosn−1 x)(sin x)

+(n −1)

∫sin2 x cosn−2 x dx

= (cosn−1 x)(sin x)

+(n −1)

∫(1− cos2 x)cosn−2 x dx

= (cosn−1x)(sin x)

+(n −1)

{∫cosn−2 x dx −

∫cosn x dx

}

i.e. In =(cosn−1 x)(sin x)+(n −1)In−2 −(n −1)In

i.e. In +(n −1)In =(cosn−1 x)(sin x)+(n −1)In−2

i.e. n In =(cosn−1 x)(sin x)+ (n −1)In−2

Thus In = 1n

cosn−1 x sinx + n−1n In−2 (5)

Problem 11. Use a reduction formula todetermine

∫cos4 x dx .

Using equation (5),∫cos4 x dx = I4 = 1

4cos3 x sin x + 3

4I2

and I2 = 1

2cos x sin x + 1

2I0

and I0 =∫

cos0 x dx

=∫

1 dx = x


Hence∫

cos4 x dx

= 1

4cos3 x sin x + 3

4

(1

2cos x sin x + 1

2x

)

= 14

cos3x sinx + 38

cosx sinx + 38

x + c

Problem 12. Determine a reduction formula

for∫ π

2

0cosn x dx and hence evaluate

∫ π2

0cos5 x dx

From equation (5),∫cosn x dx = 1

ncosn−1 x sin x + n −1

nIn−2

and hence∫ π2

0cosn x dx =

[1

ncosn−1 x sin x

] π2

0

+ n −1

nIn−2

= [0−0]+ n −1

nIn−2

i.e.∫ π

2

0cosnx dx= In = n−1

nIn−2 (6)

(Note that this is the same reduction formula as for∫ π2

0sinn x dx (in Problem 10) and the result is usually

known as Wallis’s formula).Thus, from equation (6),

∫ π2

0cos5 x dx = 4

5I3, I3 = 2

3I1

and I1 =∫ π

2

0cos1 x dx

= [sin x]π20 =(1−0)=1

Hence∫ π

2

0cos5 x dx = 4

5I3 = 4

5

[2

3I1

]

= 4

5

[2

3(1)

]= 8

15


Exercise 171 Further problems onreduction formulae for integrals of the form∫

sinn x dx and∫

cosn x dx

1. Use a reduction formula to determine∫sin7 x dx .⎡

⎢⎣− 1

7sin6 x cos x − 6

35sin4 x cos x

− 8

35sin2 x cos x − 16

35cos x +c

⎤⎥⎦

2. Evaluate∫ π

0 3 sin3 x dx using a reductionformula. [4]

3. Evaluate∫ π

2

0sin5 x dx using a reduction

formula.

[8

15

]

4. Determine, using a reduction formula,∫cos6 x dx .⎡

⎢⎣1

6cos5 x sin x + 5

24cos3 x sin x

+ 5

16cos x sin x + 5

16x +c

⎤⎥⎦

5. Evaluate∫ π

2

0cos7 x dx .

[16

35

]

44.5 Further reduction formulae

The following worked problems demonstrate furtherexamples where integrals can be determined usingreduction formulae.

Problem 13. Determine a reduction formula for∫tann x dx and hence find

∫tan7 x dx .

Let In =∫

tann x dx ≡∫

tann−2 x tan2 x dx

by the laws of indices

=∫

tann−2 x(sec2 x −1)dx

since 1+ tan2 x = sec2 x

=∫

tann−2 x sec2 x dx −∫

tann−2 x dx


=∫

tann−2 x sec2 x dx − In−2

i.e. In = tann−1 xn−1

−In−2

When n =7,

I7 =∫

tan7 x dx = tan6 x

6− I5

I5 = tan4 x

4− I3 and I3 = tan2 x

2− I1

I1 =∫

tan x dx = ln(sec x)

from Problem 9, Chapter 39, page 394

Thus∫tan7 x dx = tan6 x

6−

[tan4 x

4

−(

tan2 x

2− ln(sec x)

)]

Hence∫

tan7 x dx

=16

tan6 x − 14

tan4 x + 12

tan2 x

− ln(secx)+ c

Problem 14. Evaluate, using a reduction formula,∫ π2

0sin2 t cos6 t dt .

∫ π2

0sin2 t cos6 t dt =

∫ π2

0(1− cos2 t)cos6 t dt

=∫ π

2

0cos6 t dt −

∫ π2

0cos8 t dt

If In =∫ π

2

0cosn t dt

then∫ π2

0sin2 t cos6 t dt = I6 − I8

and from equation (6),

I6 = 5

6I4 = 5

6

[3

4I2

]

= 5

6

[3

4

(1

2I0

)]

and I0 =∫ π

2

0cos0 t dt

=∫ π

2

01 dt = [x]

π20 = π

2

Hence I6 = 5

6· 3

4· 1

2· π

2

= 15π

96or

5π

32

Similarly, I8 = 7

8I6 = 7

8· 5π

32Thus ∫ π

2

0sin2 t cos6 t dt = I6 − I8

= 5π

32− 7

8· 5π

32

= 1

8· 5π

32= 5π

256

Problem 15. Use integration by parts todetermine a reduction formula for

∫(ln x)n dx .

Hence determine∫(ln x)3 dx .

Let In = ∫(ln x)n dx .

Using integration by parts, let u =(ln x)n , from which,

du

dx= n(ln x)n−1

(1

x

)

and du = n(ln x)n−1(

1

x

)dx

and let dv=dx , from which, v= ∫dx = x

Then In =∫

(ln x)n dx

= (ln x)n(x)−∫

(x)n(ln x)n−1(

1

x

)dx


= x(ln x)n −n∫

(ln x)n−1 dx

i.e. In = x(lnx)n − nIn−1

When n =3,∫(ln x)3 dx = I3 = x(ln x)3 − 3I2

I2 = x(ln x)2 −2I1 and I1 = ∫ln x dx = x(ln x −1) from

Problem 7, page 422.

Hence∫(ln x)3 dx = x(ln x)3 − 3[x(ln x)2 − 2I1] + c

= x(ln x)3 − 3[x(ln x)2

− 2[x(ln x − 1)]] + c

= x(ln x)3 − 3[x(ln x)2

− 2x ln x + 2x] + c

= x(ln x)3 − 3x(ln x)2

+ 6x ln x − 6x + c

= x[(lnx)3 − 3(lnx)2

+6 lnx −6] + c


Exercise 172 Further problems onreduction formulae

1. Evaluate∫ π

2

0cos2 x sin5 x dx .

[8

105

]

2. Determine∫

tan6 x dx by using reduction for-

mulae and hence evaluate∫ π

4

0tan6 x dx .[

13

15− π

4

]

3. Evaluate∫ π

2

0cos5 x sin4 x dx .

[8

315

]4. Use a reduction formula to determine∫

(ln x)4 dx .[x(ln x)4 − 4x(ln x)3 + 12x(ln x)2

−24x ln x + 24x + c

]

5. Show that∫ π

2

0sin3 θ cos4 θ dθ = 2

35

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