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TUGAS 3
NAMA KELOMPOK : 1.SUSANDI 2.RAFIS 3.GERIAN
KELAS : 1EA
LATIHAN 7.1
1. ∫ 100 𝑑π‘₯ = 100π‘₯ + 𝑐
2. ∫ 6π‘₯ 𝑑π‘₯ = 3π‘₯2
+ 𝑐
3. ∫( 3π‘₯2
+ 4π‘₯ βˆ’ 5) 𝑑π‘₯ = π‘₯3
+ 2π‘₯ βˆ’
5π‘₯ + 𝑐
4. ∫( π‘₯2
+ 1)√ π‘₯ 𝑑π‘₯ =
2
7
π‘₯2
1
+
2
3
π‘₯2
3
+ 𝑐
5. ∫( π‘₯ 𝑒
+ 𝑒 π‘₯ ) 𝑑π‘₯ =
π‘₯ 𝑒+1
𝑒+1
+ 𝑒 π‘₯
+ 𝑐
6. ∫(10π‘₯ + 30)3
10 𝑑π‘₯ =
(10π‘₯+30)4
4
+ 𝑐
7. ∫(π‘₯2
βˆ’ 3)4
2π‘₯ 𝑑π‘₯ =
(π‘₯2βˆ’3)5
5
+ 𝑐
8. ∫(𝑠𝑖𝑛2
π‘₯cos π‘₯) 𝑑π‘₯ =
𝑠𝑖𝑛3
3
π‘₯ + 𝑐
9. ∫ π‘₯2
βˆ’ 𝑠𝑖𝑛 π‘₯3
𝑑π‘₯ =
βˆ’ cos π‘₯3
3
+ 𝑐
10. ∫ 𝐼𝑛 π‘₯ 𝑑π‘₯ = π‘₯ 𝐼𝑛 π‘₯ = π‘₯ 𝐼𝑛 π‘₯ βˆ’ π‘₯ +
𝑐
penyelesaian :
1. =
𝑑
𝑑π‘₯
(100π‘₯ + 𝑐) = 100π‘₯ + 𝑐
2. =
𝑑
𝑑π‘₯
(3π‘₯2
+ 𝑐) = 6π‘₯ + 0 = 100
3. =
𝑑
𝑑π‘₯
(π‘₯3
+ 2π‘₯2
βˆ’ 5π‘₯ + 𝑐) = 3π‘₯2
+ 4π‘₯ βˆ’ 5 + 0 = 3π‘₯2
+4x-5
4. =
𝑑
𝑑π‘₯
(
2π‘₯
1
2
7
+
2π‘₯
3
2
3
+ c)=
2π‘₯
βˆ’1
2
14
+
6π‘₯
1
2
6
+ 0 =
1π‘₯
βˆ’1
2
7
+ π‘₯
1
2
5. =
𝑑
𝑑π‘₯
(
π‘₯ 𝑒+1
𝑒+1
+ 𝑒 π‘₯
+ 𝑐) =
𝑒+1.π‘₯ 𝑒+1βˆ’1
1𝑒1βˆ’1+0
+ π‘₯𝑒 π‘₯βˆ’1
+ 𝑐 =
𝑒+π‘₯ 𝑒
1
+ π‘₯𝑒 π‘₯βˆ’1
+ 0 = 𝑒 + π‘₯ 𝑒
+
π‘₯𝑒 π‘₯βˆ’1
6. =
𝑑
𝑑π‘₯
(10π‘₯+30)4
4
+ 𝑐 =
10π‘₯4
4
+
304
4
+ 𝑐 =
40π‘₯3
4
+ 0 = 10π‘₯3
7. =
𝑑
𝑑π‘₯
(π‘₯2βˆ’3)5
5
+ c =
π‘₯10
5
βˆ’
35
5
+ 𝑐 =
10π‘₯9
5
+ 0 = 2π‘₯9
8. =
𝑑
𝑑π‘₯
(
𝑠𝑖𝑛3
3
π‘₯ + 𝑐) =
𝑠𝑖𝑛2 π‘₯
3
.
sin π‘₯
3
+ 0 =
(1βˆ’π‘π‘œπ‘ 2 π‘₯) .
3
sin π‘₯
3
9. =
𝑑
𝑑π‘₯
(
βˆ’ cos π‘₯3
3
+ 𝑐) =
βˆ’3 sin π‘₯2
3
+ 0 = βˆ’sin π‘₯2
10. =
𝑑
𝑑π‘₯
( π‘₯ 𝐼𝑛 π‘₯ βˆ’ π‘₯ + 𝑐) = π‘₯.
𝑑π‘₯
π‘₯
βˆ’ 1 =
π‘₯βˆ’1𝑑π‘₯
π‘₯
LATIHAN 7.2
1. ∫8 𝑑π‘₯
2. ∫
3
4
𝑑π‘₯
3. ∫9.75 𝑑π‘₯
4. ∫√3𝑑π‘₯
5. ∫(
√40
3
√10+15
)𝑑π‘₯
6. ∫16 √2 𝑑𝑑
7. ∫ 𝑒2
𝑑π‘₯
8. ∫2πœ‹ π‘‘π‘Ÿ
9. βˆ«βˆ’21𝑑𝑒
10.∫
6
𝑒
𝑑π‘₯
Penyelesaian :
1. = 8x+c
2. =
3
4
π‘₯ + 𝑐
3. = 9π‘₯. 75π‘₯ + 𝑐
4. = √3 x+c
5. =
40π‘₯
2
3
10π‘₯
1
2+15π‘₯
+ 𝑐
6. = 16𝑑. √2 𝑑 + c
7. = 𝑒π‘₯2
+ c
8. = 2π‘Ÿ. πœ‹π‘Ÿ + c
9. = -21 u + c
10.=
6π‘₯
𝑒π‘₯
+ c
LATIHAN 7.3
1. ∫ π‘₯5
𝑑π‘₯
2. ∫ √ π‘₯34
𝑑π‘₯
3. ∫ π‘₯√2
𝑑π‘₯
4. ∫
1
π‘₯2
𝑑π‘₯
5. ∫ 𝑑100
𝑑𝑑
6. ∫ 𝑒2πœ‹
𝑑𝑒
7. ∫
1
√π‘₯
𝑑π‘₯
8. ∫
π‘₯5
π‘₯2
𝑑π‘₯
9. ∫ π‘Ÿβˆ’1
π‘‘π‘Ÿ
10.∫
1
𝑑
𝑑𝑑
Penyelesaian :
1. =
π‘₯6
6
+ 𝑐
2. = ∫ π‘₯
3
4 𝑑π‘₯ =
π‘₯
7
4
7
4
+ c =
4π‘₯
7
4
7
+ 𝑐
3. =
π‘₯√2+1
√2+1
+ 𝑐
4. =∫ π‘₯βˆ’2
𝑑π‘₯ =
π‘₯βˆ’1
βˆ’1
+ 𝑐 = βˆ’
1
π‘₯
+ 𝑐
5. =
𝑑101
101
+ 𝑐
6. =
42πœ‹+1
2πœ‹+1
+ 𝑐
7. = ∫ π‘₯
βˆ’1
2 𝑑π‘₯ =
π‘₯
1
2
1
2
+ 𝑐 =
2π‘₯
1
2
1
+ 𝑐
8. =
π‘₯6
π‘₯3
+ 𝑐
9. =
π‘Ÿβˆ’1+1
βˆ’1+1
+ 𝑐 = ∞
10.∫ π‘‘βˆ’1
𝑑𝑑 =
π‘‘βˆ’1+1
βˆ’1+1
+ 𝑐 = ∞
LATIHAN 7.4
1. ∫ 𝑒 𝑑
𝑑𝑑
2. ∫ 𝑒20π‘₯
𝑑π‘₯
3. ∫ 𝑒 πœ‹π‘₯
𝑑π‘₯
4. ∫ 𝑒0,25π‘₯
𝑑π‘₯
5. ∫ 𝑒
π‘₯
5 𝑑π‘₯
6. ∫ π‘’βˆš3π‘₯
𝑑π‘₯
7. ∫4 π‘₯
𝑑π‘₯
8. ∫23π‘₯
𝑑π‘₯
9. ∫1000,25π‘₯
𝑑π‘₯
10.∫ πœ‹
π‘₯
5 𝑑π‘₯
Penyelesaian :
1. = 𝑒 𝑑
+ 𝑐
2. =
1
20
𝑒20π‘₯
+ 𝑐 =
𝑒20π‘₯
5
+ 𝑐
3. =
1
πœ‹
𝑒 πœ‹π‘₯
+ 𝑐 =
𝑒 πœ‹π‘₯
πœ‹
+ 𝑐
4. =
1
0,25
𝑒0,25π‘₯
+ 𝑐 =
𝑒0,25π‘₯
0,25
+ 𝑐
5. =
1
π‘₯
5
𝑒
π‘₯
5 + 𝑐 =
5
π‘₯
𝑒
π‘₯
5 + 𝑐
6. =
1
√3
π‘’βˆš3π‘₯
+ 𝑐 =
π‘’βˆš3π‘₯
√3
+ c
7. =
1
𝐼𝑛 4
4 π‘₯
+ 𝑐 =
4 π‘₯
𝐼𝑛 4
+ 𝑐
8. =
1
3 𝐼𝑛 2
23π‘₯
+ 𝑐 =
23π‘₯
3 𝐼𝑛 2
+ 𝑐
9. =
1
0,25 𝐼𝑛 100
1000,25π‘₯
+ 𝑐 =
1000,25π‘₯
0,25 𝐼𝑛 100
+ 𝑐
10.=
1
π‘₯
5
πœ‹
π‘₯
5 + 𝑐 =
5
π‘₯
πœ‹
π‘₯
5 + 𝑐
LATIHAN 7.5
1. ∫cos 𝑣 𝑑𝑣
2. ∫sin(
1
2
πœ‹π‘₯)𝑑π‘₯
3. ∫cos(18) 𝑑π‘₯
4. ∫ 𝑠𝑒𝑐2
(√3π‘₯)𝑑π‘₯
5. ∫ 𝑐𝑠𝑐2
(2,5) 𝑑π‘₯
6. ∫sec (
5
6
π‘₯)tan(
5
6
π‘₯)𝑑π‘₯
7. ∫csc
x
3
π‘π‘œπ‘‘
π‘₯
3
𝑑π‘₯
8. ∫csc(ex) cot(𝑒π‘₯)𝑑π‘₯
9. ∫sin 3πœƒ π‘‘πœƒ
10.∫cos(25πœ‹π‘₯) 𝑑π‘₯
Penyelesaian :
1. = Sin v + c
2. = βˆ’
1
1
2πœ‹
cos (
1
2
πœ‹π‘₯) + 𝑐 = βˆ’2πœ‹cos (
1
2
πœ‹π‘₯) + 𝑐
3. =
1
8
sin(18π‘₯) + 𝑐
4. =
1
√3
tan(√3π‘₯) + 𝑐 =
tan(√3π‘₯)
√3
+ 𝑐
5. = βˆ’
1
2,5
cot(2,5 π‘₯) + 𝑐 =
βˆ’cot(2,5π‘₯)
2,5
+ 𝑐
6. =
1
5
6
sec(
5
6
π‘₯) + 𝑐 =
6
5
sec (
5
6
π‘₯) + 𝑐
7. =
1
1
3
csc(
x
3
) + c = 3 csc (
π‘₯
3
) + 𝑐
8. = βˆ’
1
𝑒
csc (ex) + c =
βˆ’ csc( 𝑒π‘₯)
𝑒
+ 𝑐
9. = βˆ’
1
3
cos(3πœƒ) + 𝑐 =
βˆ’cos(3πœƒ)
3
+ c
10.=
1
25πœ‹
sin(25πœ‹π‘₯) + 𝑐 =
sin(25 πœ‹π‘₯)
25πœ‹
+ c
LATIHAN 7.6
1. ∫
1
1+πœƒ2
π‘‘πœƒ
2. ∫
𝑑π‘₯
√16βˆ’π‘₯2
3. ∫
1
49+π‘₯2
𝑑π‘₯
4. ∫
𝑑𝑑
0,25+𝑑2
5. ∫
𝑑𝑒
βˆšπ‘’2(𝑒2βˆ’1)
6. ∫
1
|π‘₯|√π‘₯2βˆ’41
𝑑π‘₯
7. ∫
1
√
81
100
βˆ’π‘₯2
𝑑π‘₯
8. ∫ πœ‹2+π‘₯2
𝑑π‘₯
9. ∫
𝑑𝑑
βˆšπ‘‘2(𝑑2βˆ’
1
4
)
10.∫
1
|π‘₯|√π‘₯2βˆ’7
𝑑π‘₯
Penyelesaian :
1. = π‘‘π‘Žπ‘›βˆ’1
πœƒ + 𝑐 = βˆ’π‘π‘œπ‘‘βˆ’1
πœƒ + 𝑐
2. = ∫
1
√42βˆ’βˆšπ‘₯2
𝑑π‘₯ = π‘ π‘–π‘›βˆ’1
(
π‘₯
4
) + 𝑐
3. = ∫
1
√492+π‘₯2
𝑑π‘₯ =
1
√49
π‘‘π‘Žπ‘›βˆ’1
(
π‘₯
√49
) + 𝑐
4. = ∫
1
√0,252+𝑑2
𝑑𝑑 =
1
√0,25
π‘‘π‘Žπ‘›βˆ’1
(
𝑑
√0,25
)+ 𝑐
5. = ∫
1
|𝑒|βˆšπ‘’2βˆ’βˆš12
𝑑𝑒 =
1
1
π‘ π‘’π‘βˆ’1
(
𝑒
1
) + 𝑐 = π‘ π‘’π‘βˆ’1( 𝑒) + 𝑐
6. = π‘ π‘’π‘βˆ’1
(
π‘₯
41
) + c = - π‘π‘ π‘βˆ’1
(
π‘₯
41
) + c
7. = ∫
1
√
(9)2
10
βˆ’π‘₯2
𝑑π‘₯ = π‘ π‘–π‘›βˆ’1
(
π‘₯
9
10
) + 𝑐 =π‘ π‘–π‘›βˆ’1
(
10π‘₯
9
) + 𝑐 = βˆ’π‘π‘œπ‘ βˆ’1
(
10π‘₯
9
) + 𝑐
8.
1
πœ‹
π‘‘π‘Žπ‘›βˆ’1 π‘₯
πœ‹
+ 𝑐 = βˆ’
1
πœ‹
π‘π‘œπ‘‘βˆ’1
(
π‘₯
πœ‹
)+c
9.= ∫
1
| 𝑑|√ 𝑑2βˆ’
(1)
2
2
𝑑π‘₯ =
1
1
2
π‘ π‘’π‘βˆ’1
(
𝑑
1
2
) + 𝑐 = 2π‘ π‘’π‘βˆ’1(2𝑑) + 𝑐
10.π‘ π‘’π‘βˆ’1
(
π‘₯
7
) + 𝑐 = βˆ’π‘π‘ π‘βˆ’1
(
π‘₯
7
) + 𝑐

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Tugas 3

  • 1. TUGAS 3 NAMA KELOMPOK : 1.SUSANDI 2.RAFIS 3.GERIAN KELAS : 1EA LATIHAN 7.1 1. ∫ 100 𝑑π‘₯ = 100π‘₯ + 𝑐 2. ∫ 6π‘₯ 𝑑π‘₯ = 3π‘₯2 + 𝑐 3. ∫( 3π‘₯2 + 4π‘₯ βˆ’ 5) 𝑑π‘₯ = π‘₯3 + 2π‘₯ βˆ’ 5π‘₯ + 𝑐 4. ∫( π‘₯2 + 1)√ π‘₯ 𝑑π‘₯ = 2 7 π‘₯2 1 + 2 3 π‘₯2 3 + 𝑐 5. ∫( π‘₯ 𝑒 + 𝑒 π‘₯ ) 𝑑π‘₯ = π‘₯ 𝑒+1 𝑒+1 + 𝑒 π‘₯ + 𝑐 6. ∫(10π‘₯ + 30)3 10 𝑑π‘₯ = (10π‘₯+30)4 4 + 𝑐 7. ∫(π‘₯2 βˆ’ 3)4 2π‘₯ 𝑑π‘₯ = (π‘₯2βˆ’3)5 5 + 𝑐 8. ∫(𝑠𝑖𝑛2 π‘₯cos π‘₯) 𝑑π‘₯ = 𝑠𝑖𝑛3 3 π‘₯ + 𝑐 9. ∫ π‘₯2 βˆ’ 𝑠𝑖𝑛 π‘₯3 𝑑π‘₯ = βˆ’ cos π‘₯3 3 + 𝑐 10. ∫ 𝐼𝑛 π‘₯ 𝑑π‘₯ = π‘₯ 𝐼𝑛 π‘₯ = π‘₯ 𝐼𝑛 π‘₯ βˆ’ π‘₯ + 𝑐 penyelesaian : 1. = 𝑑 𝑑π‘₯ (100π‘₯ + 𝑐) = 100π‘₯ + 𝑐 2. = 𝑑 𝑑π‘₯ (3π‘₯2 + 𝑐) = 6π‘₯ + 0 = 100 3. = 𝑑 𝑑π‘₯ (π‘₯3 + 2π‘₯2 βˆ’ 5π‘₯ + 𝑐) = 3π‘₯2 + 4π‘₯ βˆ’ 5 + 0 = 3π‘₯2 +4x-5 4. = 𝑑 𝑑π‘₯ ( 2π‘₯ 1 2 7 + 2π‘₯ 3 2 3 + c)= 2π‘₯ βˆ’1 2 14 + 6π‘₯ 1 2 6 + 0 = 1π‘₯ βˆ’1 2 7 + π‘₯ 1 2 5. = 𝑑 𝑑π‘₯ ( π‘₯ 𝑒+1 𝑒+1 + 𝑒 π‘₯ + 𝑐) = 𝑒+1.π‘₯ 𝑒+1βˆ’1 1𝑒1βˆ’1+0 + π‘₯𝑒 π‘₯βˆ’1 + 𝑐 = 𝑒+π‘₯ 𝑒 1 + π‘₯𝑒 π‘₯βˆ’1 + 0 = 𝑒 + π‘₯ 𝑒 + π‘₯𝑒 π‘₯βˆ’1 6. = 𝑑 𝑑π‘₯ (10π‘₯+30)4 4 + 𝑐 = 10π‘₯4 4 + 304 4 + 𝑐 = 40π‘₯3 4 + 0 = 10π‘₯3 7. = 𝑑 𝑑π‘₯ (π‘₯2βˆ’3)5 5 + c = π‘₯10 5 βˆ’ 35 5 + 𝑐 = 10π‘₯9 5 + 0 = 2π‘₯9 8. = 𝑑 𝑑π‘₯ ( 𝑠𝑖𝑛3 3 π‘₯ + 𝑐) = 𝑠𝑖𝑛2 π‘₯ 3 . sin π‘₯ 3 + 0 = (1βˆ’π‘π‘œπ‘ 2 π‘₯) . 3 sin π‘₯ 3 9. = 𝑑 𝑑π‘₯ ( βˆ’ cos π‘₯3 3 + 𝑐) = βˆ’3 sin π‘₯2 3 + 0 = βˆ’sin π‘₯2 10. = 𝑑 𝑑π‘₯ ( π‘₯ 𝐼𝑛 π‘₯ βˆ’ π‘₯ + 𝑐) = π‘₯. 𝑑π‘₯ π‘₯ βˆ’ 1 = π‘₯βˆ’1𝑑π‘₯ π‘₯
  • 2. LATIHAN 7.2 1. ∫8 𝑑π‘₯ 2. ∫ 3 4 𝑑π‘₯ 3. ∫9.75 𝑑π‘₯ 4. ∫√3𝑑π‘₯ 5. ∫( √40 3 √10+15 )𝑑π‘₯ 6. ∫16 √2 𝑑𝑑 7. ∫ 𝑒2 𝑑π‘₯ 8. ∫2πœ‹ π‘‘π‘Ÿ 9. βˆ«βˆ’21𝑑𝑒 10.∫ 6 𝑒 𝑑π‘₯ Penyelesaian : 1. = 8x+c 2. = 3 4 π‘₯ + 𝑐 3. = 9π‘₯. 75π‘₯ + 𝑐 4. = √3 x+c 5. = 40π‘₯ 2 3 10π‘₯ 1 2+15π‘₯ + 𝑐 6. = 16𝑑. √2 𝑑 + c 7. = 𝑒π‘₯2 + c 8. = 2π‘Ÿ. πœ‹π‘Ÿ + c 9. = -21 u + c 10.= 6π‘₯ 𝑒π‘₯ + c
  • 3. LATIHAN 7.3 1. ∫ π‘₯5 𝑑π‘₯ 2. ∫ √ π‘₯34 𝑑π‘₯ 3. ∫ π‘₯√2 𝑑π‘₯ 4. ∫ 1 π‘₯2 𝑑π‘₯ 5. ∫ 𝑑100 𝑑𝑑 6. ∫ 𝑒2πœ‹ 𝑑𝑒 7. ∫ 1 √π‘₯ 𝑑π‘₯ 8. ∫ π‘₯5 π‘₯2 𝑑π‘₯ 9. ∫ π‘Ÿβˆ’1 π‘‘π‘Ÿ 10.∫ 1 𝑑 𝑑𝑑 Penyelesaian : 1. = π‘₯6 6 + 𝑐 2. = ∫ π‘₯ 3 4 𝑑π‘₯ = π‘₯ 7 4 7 4 + c = 4π‘₯ 7 4 7 + 𝑐 3. = π‘₯√2+1 √2+1 + 𝑐 4. =∫ π‘₯βˆ’2 𝑑π‘₯ = π‘₯βˆ’1 βˆ’1 + 𝑐 = βˆ’ 1 π‘₯ + 𝑐 5. = 𝑑101 101 + 𝑐 6. = 42πœ‹+1 2πœ‹+1 + 𝑐 7. = ∫ π‘₯ βˆ’1 2 𝑑π‘₯ = π‘₯ 1 2 1 2 + 𝑐 = 2π‘₯ 1 2 1 + 𝑐 8. = π‘₯6 π‘₯3 + 𝑐 9. = π‘Ÿβˆ’1+1 βˆ’1+1 + 𝑐 = ∞
  • 4. 10.∫ π‘‘βˆ’1 𝑑𝑑 = π‘‘βˆ’1+1 βˆ’1+1 + 𝑐 = ∞ LATIHAN 7.4 1. ∫ 𝑒 𝑑 𝑑𝑑 2. ∫ 𝑒20π‘₯ 𝑑π‘₯ 3. ∫ 𝑒 πœ‹π‘₯ 𝑑π‘₯ 4. ∫ 𝑒0,25π‘₯ 𝑑π‘₯ 5. ∫ 𝑒 π‘₯ 5 𝑑π‘₯ 6. ∫ π‘’βˆš3π‘₯ 𝑑π‘₯ 7. ∫4 π‘₯ 𝑑π‘₯ 8. ∫23π‘₯ 𝑑π‘₯ 9. ∫1000,25π‘₯ 𝑑π‘₯ 10.∫ πœ‹ π‘₯ 5 𝑑π‘₯ Penyelesaian : 1. = 𝑒 𝑑 + 𝑐 2. = 1 20 𝑒20π‘₯ + 𝑐 = 𝑒20π‘₯ 5 + 𝑐 3. = 1 πœ‹ 𝑒 πœ‹π‘₯ + 𝑐 = 𝑒 πœ‹π‘₯ πœ‹ + 𝑐 4. = 1 0,25 𝑒0,25π‘₯ + 𝑐 = 𝑒0,25π‘₯ 0,25 + 𝑐 5. = 1 π‘₯ 5 𝑒 π‘₯ 5 + 𝑐 = 5 π‘₯ 𝑒 π‘₯ 5 + 𝑐 6. = 1 √3 π‘’βˆš3π‘₯ + 𝑐 = π‘’βˆš3π‘₯ √3 + c 7. = 1 𝐼𝑛 4 4 π‘₯ + 𝑐 = 4 π‘₯ 𝐼𝑛 4 + 𝑐 8. = 1 3 𝐼𝑛 2 23π‘₯ + 𝑐 = 23π‘₯ 3 𝐼𝑛 2 + 𝑐 9. = 1 0,25 𝐼𝑛 100 1000,25π‘₯ + 𝑐 = 1000,25π‘₯ 0,25 𝐼𝑛 100 + 𝑐 10.= 1 π‘₯ 5 πœ‹ π‘₯ 5 + 𝑐 = 5 π‘₯ πœ‹ π‘₯ 5 + 𝑐
  • 5. LATIHAN 7.5 1. ∫cos 𝑣 𝑑𝑣 2. ∫sin( 1 2 πœ‹π‘₯)𝑑π‘₯ 3. ∫cos(18) 𝑑π‘₯ 4. ∫ 𝑠𝑒𝑐2 (√3π‘₯)𝑑π‘₯ 5. ∫ 𝑐𝑠𝑐2 (2,5) 𝑑π‘₯ 6. ∫sec ( 5 6 π‘₯)tan( 5 6 π‘₯)𝑑π‘₯ 7. ∫csc x 3 π‘π‘œπ‘‘ π‘₯ 3 𝑑π‘₯ 8. ∫csc(ex) cot(𝑒π‘₯)𝑑π‘₯ 9. ∫sin 3πœƒ π‘‘πœƒ 10.∫cos(25πœ‹π‘₯) 𝑑π‘₯ Penyelesaian : 1. = Sin v + c 2. = βˆ’ 1 1 2πœ‹ cos ( 1 2 πœ‹π‘₯) + 𝑐 = βˆ’2πœ‹cos ( 1 2 πœ‹π‘₯) + 𝑐 3. = 1 8 sin(18π‘₯) + 𝑐 4. = 1 √3 tan(√3π‘₯) + 𝑐 = tan(√3π‘₯) √3 + 𝑐 5. = βˆ’ 1 2,5 cot(2,5 π‘₯) + 𝑐 = βˆ’cot(2,5π‘₯) 2,5 + 𝑐 6. = 1 5 6 sec( 5 6 π‘₯) + 𝑐 = 6 5 sec ( 5 6 π‘₯) + 𝑐 7. = 1 1 3 csc( x 3 ) + c = 3 csc ( π‘₯ 3 ) + 𝑐 8. = βˆ’ 1 𝑒 csc (ex) + c = βˆ’ csc( 𝑒π‘₯) 𝑒 + 𝑐 9. = βˆ’ 1 3 cos(3πœƒ) + 𝑐 = βˆ’cos(3πœƒ) 3 + c 10.= 1 25πœ‹ sin(25πœ‹π‘₯) + 𝑐 = sin(25 πœ‹π‘₯) 25πœ‹ + c
  • 6. LATIHAN 7.6 1. ∫ 1 1+πœƒ2 π‘‘πœƒ 2. ∫ 𝑑π‘₯ √16βˆ’π‘₯2 3. ∫ 1 49+π‘₯2 𝑑π‘₯ 4. ∫ 𝑑𝑑 0,25+𝑑2 5. ∫ 𝑑𝑒 βˆšπ‘’2(𝑒2βˆ’1) 6. ∫ 1 |π‘₯|√π‘₯2βˆ’41 𝑑π‘₯ 7. ∫ 1 √ 81 100 βˆ’π‘₯2 𝑑π‘₯ 8. ∫ πœ‹2+π‘₯2 𝑑π‘₯ 9. ∫ 𝑑𝑑 βˆšπ‘‘2(𝑑2βˆ’ 1 4 ) 10.∫ 1 |π‘₯|√π‘₯2βˆ’7 𝑑π‘₯ Penyelesaian : 1. = π‘‘π‘Žπ‘›βˆ’1 πœƒ + 𝑐 = βˆ’π‘π‘œπ‘‘βˆ’1 πœƒ + 𝑐 2. = ∫ 1 √42βˆ’βˆšπ‘₯2 𝑑π‘₯ = π‘ π‘–π‘›βˆ’1 ( π‘₯ 4 ) + 𝑐 3. = ∫ 1 √492+π‘₯2 𝑑π‘₯ = 1 √49 π‘‘π‘Žπ‘›βˆ’1 ( π‘₯ √49 ) + 𝑐 4. = ∫ 1 √0,252+𝑑2 𝑑𝑑 = 1 √0,25 π‘‘π‘Žπ‘›βˆ’1 ( 𝑑 √0,25 )+ 𝑐 5. = ∫ 1 |𝑒|βˆšπ‘’2βˆ’βˆš12 𝑑𝑒 = 1 1 π‘ π‘’π‘βˆ’1 ( 𝑒 1 ) + 𝑐 = π‘ π‘’π‘βˆ’1( 𝑒) + 𝑐 6. = π‘ π‘’π‘βˆ’1 ( π‘₯ 41 ) + c = - π‘π‘ π‘βˆ’1 ( π‘₯ 41 ) + c 7. = ∫ 1 √ (9)2 10 βˆ’π‘₯2 𝑑π‘₯ = π‘ π‘–π‘›βˆ’1 ( π‘₯ 9 10 ) + 𝑐 =π‘ π‘–π‘›βˆ’1 ( 10π‘₯ 9 ) + 𝑐 = βˆ’π‘π‘œπ‘ βˆ’1 ( 10π‘₯ 9 ) + 𝑐 8. 1 πœ‹ π‘‘π‘Žπ‘›βˆ’1 π‘₯ πœ‹ + 𝑐 = βˆ’ 1 πœ‹ π‘π‘œπ‘‘βˆ’1 ( π‘₯ πœ‹ )+c
  • 7. 9.= ∫ 1 | 𝑑|√ 𝑑2βˆ’ (1) 2 2 𝑑π‘₯ = 1 1 2 π‘ π‘’π‘βˆ’1 ( 𝑑 1 2 ) + 𝑐 = 2π‘ π‘’π‘βˆ’1(2𝑑) + 𝑐 10.π‘ π‘’π‘βˆ’1 ( π‘₯ 7 ) + 𝑐 = βˆ’π‘π‘ π‘βˆ’1 ( π‘₯ 7 ) + 𝑐