dadang amir hamzah · outline 1 integration by parts 2 trigonometric integrals 3 trigonometric...

Matematika II : Technique of Integration

Dadang Amir Hamzah

sumber : http://www.whsd.org/uploaded/faculty/tmm/calc front image.jpg

2016

Dadang Amir Hamzah Matematika II Semester II 2016 1 / 36

Outline

1 Integration by parts

2 Trigonometric Integrals

3 Trigonometric Substitutions

4 Integration of Rational Functions by Partial Fractions


Outline






Teknik Integral Parsial

Dalam turunan kita mengenal aturan turunan perkalian fungsi, yaknijika fungsi f dan g differensiabel maka

d

dx[f(x)g(x)] = f(x)g′(x) + g(x)f ′(x)

ini dapat ditulis dalam integral tak tentu yakni,∫[f(x)g′(x) + g(x)f ′(x)]dx = f(x)g(x)

atau ∫f(x)g′(x)dx+

∫g(x)f ′(x)dx = f(x)g(x)

dari sini didapat∫f(x)g′(x)dx = f(x)g(x)−

∫g(x)f ′(x)dx (1)


Teknik Integral Parsial

Persamaan (1) dinamakan Rumus Integral Parsial.Misalkan u = f(x) dan v = g(x)artinya du = f ′(x)dx dan dv = g′(x)dx.Akibatnya persamaan (1) dapat ditulis menjadi∫

udv = uv −∫vdu (2)

bentuk yang mudah diingat.


Contoh 1

Tentukan ∫x sin(x) dx


Contoh 1


Solusi :Misalkan

u = x, → du = dxdv = sin(x)dx → v =

∫sin(x)dx = − cos(x)


Contoh 1


Solusi :Misalkan

u = x, → du = dxdv = sin(x)dx → v =

∫sin(x)dx = − cos(x)

Jadi : ∫x sin(x)dx = −x cos(x) +

∫cos(x)dx

= −x cos(x) + sin(x) + C


Contoh 2

Tentukan ∫x sec2(x) dx


Contoh 2


Solusi :Misalkan

u = x, → du = dxdv = sec2(x)dx → v =

∫sec2(x)dx = tan(x)


Contoh 2


Solusi :Misalkan

u = x, → du = dxdv = sec2(x)dx → v =

∫sec2(x)dx = tan(x)

Jadi : ∫x sec2(x)dx = x tan(x)−

∫tan(x)dx

= x tan(x) + ln | cos(x)|+ C

catatan: Hitung∫tan(x)dx dengan metode substitusi


Outline






Contoh 1

Tentukan∫cos3(x) dx


Contoh 1

Tentukan∫cos3(x) dx

Solusi :cos3(x) = cos2(x) · cos(x) = (1− sin2(x)) cos(x)∫

cos3(x)dx =∫cos2(x) · cos(x)dx

=∫(1− sin2(x)) cos(x)dx

Misalkan u = sin(x), du = cos(x)dx

=∫(1− u2)du

= u− 13u

3 + C

= sin(x)− 13 sin

3(x) + C


Contoh 2

Tentukan∫sin5(x) cos2(x) dx


Contoh 2

Tentukan∫sin5(x) cos2(x) dx

Solusi :sin5(x) cos(x) =

(sin2(x)

)2cos(x) sin(x) =

(1− cos2(x)

)2cos(x) sin(x)∫

sin5(x) cos(x)dx =∫ (

1− cos2(x))2

cos(x) sin(x)dx

Misalkan u = cos(x), du = − sin(x)dx

=∫(1− u2)2u2(−du)

= −∫(u2 − 2u4 + u6)du

= −(u3

3 − 2u5

5 + u7

7

)+ C

= −13 cos

3(x) + 23 cos

5(x)− 17 cos

7(x) + C


Teknik Integrasi

Dari contoh 1 dan 2 kita selalu bisa mengubah integran ke bentukpangkat ganjil - genap atau sebaliknya.


Teknik Integrasi

Dari contoh 1 dan 2 kita selalu bisa mengubah integran ke bentukpangkat ganjil - genap atau sebaliknya.Apabila pangkatnya genap atau keduanya genap kita gunakan

sin2(x) =1

2(1− cos(2x)) dan cos2(x) =

1

2(1 + cos(2x))


Teknik Integrasi

Dari contoh 1 dan 2 kita selalu bisa mengubah integran ke bentukpangkat ganjil - genap atau sebaliknya.Apabila pangkatnya genap atau keduanya genap kita gunakan

sin2(x) =1

2(1− cos(2x)) dan cos2(x) =

1

2(1 + cos(2x))

Contoh 3 : Tentukanπ∫0

sin2(x)dx


Contoh 3

Tentukanπ∫0

sin2(x)dx


Contoh 3

Tentukanπ∫0

sin2(x)dx

Solusi :∫ π0 sin2(x)dx = 1

2

∫ π0 (1− cos(2x))dx

=[12(x−

12 sin(2x))

]π0

= 12

(π − 1

2 sin(2π))− 1

2

(0− 1

2 sin(0))= 1

2π


Contoh 4

Tentukan∫sin4(x)dx


Contoh 4

Tentukan∫sin4(x)dx

Solusi: ∫sin4(x)dx =

∫(sin2(x))2dx

=∫ (1−cos(2x)

2

)2

dx

= 14

∫(1− 2 cos(2x) + cos2(2x))dx

ingat : cos2(2x) = 12(1 + cos(4x))

= 14

∫(1− 2 cos(2x) + 1

2(1 + cos(4x)))dx

= 14

(32x− sin(2x) + 1

8 sin(4x)

)+ C


Strategi mengevaluasi∫sinm(x) cosn(x)dx


Contoh 5

Tentukan∫tan6(x) sec4(x) dx


Contoh 5


Solusi :sec2(x) = 1 + tan2(x)∫

tan6(x) sec4(x) dx =∫tan6(x) sec2(x) sec2(x)dx

=∫tan6(x)(1 + tan2(x)) sec2(x)dx

Misalkan u = tan(x), du = sec2(x)dx

=∫u6(1 + u2)du

= u7

7 + u9

9 + C

= 17 tan

7(x) + 19 tan

9(x) + C


Contoh 6



Contoh 6


Solusi :tan2(x) = sec2(x)− 1∫

tan5(x) sec7(x) dx =∫tan4(x) sec6(x) sec(x) tan(x)dx

=∫(sec2(x)− 1)2 sec6(x) sec(x) tan(x)dx

Misalkan u = sec(x), du = sec(x) tan(x)dx

=∫(u2 − 1)2u6du

=∫(u10 − 2u8 + u6)du

= u11

11 − 2u9

9 + u7

7 + C

= 111 sec

11(x)− 29 sec

9(x) + 17 sec

7(x) + C


Strategi mengevaluasi∫tanm(x) secn(x)dx


Contoh 7

∫sec(x)dx = ln | sec(x) + tan(x)|+ C


Contoh 7

∫sec(x)dx = ln | sec(x) + tan(x)|+ C

Bukti :

∫sec(x)dx =

∫sec(x) sec(x)+tan(x)

sec(x)+tan(x)dx

=∫ sec2(x)+sec(x) tan(x)

sec(x)+tan(x) dx

Misalkan u = sec(x) + tan(x), du = (sec(x) tan(x) + sec2(x))dx

=∫

1udu

= ln |u|+ C

= ln | sec(x) + tan(x)|+ C


Contoh 8

Tentukan∫sec3(x)dx


Contoh 8

Tentukan∫sec3(x)dx

Dengan integral parsial

u = sec(x) dv = sec2(x)dxdu = sec(x) tan(x)dx v = tan(x)∫

sec3(x)dx = sec(x) tan(x)−∫sec(x) tan2(x)dx

= sec(x) tan(x)−∫sec(x)(sec2(x)− 1)dx

= sec(x) tan(x)−∫sec3(x)dx+

∫sec(x)dx

2∫sec3(x)dx = sec(x) tan(x) +

∫sec(x)dx∫

sec3(x)dx = 12

[sec(x) tan(x) +

∫sec(x)dx

]= 1

2

[sec(x) tan(x) + ln | sec(x) + tan(x)|

]+ C


Strategi Integrasi

Untuk mengevaluasi integral bentuk :1∫sin(mx) cos(nx)dx

2∫sin(mx) sin(nx)dx

3∫cos(mx) cos(nx)dx

Gunakan rumusa. sin(A) cos(B) = 1

2 [sin(A−B) + sin(A+B)]

b. sin(A) sin(B) = 12 [cos(A−B)− cos(A+B)]

c. cos(A) cos(B) = 12 [cos(A−B) + cos(A+B)]


Contoh 9

Tentukan∫sin(4x) cos(5x) dx


Contoh 9

Tentukan∫sin(4x) cos(5x) dx

Solusi : ∫sin(4x) cos(5x)dx =

∫12 [sin(−x) + sin(9x)]dx

= 12

∫(− sin(x) + sin(9x))dx

= 12

(cos(x)− 1

9 cos(9x)

)+ C


Outline






Integrands involving√a2 − x2,

√a2 + x2 and

√x2 − a2

Radical Substitusi Batas t

1.√a2 − x2 x = a sin(θ) −π/2 ≤ θ ≤ π/2

2.√a2 + x2 x = a tan(θ) −π/2 ≤ θ ≤ π/2

3.√x2 − a2 x = a sec(θ) 0 ≤ θ ≤ π, θ 6= π/2



√a2 + x2 and

√x2 − a2


1.√a2 − x2 x = a sin(θ) −π/2 ≤ θ ≤ π/2

2.√a2 + x2 x = a tan(θ) −π/2 ≤ θ ≤ π/2

3.√x2 − a2 x = a sec(θ) 0 ≤ θ ≤ π, θ 6= π/2

1. sin2(θ) + cos2(θ) = 12. tan2(θ) + 1 = sec2(θ)3.1 + cot2(θ) = csc2(θ)



√a2 + x2 and

√x2 − a2


1.√a2 − x2 x = a sin(θ) −π/2 ≤ θ ≤ π/2

2.√a2 + x2 x = a tan(θ) −π/2 ≤ θ ≤ π/2

3.√x2 − a2 x = a sec(θ) 0 ≤ θ ≤ π, θ 6= π/2

1. sin2(θ) + cos2(θ) = 12. tan2(θ) + 1 = sec2(θ)3.1 + cot2(θ) = csc2(θ)

Akibat :

1.√a2 − x2 =

√a2 − a2 sin2(θ) =

√a2 cos2(θ) = a cos(θ)

2.√a2 + x2 =

√a2 + a2 tan2(θ) =

√a2 sec2(θ) = a sec(θ)

3.√x2 − a2 =

√a2 sec2(θ)− a2 =

√a2 tan2(θ) = a tan(θ)


Contoh 1

Tetukan∫ √

a2 − x2dx


Contoh 2


Contoh 2

Tentukan∫

dx√9+x2


Contoh 3

Hitung∫ 42

√x2−4x dx


Contoh 4

Tentukana.∫

dx√x2+2x+26

b.∫

2x√x2+2x+26

dx


Outline






Dekomposisi Parsial Bentuk Pembagian

Perhatikan operasi berikut

2x−1 + 3

x+1




2x−1 + 3

x+1 = 2(x+1)+3(x−1)(x−1)(x+1)




2x−1 + 3

x+1 = 2(x+1)+3(x−1)(x−1)(x+1) = 5x−1

(x−1)(x+1)




2x−1 + 3

x+1 = 2(x+1)+3(x−1)(x−1)(x+1) = 5x−1

(x−1)(x+1) = 5x−1x2−1




2x−1 + 3

x+1 = 2(x+1)+3(x−1)(x−1)(x+1) = 5x−1

(x−1)(x+1) = 5x−1x2−1

Akibatnya

∫5x−1x2−1 dx =

∫ (2

x−1 + 3x+1

)dx

=∫

2x−1 dx+

∫3

x+1 dx

= 2 ln |x− 1|+ 3 ln |x+ 1|+K


Bagaimana mendapatkan 2x−1 + 3

x+1 dari 5x−1x2−1 ?


5x−1x2−1 = 5x−1

(x−1)(x+1) = A(x−1) +

B(x+1)


5x−1x2−1 = 5x−1

(x−1)(x+1) = A(x−1) +

B(x+1)

Kalikan kedua ruas dengan (x+ 1)(x− 1) didapat


5x−1x2−1 = 5x−1

(x−1)(x+1) = A(x−1) +

B(x+1)


5x− 1 = (A+B)x+ (B −A)


5x−1x2−1 = 5x−1

(x−1)(x+1) = A(x−1) +

B(x+1)


5x− 1 = (A+B)x+ (B −A)

A+B = 5A−B = −1


5x−1x2−1 = 5x−1

(x−1)(x+1) = A(x−1) +

B(x+1)


5x− 1 = (A+B)x+ (B −A)

A+B = 5A−B = −1 ⇒


5x−1x2−1 = 5x−1

(x−1)(x+1) = A(x−1) +

B(x+1)


5x− 1 = (A+B)x+ (B −A)

A+B = 5A−B = −1 ⇒ A = 2, B = 3


5x−1x2−1 = 5x−1

(x−1)(x+1) = A(x−1) +

B(x+1)


5x− 1 = (A+B)x+ (B −A)

A+B = 5A−B = −1 ⇒ A = 2, B = 3

Jadi :5x− 1

x2 − 1=

2

(x− 1)+

3

(x+ 1)


Contoh 1 : Faktor linear beda

Tentukan∫

3x−1(x2−x−6) dx



Tentukan∫

3x−1(x2−x−6) dx

3x−1(x2−x−6)



Tentukan∫

3x−1(x2−x−6) dx

3x−1(x2−x−6) = 3x−1

(x+2)(x−3)



Tentukan∫

3x−1(x2−x−6) dx

3x−1(x2−x−6) = 3x−1

(x+2)(x−3) = Ax+2 + B

x−3



Tentukan∫

3x−1(x2−x−6) dx

3x−1(x2−x−6) = 3x−1

(x+2)(x−3) = Ax+2 + B

x−3

kalikan dua ruas dgn (x+ 2)(x− 3)



Tentukan∫

3x−1(x2−x−6) dx

3x−1(x2−x−6) = 3x−1

(x+2)(x−3) = Ax+2 + B

x−3


3x− 1 = A(x− 3) +B(x+ 2)



Tentukan∫

3x−1(x2−x−6) dx

3x−1(x2−x−6) = 3x−1

(x+2)(x−3) = Ax+2 + B

x−3


3x− 1 = A(x− 3) +B(x+ 2)

= (A+B)x+ (2B − 3A)

A+B = 3



Tentukan∫

3x−1(x2−x−6) dx

3x−1(x2−x−6) = 3x−1

(x+2)(x−3) = Ax+2 + B

x−3


3x− 1 = A(x− 3) +B(x+ 2)

= (A+B)x+ (2B − 3A)

A+B = 32B − 3A = −1



Tentukan∫

3x−1(x2−x−6) dx

3x−1(x2−x−6) = 3x−1

(x+2)(x−3) = Ax+2 + B

x−3


3x− 1 = A(x− 3) +B(x+ 2)

= (A+B)x+ (2B − 3A)

A+B = 32B − 3A = −1

A = 75 B = 8

5


∫3x−1

(x2−x−6) dx =∫

75(x+2) dx+

∫8

5(x−3) dx

= 75 ln |x+ 2|+ 8

5 ln |x− 3|+ C


Contoh 2 : Faktor linear berulang

Tentukan∫

x(x−3)2 dx



Tentukan∫

x(x−3)2 dx

x(x−3)2



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x = A(x− 3) +B



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x = A(x− 3) +B

= Ax+ (B − 3)



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x = A(x− 3) +B

= Ax+ (B − 3)

A



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x = A(x− 3) +B

= Ax+ (B − 3)

A = 1



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x = A(x− 3) +B

= Ax+ (B − 3)

A = 1B − 3



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x = A(x− 3) +B

= Ax+ (B − 3)

A = 1B − 3 = 0



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x = A(x− 3) +B

= Ax+ (B − 3)

A = 1B − 3 = 0 → B = 3



Tentukan∫

x(x−3)2 dx

x(x−3)2 = A

x−3 + B(x−3)2

x = A(x− 3) +B

= Ax+ (B − 3)

A = 1B − 3 = 0 → B = 3∫

x

(x− 3)2dx =

∫1

(x− 3)dx+

∫3

(x− 3)2dx


Contoh 3: Faktor linear campuran

Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13 = A(x− 1)2 +B(x+ 3)(x− 1) + C(x+ 3)



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13 = A(x− 1)2 +B(x+ 3)(x− 1) + C(x+ 3)

= (A+B)x2 + (−2A+ 2B + C)x+ (A− 3B + 3C)



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13 = A(x− 1)2 +B(x+ 3)(x− 1) + C(x+ 3)

= (A+B)x2 + (−2A+ 2B + C)x+ (A− 3B + 3C)

A



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13 = A(x− 1)2 +B(x+ 3)(x− 1) + C(x+ 3)

= (A+B)x2 + (−2A+ 2B + C)x+ (A− 3B + 3C)

A = 4



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13 = A(x− 1)2 +B(x+ 3)(x− 1) + C(x+ 3)

= (A+B)x2 + (−2A+ 2B + C)x+ (A− 3B + 3C)

A = 4B



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13 = A(x− 1)2 +B(x+ 3)(x− 1) + C(x+ 3)

= (A+B)x2 + (−2A+ 2B + C)x+ (A− 3B + 3C)

A = 4B = −1



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13 = A(x− 1)2 +B(x+ 3)(x− 1) + C(x+ 3)

= (A+B)x2 + (−2A+ 2B + C)x+ (A− 3B + 3C)

A = 4B = −1C



Tentukan∫

3x2−8x+13(x+3)(x−1)2 dx

3x2−8x+13(x+3)(x−1)2 = A

x+3 + Bx−1 + C

(x−1)2

3x2 − 8x+ 13 = A(x− 1)2 +B(x+ 3)(x− 1) + C(x+ 3)

= (A+B)x2 + (−2A+ 2B + C)x+ (A− 3B + 3C)

A = 4B = −1C = 2


Contoh 4 : Faktor kuadrat berulang

Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2

A



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2

A = 1



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2

A = 1B



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2

A = 1B = −1



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2

A = 1B = −1C



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2

A = 1B = −1C = 3



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2

A = 1B = −1C = 3E



Tentukan∫

6x2−15x+22(x+3)(x2+2)2

dx

6x2−15x+22(x+3)(x2+2)2

= A(x+3) +

Bx+Cx2+2

+ Dx+E(x2+2)2

A = 1B = −1C = 3E = 0


dadang amir hamzah · outline 1 integration by parts 2 trigonometric integrals 3 trigonometric...

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