Dari contoh-contoh tersebut dapat dibuat rumus integral dengan cara substitusi sbb
∫
( ax + b )n dx =
∫
∫
1 ( ax + b )n+1 + c a(n + 1)
cos ( ax + b ) dx = sin ( ax + b ) dx = -
1 sin ( ax + b ) + c a 1 cos ( ax + b )n+1 + c a
Keterangan : Rumus no.1 di atas hanyalah penjabaran dari rumus baku yang sudah kita pelajari, yaitu :
∫
xn dx =
1 xn+1 + c n +1
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Integral trigonometri dan integral tak tentu
Pembuktian : Hitunglah
∫
4x2 dx
1. Dikerjakan dengan rumus baku
∫
4x2 dx = 4
∫
x2dx = 4.
1 3 4 x + c = x3 + c 3 3
2. Dikerjakan dengan rumus 1 di atas
∫
4x2 dx =
∫
( 2x )2 dx =
∫
( 2x + 0 )2 dx
dari rumus diketahui :
∫
( ax + b )n dx =
1 ( ax + b )n+1 + c a(n + 1)
∫
( 2x + 0 )2 dx =
1 ( 2x + 0 )2+1 + c 2(2 + 1)
=
1 ( 2x )3 + c 6
1 = .23.x3 + c 6 1 = .8.x3 + c 6 8 = .x3 + c 6 4 = .x3 + c 3
Jadi terbukti bahwa rumus no. 1 tersebut merupakan penjabaran dari rumus bakunya.
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Integral trigonometri dan integral tak tentu
C. INTEGRAL TRIGONOMETRI Rumus-rumus penunjang untuk mengerjakan integral trigonometri adalah sbb: 1. sin2 x + cos2 x = 1 2. 1 + tg2 x = sec2x 3. 1 + ctg2 x = cosec2 x 4. sin2 x = ½ ( 1 – cos 2x ) 5. cos2 x = ½ ( 1 + cos 2x ) 6. sin x. cos x = ½ sin 2x 7. sin x. cos y = ½ [sin( x + y ) + sin( x − y )] 8. sin x. sin y = ½ [cos( x + y ) − cos( x + y )] 9. cos x. cos y = ½ [cos( x − y ) + cos( x + y )] 10. 1 – cos x = 2 sin2
1 x 2
11. 1 + cos x = 2 cos2
1 x 2
contoh 1.
∫ sin
2
x dx
= ∫ 1 / 2 ( 1 - cos 2x ) dx → =
∫
rumus no. 4
(1/2 - 1/2 cos 2x ) dx
= ∫ 1 / 2 dx - ∫ 1 / 2 cos 2x dx = ∫ 1 / 2 dx - ∫ 1 / 2 cos 2x 1/2 d ( 2x ) = 1/2 x – ¼ sin 2x + c
ingat
d (2 x) = 2, sehingga dx = ½ d ( 2x ) dx
Integral + Int Trigonometri
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Integral trigonometri dan integral tak tentu
contoh 2.
∫ cos
2
3x dx
= ∫ 1 / 2 ( 1 + cos 6x ) dx → =
∫
rumus no. 5
( ½ + ½ cos 6x ) dx
= ∫ 1 / 2 dx + ∫ 1 / 2 cos 6x dx =
∫1 / 2 dx + ∫1 / 2 cos 6x 1/6 d (6x)
=½
∫
dx + 1/12
∫
cos 6x d ( 6x )
= ½ x + 1/12 sin 6x + c ingat
d (6 x ) =6 → dx
dx = 1/6 d ( 6x )
D. Integral dengan bentuk f1 ( x ) / f ( x )
Contoh
∫
dan f1 ( x ). f ( x )
f1 ( x ) / f ( x ):
1. Tentukan harga dari Jawab :
∫ (x
(2 x + 3) dx + 3x − 5)
2
misal z = ( x2 + 3x – 5 ) dz = 2x + 3 dx sehingga dz
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