Integral Fungsi Trigonometri
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Materi Matematika Peminatan SMA
![Matematika SMA : Integral Page 1
π π¬π’π§ π ππ¨π¬ π = π¬π’π§( π + π) + π¬π’π§( π β π)
π ππ¨π¬ π π¬π’π§ π = π¬π’π§( π + π) β π¬π’π§( π β π)
π ππ¨π¬ π ππ¨π¬ π = ππ¨π¬( π + π) + ππ¨π¬( π β π)
π π¬π’π§ π π¬π’π§ π = ππ¨π¬( π β π) β ππ¨π¬( π + π)
INTEGRAL FUNGSI TRIGONOMETRI
Rumus Dasar:
ο± β« sin( ππ₯ + π) ππ₯ = β
1
π
cos( ππ₯ + π) + πΆ
ο± β« cos( ππ₯ + π) ππ₯ =
1
π
sin( ππ₯ + π) + πΆ
ο± β« sec2( ππ₯ + π) ππ₯ =
1
π
tan( ππ₯ + π) + πΆ
ο± β« cosec2( ππ₯ + π) ππ₯ = β
1
π
cotan( ππ₯ + π) + πΆ
ο± β« sec( ππ₯ + π)tan( ππ₯ + π) ππ₯ =
1
π
sec( ππ₯ + π) + πΆ
ο± β« cosec( ππ₯ + π) cotan( ππ₯ + π) ππ₯ = β
1
π
cosec( ππ₯ + π) + πΆ
Contoh
1. β« sin(3π₯ + 2) ππ₯ = β
1
3
cos(3π₯ + 2) + πΆ
2. β« sec4π₯ tan4π₯ ππ₯ =
1
4
sec 4π₯ + πΆ
3. β« 2cos3π₯ sin π₯ ππ₯ = β«(sin 4π₯ β sin 2π₯) ππ₯
= β
1
4
cos4π₯ β (β
1
2
cos 2π₯) + πΆ
=
1
2
cos2π₯ β
1
4
cos4π₯ + πΆ
4. β« sin5
π₯ cos π₯ ππ₯
Dengan menggunakan teknik substitusi, maka
Misal π’ = sin π₯
ππ’ = cos π₯ ππ₯
β« sin5
π₯ cos π₯ ππ₯ = β« π’5
ππ’ =
1
6
π’6
+ πΆ =
1
6
sin6
π₯ + πΆ
5.
β« sin 6π₯ ππ₯
π
0
= [β
1
6
cos6π₯]
0
π
= (β
1
6
cos6π) β (β
1
6
cos0)
= (β
1
6
β 1) β (β
1
6
β 1)
= 0](https://image.slidesharecdn.com/integralfungsitrigonometri-170924000746/85/Integral-Fungsi-Trigonometri-1-320.jpg)
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Integral Fungsi Trigonometri
- 1. Matematika SMA : Integral Page 1 π π¬π’π§ π ππ¨π¬ π = π¬π’π§( π + π) + π¬π’π§( π β π) π ππ¨π¬ π π¬π’π§ π = π¬π’π§( π + π) β π¬π’π§( π β π) π ππ¨π¬ π ππ¨π¬ π = ππ¨π¬( π + π) + ππ¨π¬( π β π) π π¬π’π§ π π¬π’π§ π = ππ¨π¬( π β π) β ππ¨π¬( π + π) INTEGRAL FUNGSI TRIGONOMETRI Rumus Dasar: ο± β« sin( ππ₯ + π) ππ₯ = β 1 π cos( ππ₯ + π) + πΆ ο± β« cos( ππ₯ + π) ππ₯ = 1 π sin( ππ₯ + π) + πΆ ο± β« sec2( ππ₯ + π) ππ₯ = 1 π tan( ππ₯ + π) + πΆ ο± β« cosec2( ππ₯ + π) ππ₯ = β 1 π cotan( ππ₯ + π) + πΆ ο± β« sec( ππ₯ + π)tan( ππ₯ + π) ππ₯ = 1 π sec( ππ₯ + π) + πΆ ο± β« cosec( ππ₯ + π) cotan( ππ₯ + π) ππ₯ = β 1 π cosec( ππ₯ + π) + πΆ Contoh 1. β« sin(3π₯ + 2) ππ₯ = β 1 3 cos(3π₯ + 2) + πΆ 2. β« sec4π₯ tan4π₯ ππ₯ = 1 4 sec 4π₯ + πΆ 3. β« 2cos3π₯ sin π₯ ππ₯ = β«(sin 4π₯ β sin 2π₯) ππ₯ = β 1 4 cos4π₯ β (β 1 2 cos 2π₯) + πΆ = 1 2 cos2π₯ β 1 4 cos4π₯ + πΆ 4. β« sin5 π₯ cos π₯ ππ₯ Dengan menggunakan teknik substitusi, maka Misal π’ = sin π₯ ππ’ = cos π₯ ππ₯ β« sin5 π₯ cos π₯ ππ₯ = β« π’5 ππ’ = 1 6 π’6 + πΆ = 1 6 sin6 π₯ + πΆ 5. β« sin 6π₯ ππ₯ π 0 = [β 1 6 cos6π₯] 0 π = (β 1 6 cos6π) β (β 1 6 cos0) = (β 1 6 β 1) β (β 1 6 β 1) = 0
- 2. Matematika SMA : Integral Page 2 ο Jika π genap, maka : ο Jika π ganjil, maka : Contoh β« sin3 π₯ ππ₯ Jawab Karena π = 3, maka sin2 π₯ = 1 β cos2 π₯ dan π’ = cos π₯ β« sin3 π₯ ππ₯ = β« sin π₯ sin2 π₯ ππ₯ = β«sin π₯ (1 β cos2 π₯) ππ₯ = β«(sin π₯ β sin π₯ cos2 π₯) ππ₯ = βcos π₯ β β« sin π₯ cos2 π₯ ππ₯ Jika π’ = cos π₯ maka ππ’ = βsin π₯ ππ₯ β« sin3 π₯ ππ₯ = β cos π₯ β β« βπ’2 ππ’ = βcos π₯ + 1 3 π’3 + πΆ = βcos π₯ + 1 3 cos3 π₯ + πΆ Contoh β« cos4 π₯ ππ₯ Jawab Karena π = 4, maka cos2 π₯ = 1 2 (1 + cos2π₯) β« cos4 π₯ ππ₯ = β«(cos2 π₯)2 ππ₯ = β«( 1 2 (1 + cos2π₯)) 2 ππ₯ = β« 1 4 (1 + 2 cos2π₯ + cos2 2π₯) ππ₯ β«sin π π₯ ππ₯ , β«cos π π₯ ππ₯ sin2 π₯ = 1 2 (1 β cos2π₯) cos2 π₯ = 1 2 (1 + cos 2π₯) sin2 π₯ = 1 β cos2 π₯ , π’ = cos π₯ cos2 π₯ = 1 β sin2 π₯ , π’ = sin π₯
- 3. Matematika SMA : Integral Page 3 = β« 1 4 (1 + 2cos2π₯ + 1 2 (1 + cos4π₯)) ππ₯ = β« 1 4 (1 + 2 cos2π₯ + 1 2 + 1 2 cos 4π₯) ππ₯ = β« 1 4 ( 3 2 + 2 cos2π₯ + 1 2 cos4π₯) ππ₯ = 1 4 ( 3 2 π₯ + 2 β 1 2 sin 2π₯ + 1 2 β 1 4 cos4π₯) + πΆ = 3 8 π₯ + 1 4 sin 2π₯ + 1 32 cos4π₯ + πΆ ο Jika π, π keduanya genap genap, maka : ο Jika π ganjil, maka : ο Jika π ganjil, maka : Contoh β« sin4 π₯ cos5 π₯ ππ₯ Jawab Karena π = 4 dan π = 5 maka gunakan cos2 π₯ = 1 β sin2 π₯ dan π’ = sin π₯ ππ’ = cos π₯ ππ₯ β« sin4 π₯ cos5 π₯ ππ₯ = β« sin4 π₯ cos4 π₯ cos π₯ ππ₯ = β« sin4 π₯ (cos2 π₯)2 cos π₯ ππ₯ = β« sin4 π₯ (1 β sin2 π₯)2 cos π₯ ππ₯ = β« π’4(1 β π’2)2 ππ’ = β« π’4(1 β 2π’2 + π’4) ππ’ = β«( π’4 β 2π’6 + π’8) ππ’ β« sin π π₯ cos π π₯ ππ₯ sin2 π₯ = 1 2 (1 β cos2π₯) cos2 π₯ = 1 2 (1 + cos 2π₯) sin2 π₯ = 1 β cos2 π₯ , π’ = cos π₯ cos2 π₯ = 1 β sin2 π₯ , π’ = sin π₯
- 4. Matematika SMA : Integral Page 4 = 1 5 π’5 β 2 β 1 7 π’7 + 1 9 π’9 + πΆ = 1 5 sin5 π₯ β 2 7 sin7 π₯ + 1 9 sin9 π₯ + πΆ Contoh β« sin3 π₯ cos2 π₯ ππ₯ Jawab Karena π = 3 dan π = 2 maka gunakan sin2 π₯ = 1 β cos2 π₯ dan π’ = cos π₯ ππ’ = β sin π₯ ππ₯ β« sin3 π₯ cos2 π₯ ππ₯ = β« sin2 π₯ cos2 π₯ sin π₯ ππ₯ = β«(1 β cos2 π₯)cos2 π₯ sin π₯ ππ₯ = β«(1 β π’2) π’2(βππ’) = β«(βπ’2 + π’4) ππ’ = β 1 3 π’3 + 1 5 π’5 + πΆ = β 1 3 sin3 π₯ + 1 5 sin5 π₯ + πΆ Contoh β« sin4 π₯ cos4 π₯ ππ₯ Jawab Karena π = 4 dan π = 4 maka gunakan cos2 π₯ = 1 2 (1 + cos2π₯),sin2 π₯ = 1 2 (1 β cos2π₯) β« sin4 π₯ cos4 π₯ ππ₯ = β«(sin2 π₯ cos2 π₯)2 ππ₯ = β« ( 1 2 (1 β cos2π₯) β 1 2 (1 + cos2π₯)) 2 ππ₯ = β« ( 1 4 (1 β cos2 2π₯)) 2 ππ₯ = β« ( 1 4 sin2 2π₯) 2 ππ₯ = β« 1 16 ( 1 2 (1 β cos 4π₯)) 2 ππ₯ = β« 1 64 (1 β 2cos4π₯ + cos2 4π₯) ππ₯
- 5. Matematika SMA : Integral Page 5 = 1 64 β«(1 β 2 cos4π₯ + 1 2 (1 + cos8π₯)) ππ₯ = 1 64 β«(1 β 2 cos4π₯ + 1 2 + 1 2 cos8π₯) ππ₯ = 1 64 β«( 3 2 β 2 cos4π₯ + 1 2 cos8π₯) ππ₯ = 1 64 ( 3 2 π₯ β 2 β 1 4 sin 4π₯ + 1 2 β 1 8 sin 8π₯) + πΆ = 3 128 π₯ β 1 128 sin 4π₯ + 1 1.024 sin 8π₯ + πΆ LATIHAN YUKS! 1. β« cos2 π₯ ππ₯ 2. β« cos3 π₯ ππ₯ 3. β« sin5 π₯ ππ₯ 4. β« sin7 3π₯ cos2 3π₯ ππ₯ 5. β« sin4 2π₯ cos4 2π₯ ππ₯