pertemuan 4

Chapter 40

Integration usingtrigonometric and

hyperbolic substitutions

40.1 Introduction

Table 40.1 gives a summary of the integrals that requirethe use of trigonometric and hyperbolic substitutionsand their application is demonstrated in Problems 1to 27.

40.2 Worked problems on integrationof sin2 x, cos2 x, tan2 x and cot2 x

Problem 1. Evaluate∫ π

4

02 cos 24t dt .

Since cos 2t =2 cos 2t − 1 (from Chapter 17),

then cos 2t = 1

2(1 + cos 2t) and

cos 24t = 1

2(1 + cos 8t)

Hence∫ π

4

02 cos 24t dt

= 2∫ π

4

0

1

2(1 + cos 8t)dt

=[

t + sin 8t

8

]π4

0

=⎡⎢⎣π

4+

sin 8(π

4

)8

⎤⎥⎦−

[0 + sin0

8

]

= π

4or 0.7854

Problem 2. Determine∫

sin23x dx .

Since cos 2x =1 − 2 sin2x (from Chapter 17),

then sin 2x = 1

2(1 − cos 2x) and

sin 23x = 1

2(1 − cos 6x)

Hence∫

sin 23x dx =∫

1

2(1 − cos 6x)dx

= 12

(x− sin 6x

6

)+c

Problem 3. Find 3∫

tan2 4x dx .

Since 1+ tan2 x = sec2 x , then tan2 x = sec2 x −1 andtan2 4x = sec2 4x −1.

Hence 3∫

tan2 4x dx = 3∫

(sec2 4x − 1)dx

= 3(

tan 4x4

−x)

+c

Integration using trigonometric and hyperbolic substitutions 399

Table 40.1 Integrals using trigonometric and hyperbolic substitutions

f (x)∫

f (x)dx Method See problem

1. cos 2x1

2

(x + sin 2x

2

)+ c Use cos 2x = 2 cos 2x − 1 1

2. sin 2x1

2

(x − sin 2x

2

)+ c Use cos 2x = 1 − 2 sin2x 2

3. tan2 x tan x − x + c Use 1 + tan2 x = sec2 x 3

4. cot2 x −cot x − x + c Use cot2 x + 1 = cosec2x 4

5. cos m x sinn x (a) If either m or n is odd (but not both), use

cos 2x + sin 2x = 1 5, 6

(b) If both m and n are even, use either

cos 2x = 2 cos 2x − 1 or cos 2x = 1 − 2 sin2x 7, 8

6. sin Acos B Use 12 [ sin(A + B)+ sin(A − B)] 9

7. cos Asin B Use 12 [ sin(A + B)− sin(A − B)] 10

8. cos Acos B Use 12 [ cos(A + B)+ cos(A − B)] 11

9. sin Asin B Use −12 [ cos(A + B)− cos(A − B)] 12

10.1√

(a2 − x2)sin−1 x

a+ c Use x = a sinθ substitution 13, 14

11.√

(a2 − x2)a2

2sin−1 x

a+ x

2

√(a2 − x2)+ c Use x = a sinθ substitution 15, 16

12.1

a2 + x2

1

atan−1 x

a+ c Use x = a tanθ substitution 17–19

13.1√

(x2 + a2)sinh−1 x

a+ c Use x = a sinhθ substitution 20–22

or ln

{x +√

(x2 + a2)

a

}+ c

14.√

(x2 + a2)a2

2sinh−1 x

a+ x

2

√(x2 + a2)+ c Use x = a sinhθ substitution 23

15.1√

(x2 − a2)cosh−1 x

a+ c Use x = a cosh θ substitution 24, 25

or ln

{x +√

(x2 − a2)

a

}+ c

16.√

(x2 − a2)x

2

√(x2 − a2)− a2

2cosh−1 x

a+ c Use x = a cosh θ substitution 26, 27

400 Higher Engineering Mathematics


3

π6

1

2cot2 2θ dθ.

Since cot2 θ +1 = cosec2θ , then cot2 θ = cosec2θ−1and cot2 2θ =cosec2 2θ −1.

Hence∫ π

3

π6

1

2cot2 2θ dθ

= 1

2

∫ π3

π6

(cosec2 2θ − 1)dθ = 1

2

[−cot 2θ

2− θ

] π3

π6

= 1

2

⎡⎢⎣⎛⎜⎝−cot 2

(π

3

)2

− π

3

⎞⎟⎠−

⎛⎜⎝−cot 2

(π

6

)2

− π

6

⎞⎟⎠⎤⎥⎦

= 1

2[(0.2887 − 1.0472)− (−0.2887 − 0.5236)]

= 0.0269

Now try the following exercise

Exercise 155 Further problems onintegration of sin2 x, cos2 x, tan2 x and cot2 x

In Problems 1 to 4, integrate with respect to thevariable.

1. sin 22x

[1

2

(x − sin 4x

4

)+ c

]

2. 3 cos 2t

[3

2

(t + sin 2t

2

)+ c

]

3. 5 tan2 3θ

[5

(1

3tan 3θ − θ

)+ c

]4. 2 cot2 2t [−(cot 2t + 2t)+ c]

In Problems 5 to 8, evaluate the definite integrals,correct to 4 significant figures.

5.∫ π

3

03 sin 23x dx

[π

2or 1.571

]

6.∫ π

4

0cos 24x dx

[π

8or 0.3927

]

7.∫ 1

02 tan2 2t dt [−4.185]

8.∫ π

3

π6

cot2 θ dθ [0.6311]

40.3 Worked problems on powers ofsines and cosines


sin5θ dθ.

Since cos 2θ + sin2θ =1 then sin 2θ =(1− cos 2θ).

Hence∫

sin 5θ dθ

=∫

sinθ(sin 2θ)2 dθ =∫

sinθ(1 − cos 2θ)2 dθ

=∫

sinθ(1 − 2 cos 2θ + cos 4θ)dθ

=∫

(sinθ − 2 sinθ cos 2θ + sinθ cos 4θ)dθ

= −cos θ + 2 cos3 θ

3− cos5 θ

5+c

Whenever a power of a cosine is multiplied by a sine ofpower 1, or vice-versa, the integral may be determinedby inspection as shown.

In general,∫

cos nθ sinθ dθ = −cos n+1θ

(n + 1)+ c

and∫

sin nθ cos θ dθ = sin n+1θ

(n + 1)+ c


2

0sin2x cos 3x dx .

∫ π2

0sin 2x cos 3x dx =

∫ π2

0sin 2x cos 2x cos x dx

=∫ π

2

0(sin 2x)(1 − sin 2x)(cos x)dx

=∫ π

2

0(sin 2x cos x − sin 4x cos x)dx

=[

sin 3x

3− sin 5x

5

]π2

0

=⎡⎢⎣

(sin

π

2

)3

3−

(sin

π

2

)5

5

⎤⎥⎦ − [0 − 0]

= 1

3− 1

5= 2

15or 0.1333


4

04 cos 4θ dθ , correct to 4

significant figures.


∫ π4

04 cos 4θ dθ = 4

∫ π4

0(cos 2θ)2 dθ

= 4∫ π

4

0

[1

2(1 + cos 2θ)

]2

dθ

=∫ π

4

0(1 + 2 cos 2θ + cos 22θ)dθ

=∫ π

4

0

[1 + 2 cos2θ + 1

2(1 + cos 4θ)

]dθ

=∫ π

4

0

(3

2+ 2 cos 2θ + 1

2cos 4θ

)dθ

=[

3θ

2+ sin 2θ + sin 4θ

8

]π4

0

=[

3

2

(π

4

)+ sin

2π

4+ sin 4(π/4)

8

]− [0]

= 3π

8+ 1 = 2.178,

correct to 4 significant figures.

Problem 8. Find∫

sin 2t cos 4t dt .

∫sin 2t cos 4t dt =

∫sin 2t (cos 2t)2 dt

=∫ (

1 − cos 2t

2

)(1 + cos2t

2

)2

dt

= 1

8

∫(1 − cos2t)(1 + 2 cos 2t + cos 22t)dt

= 1

8

∫(1 + 2 cos2t + cos 22t − cos 2t

−2 cos 22t − cos 32t)dt

= 1

8

∫(1 + cos2t − cos 22t − cos 32t)dt

= 1

8

∫ [1 + cos 2t −

(1 + cos 4t

2

)− cos 2t (1 − sin 22t)

]dt

= 1

8

∫ (1

2− cos 4t

2+ cos 2t sin 22t

)dt

= 18

(t2

− sin4t8

+ sin32t6

)+c


Exercise 156 Further problems onintegration of powers of sines and cosines


1. sin 3θ

[(a)−cos θ + cos 3θ

3+ c

]

2. 2 cos 32x

[sin 2x − sin 32x

3+ c

]

3. 2 sin3t cos 2t [−2

3cos 3t + 2

5cos 5t + c

]

4. sin 3x cos 4x

[−cos 5x

5+ cos 7x

7+ c

]

5. 2 sin42θ [3θ

4− 1

4sin 4θ + 1

32sin 8θ + c

]

6. sin 2t cos 2t

[t

8− 1

32sin 4t + c

]

40.4 Worked problems on integrationof products of sines and cosines


sin 3t cos2t dt .∫sin 3t cos 2t dt

=∫

1

2[sin(3t + 2t)+ sin (3t − 2t)]dt,

from 6 of Table 40.1, which follows from Section 17.4,page 170,

= 1

2

∫(sin 5t + sin t)dt

= 12

(−cos 5t5

−cos t)

+c

Problem 10. Find∫

1

3cos 5x sin2x dx .


∫1

3cos5x sin 2x dx

= 1

3

∫1

2[sin(5x + 2x)− sin (5x − 2x)]dx ,

from 7 of Table 40.1

= 1

6

∫(sin 7x − sin 3x)dx

= 16

(−cos 7x7

+ cos 3x3

)+c

Problem 11. Evaluate∫ 1

02 cos6θ cos θ dθ ,

correct to 4 decimal places.

∫ 1

02 cos6θ cos θ dθ

= 2∫ 1

0

1

2[cos (6θ + θ)+ cos (6θ − θ)]dθ ,


=∫ 1

0(cos 7θ + cos 5θ)dθ =

[sin 7θ

7+ sin 5θ

5

]1

0

=(

sin 7

7+ sin 5

5

)−

(sin 0

7+ sin 0

5

)‘sin 7’ means ‘the sine of 7 radians’ (≡401◦4′) andsin 5≡286◦29′.

Hence∫ 1

02 cos6θ cos θ dθ

= (0.09386 + (−0.19178))− (0)

= −0.0979, correct to 4 decimal places.

Problem 12. Find 3∫

sin 5x sin 3x dx .

3∫

sin5x sin 3x dx

= 3∫

−1

2[cos (5x + 3x)− cos (5x − 3x)]dx ,


= −3

2

∫(cos 8x − cos 2x)dx

= −32

(sin 8

8− sin 2x

2

)+c or

316

(4 sin 2x−sin 8x)+c


Exercise 157 Further problems onintegration of products of sines and cosines


1. sin 5t cos 2t

[−1

2

(cos 7t

7+ cos 3t

3

)+ c

]

2. 2 sin3x sin x

[sin2x

2− sin 4x

4+ c

]3. 3 cos 6x cos x [

3

2

(sin 7x

7+ sin 5x

5

)+ c

]

4.1

2cos 4θ sin 2θ [

1

4

(cos2θ

2− cos 6θ

6

)+ c

]In Problems 5 to 8, evaluate the definite integrals.

5.∫ π

2

0cos 4x cos 3x dx

[(a)

3

7or 0.4286

]

6.∫ 1

02 sin7t cos 3t dt [0.5973]

7. −4∫ π

3

0sin 5θ sin2θ dθ [0.2474]

8.∫ 2

13 cos 8t sin 3t dt [−0.1999]

40.5 Worked problems on integrationusing the sin θ substitution


1√(a2 − x2)

dx .

Let x =a sinθ , thendx

dθ=a cos θ and dx =a cos θ dθ .

Hence∫

1√(a2 − x2)

dx

=∫

1√(a2 − a2 sin 2θ)

a cosθ dθ

=∫

a cosθ dθ√[a2(1 − sin 2θ)]


=∫

a cos θ dθ√(a2 cos 2θ)

, since sin2θ + cos 2θ = 1

=∫

a cos θ dθ

a cos θ=

∫dθ = θ + c

Since x =a sinθ , then sinθ = x

aand θ =sin−1 x

a.

Hence∫

1√(a2 − x2)

dx = sin−1 xa

+c


0

1√(9 − x2)

dx .

From Problem 13,∫ 3

0

1√(9 − x2)

dx

=[sin−1 x

3

]3

0, since a = 3

= (sin−1 1 − sin−1 0) = π

2or 1.5708

Problem 15. Find∫ √

(a2 − x2)dx .

Let x =a sinθ thendx

dθ=a cosθ and dx =a cos θ dθ.

Hence∫ √

(a2 − x2)dx

=∫ √

(a2 − a2 sin 2θ)(a cosθ dθ)

=∫ √

[a2(1 − sin 2θ)] (a cos θ dθ)

=∫ √

(a2 cos 2θ)(a cosθ dθ)

=∫

(a cos θ)(a cos θ dθ)

= a2∫

cos 2θ dθ = a2∫ (

1 + cos2θ

2

)dθ

(since cos 2θ = 2 cos 2θ − 1)

= a2

2

(θ + sin2θ

2

)+ c

= a2

2

(θ + 2 sinθ cos θ

2

)+ c

since from Chapter 17, sin 2θ = 2 sinθ cos θ

= a2

2[θ + sinθ cosθ] + c

Since x =a sinθ , then sinθ = x

aand θ =sin−1 x

a

Also, cos 2θ + sin 2θ = 1, from which,

cos θ =√

(1 − sin 2θ) =√[

1 −( x

a

)2]

=√(

a2 − x2

a2

)=

√(a2 − x2)

a

Thus∫ √

(a2 − x2)dx = a2

2[θ + sinθ cos θ]

= a2

2

[sin−1 x

a+

( x

a

) √(a2 − x2)

a

]+ c

= a2

2sin−1 x

a+ x

2

√(a2 −x2) + c


0

√(16−x2)dx .


0

√(16−x2)dx

=[

16

2sin−1 x

4+ x

2

√(16 − x2)

]4

0

=[8 sin−11 + 2

√(0)

]− [8 sin−1 0 + 0]

= 8 sin−11 = 8(π

2

)= 4π or 12.57


Exercise 158 Further problems onintegration using the sine θ substitution

1. Determine∫

5√(4 − t2)

dt . [5 sin−1 x

2+ c

]2. Determine

∫3√

(9 − x2)dx . [

3 sin−1 x

3+ c

]3. Determine

∫ √(4 − x2)dx .[

2 sin−1 x

2+ x

2

√(4 − x2)+ c

]


4. Determine∫ √

(16 − 9t2)dt .[8

3sin−1 3t

4+ t

2

√(16 − 9t2)+ c

]

5. Evaluate∫ 4

0

1√(16 − x2)

dx .[π

2or 1.571

]

6. Evaluate∫ 1

0

√(9 − 4x2)dx . [2.760]

40.6 Worked problems on integrationusing tanθ substitution


1

(a2 + x2)dx .

Letx = a tanθ thendx

dθ= a sec2 θ and dx = a sec2 θ dθ .

Hence∫

1

(a2 + x2)dx

=∫

1

(a2 + a2 tan2 θ)(a sec2 θ dθ)

=∫

a sec2 θ dθ

a2(1 + tan2 θ)

=∫

a sec2 θ dθ

a2 sec2 θ, since 1+tan2 θ =sec2 θ

=∫

1

adθ = 1

a(θ)+ c

Since x =a tanθ , θ = tan−1 x

a

Hence∫

1(a2 +x2)

dx= 1a

tan−1 xa

+c


0

1

(4 + x2)dx .


0

1

(4 + x2)dx

= 1

2

[tan−1 x

2

]2

0since a = 2

= 1

2(tan−1 1− tan−1 0) = 1

2

(π

4− 0

)= π

8or 0.3927


0

5

(3 + 2x2)dx , correct

to 4 decimal places.∫ 1

0

5

(3 + 2x2)dx =

∫ 1

0

5

2[(3/2)+ x2]dx

= 5

2

∫ 1

0

1

[√

(3/2)]2 + x2dx

= 5

2

[1√

(3/2)tan−1 x√

(3/2)

]1

0

= 5

2

√(2

3

)[tan−1

√(2

3

)− tan−1 0

]

= (2.0412)[0.6847 − 0]

= 1.3976, correct to 4 decimal places.


Exercise 159 Further problems onintegration using the tan θ substitution

1. Determine∫

3

4 + t2 dt .

[3

2tan−1 t

2+ c

]

2. Determine∫

5

16 + 9θ2dθ .[

5

12tan−1 3θ

4+ c

]

3. Evaluate∫ 1

0

3

1 + t2dt . [2.356]

4. Evaluate∫ 3

0

5

4 + x2 dx . [2.457]

40.7 Worked problems on integrationusing the sinhθ substitution


1√(x2 + a2)

dx .

Let x =a sinhθ , thendx

dθ=a cosh θ and

dx = a coshθ dθ


Hence∫

1√(x2 + a2)

dx

=∫

1√(a2 sinh2θ + a2)

(a cosh θ dθ)

=∫

a coshθ dθ√(a2 cosh2 θ)

,

since cosh2 θ − sinh2 θ = 1

=∫

a cosh θ

a cosh θdθ =

∫dθ = θ + c

= sinh−1 xa

+c, since x = a sinhθ

It is shown on page 339 that

sinh−1 x

a= ln

{x +√

(x2 + a2)

a

},

which provides an alternative solution to

∫1√

(x2 + a2)dx


0

1√(x2 + 4)

dx , correct

to 4 decimal places.

∫ 2

0

1√(x2 + 4)

dx =[sinh−1 x

2

]2

0or

[ln

{x +√

(x2 + 4)

2

}]2

0

from Problem 20, where a =2

Using the logarithmic form,∫ 2

0

1√(x2 + 4)

dx

=[

ln

(2 +√

8

2

)− ln

(0 +√

4

2

)]

= ln2.4142 − ln 1 = 0.8814,correct to 4 decimal places.


1

2

x2√

(1 + x2)dx ,

correct to 3 significant figures.

Since the integral contains a term of the form√(a2 + x2), then let x = sinhθ , from which

dx

dθ= coshθ and dx = cosh θ dθ

Hence∫

2

x2√

(1 + x2)dx

=∫

2(cosh θ dθ)

sinh2 θ√

(1 + sinh2 θ)

= 2∫

cosh θ dθ

sinh2 θ coshθ,


= 2∫

dθ

sinh2 θ= 2

∫cosech2 θ dθ

= −2 cothθ + c

cothθ = coshθ

sinhθ=

√(1 + sinh2 θ)

sinhθ=

√(1 + x2)

x

Hence∫ 2

1

2

x2√

1 + x2)dx

= −[2 cothθ]21 = −2

[√(1 + x2)

x

]2

1

= −2

[√5

2−

√2

1

]= 0.592,

correct to 3 significant figures

Problem 23. Find∫ √

(x2 + a2)dx .

Let x =a sinhθ thendx

dθ=a coshθ and

dx =a coshθ dθ

Hence∫ √

(x2 + a2)dx

=∫ √

(a2 sinh2 θ + a2)(a cosh θ dθ)

=∫ √

[a2(sinh2 θ + 1)](a cosh θ dθ)

=∫ √

(a2 cosh2 θ)(a cosh θ dθ),


=∫

(a cosh θ)(a cosh θ)dθ = a2∫

cosh2 θ dθ

= a2∫ (

1 + cosh 2θ

2

)dθ


= a2

2

(θ + sinh2θ

2

)+ c

= a2

2[θ + sinhθ coshθ] + c,

since sinh 2θ = 2 sinhθ cosh θ

Since x =a sinhθ , then sinhθ = x

aand θ =sinh−1 x

a

Also since cosh2 θ − sinh2 θ =1

then coshθ =√

(1 + sinh2 θ)

=√[

1 +(x

a

)2]

=√(

a2 + x2

a2

)

=√

(a2 + x2)

a

Hence∫ √

(x2 + a2)dx

= a2

2

[sinh−1 x

a+

( x

a

) √(x2 + a2)

a

]+ c

= a2

2sinh−1 x

a+ x

2

√(x2 +a2) + c


Exercise 160 Further problems onintegration using the sinhθ substitution

1. Find∫

2√(x2 + 16)

dx .[2 sinh−1 x

4+ c

]

2. Find∫

3√(9 + 5x2)

dx .[3√5

sinh−1

√5

3x + c

]

3. Find∫ √

(x2 + 9)dx .[9

2sinh−1 x

3+ x

2

√(x2 + 9)+ c

]

4. Find∫ √

(4t2 + 25)dt .[25

4sinh−1 2t

5+ t

2

√(4t2 + 25)+ c

]

5. Evaluate∫ 3

0

4√(t2 + 9)

dt . [3.525]

6. Evaluate∫ 1

0

√(16 + 9θ2)dθ . [4.348]

40.8 Worked problems on integrationusing the coshθ substitution


1√(x2 − a2)

dx .

Let x =a cosh θ thendx

dθ= a sinhθ and

dx =a sinhθ dθ

Hence∫

1√(x2 − a2)

dx

=∫

1√(a2 cosh2 θ − a2)

(a sinhθ dθ)

=∫

a sinhθ dθ√[a2(cosh2 θ − 1)]

=∫

a sinhθ dθ√(a2 sinh2 θ)

,


=∫

a sinhθ dθ

a sinhθ=

∫dθ = θ + c

= cosh−1 xa

+c, since x = a cosh θ

It is shown on page 339 that

cosh−1 xa

= ln

{x+

√(x2 −a2)a

}

which provides as alternative solution to

∫1√

(x2 − a2)dx


2x −3√(x2 −9)

dx .


∫2x − 3√(x2 − 9)

dx =∫

2x√(x2 − 9)

dx

−∫

3√(x2 − 9)

dx

The first integral is determined using the algebraic sub-stitution u =(x2 −9), and the second integral is of the

form∫

1√(x2 − a2)

dx (see Problem 24)

Hence∫

2x√(x2 − 9)

dx −∫

3√(x2 − 9)

dx

= 2√

(x2 −9)−3 cosh−1 x3

+c

Problem 26.∫ √

(x2 − a2)dx .

Let x =a coshθ thendx

dθ=a sinhθ and

dx =a sinhθ dθ

Hence∫ √

(x2 − a2)dx

=∫ √

(a2 cosh2 θ − a2)(a sinhθ dθ)

=∫ √

[a2(cosh2 θ − 1)] (a sinh θ dθ)

=∫ √

(a2 sinh2 θ)(a sinh θ dθ)

= a2∫

sinh2 θ dθ = a2∫ (

cosh 2θ − 1

2

)dθ

since cosh 2θ = 1 + 2 sinh2 θ

from Table 5.1, page 45,

= a2

2

[sinh 2θ

2− θ

]+ c

= a2

2[sinhθ coshθ − θ] + c,

since sinh2θ = 2 sinhθ coshθ

Since x =a coshθ then coshθ = x

aand

θ =cosh−1 x

a

Also, since cosh2 θ − sinh2 θ =1, then

sinhθ =√

(cosh2 θ − 1)

=√[( x

a

)2 − 1

]=

√(x2 − a2)

a

Hence∫ √

(x2 − a2)dx

= a2

2

[√(x2 − a2)

a

(x

a

)− cosh−1 x

a

]+ c

= x2

√(x2 −a2)− a2

2cosh−1 x

a+c


2

√(x2 − 4)dx .

∫ 3

2

√(x2 − 4)dx =

[x

2

√(x2 − 4) − 4

2cosh−1 x

2

]3

2

from Problem 26, when a = 2,

=(

3

5

√5 − 2 cosh−1 3

2

)−(0 − 2 cosh−1 1)

Since cosh−1 x

a= ln

{x +√

(x2 − a2)

a

}then

cosh−1 3

2= ln

{3 +√

(32 − 22)

2

}

= ln 2.6180 = 0.9624

Similarly, cosh−11=0

Hence∫ 3

2

√(x2 − 4)dx

=[

3

2

√5 − 2(0.9624)

]− [0]

= 1.429, correct to 4 significant figures.


Exercise 161 Further problems onintegration using the cosh θ substitution

1. Find∫

1√(t2 − 16)

dt .[cosh−1 x

4+ c

]


2. Find∫

3√(4x2 − 9)

dx .

[3

2cosh−1 2x

3+ c

]

3. Find∫ √

(θ2 − 9)dθ .[θ

2

√(θ2 − 9) − 9

2cosh−1 θ

3+ c

]

4. Find∫ √

(4θ2 − 25)dθ .

[θ

√(θ2 − 25

4

)− 25

4cosh−1 2θ

5+ c

]

5. Evaluate∫ 2

1

2√(x2 − 1)

dx . [2.634]

6. Evaluate∫ 3

2

√(t2 − 4)dt . [1.429]

pertemuan 4

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