A. Limit Fungsi Trigonometri - Limit, Turunan, dan Integral Fungsi Trigonometri
Limit, Turunan, dan Integral Fungsi Trigonometri
A. Limit Fungsi Trigonometri
BC
sin x=
OB BC=OB ⋅sin x ( )
BC=r ⋅sin x ⋯
1 AD tan x=
OA AD=OA ⋅tan x ( )
AD=r ⋅tan x ⋯
2 L
x juring O AD
2
2 π =
π r
2
2 π x
⋅ L = ⋅ π r juringOAD
2 x ⋅ π r
L = juring OAD
2 π
2 x ⋅ r
L = juring OAD
2
1
2 L ⋯
3 = ( )
juring OAD
2 ⋅ x ⋅r
L < L < L ⊿ OBC juringOAD ⊿ OAD
1
1
2
1 < 2 ⋅OC ⋅BC< 2 ⋅ x ⋅r 2 ⋅OA⋅ AD
1
1
2
1
1 <
2
2 ⋅OC ⋅r ⋅ sin x < 2 ⋅ x ⋅ r 2 ⋅OA ⋅r ⋅ tan x∨: 2 r
OC ⋅sin x OA ⋅tan x x<
<
r r OC OA r ⋅sin x< x< r ⋅ tan x
cos x
⋅sin x<x < rr ⋅tan x
cos x ⋅sin x<x <tan x ∨:sin x cos x x tan x
⋅sin x
< sin x sin x < sin x sin x cos x x cos x
⋅sin x
< sin x sin x < sin x
x
1 cos x < sin x < cos x
x
1 lim cos x<lim sin x <lim cos x
x →0 x → 0 x→ 0 x
1 cos0 <lim sin x < cos0
x→ 0 x
1 1<lim
x→ 0 sin x <
1
x
1<lim sin x <1, maka
x→ 0 x
lim
1
( )
sin x =1⋯
x →0
cos x
⋅sin x<x <tan x ∨:tan x
cos x x tan x
⋅sin x
< tan x tan x < tan x cos x x
⋅sin x
< sin x tan x <1 cos x
x
cos x ⋅sin x ⋅ cos x sin x < tan x <1
x
cos x
⋅ cos x
1 < tan x <1
2 x
cos x < tan x <1
2 x
lim cos x<lim
1
x →0 x → 0 tan x <lim x →0 2 x
cos 0<lim tan x <1
x → 0 x
1<lim tan x <1, maka
x→ 0 x
lim ( 2 )
x →0 tan x =1 ⋯ Rumus Limit Fungsi Trigonometri x
lim sin x =1
x →0 x
lim tan x =1
x →0
sin x lim
x =1 x →0
tan x lim
x =1 x →0 ax
lim
x →0 sin ax =1 ax
lim tan ax =1
x →0
sin ax lim
ax =1 x →0
tan ax lim
x →0 ax =1
Contoh: Hitunglah sin 2 x lim
x →0 3 x
Jawab: sin 2 x sin 2 x lim 3 x =lim 3 x ⋅1
x →0 x → 0
sin 2 x 2 x lim
¿
3 x ⋅ 2 x
x→ 0
sin 2 x 2 x
¿ lim x→ 0 2 x ⋅ 3 x
¿
1
⋅ 23
2
¿
3 Jadi, sin 2 x 2 lim
3 x =
3
x →0
B. Turunan Fungsi Trigonometri
( ) x =
Misal: f sin x, maka
( ) ( )
' f x+ h − f x f ( x ) = limh
h → 0sin ( x+h ) sin x −
¿ lim h→ 0 h
1 2 ( x+h+ x ) sin ( x +h−x )
⋅cos 12
2 lim
¿ h h→ 0
1 2 ( 2 x +h ) sin ( h )
⋅cos 12
2 lim
¿ h h→ 0
1
1
1 2 ⋅cos 2 x + sin ( h ) 2 2 h
2
( ) ( )
¿ lim h h→ 0
1
1 2 x + sin
⋅cos
2 h 2 h
( ) ¿ lim h h→ 0
1
1 2 ⋅cos x + sin 2 h 2 h
( )
lim
¿ ⋅1 h h→ 0
1
1
1 2 x + sin
⋅cos
2 h 2 h
( )
2
¿ lim ⋅ h
1
h→ 0
2
1
1 cos x + sin
( 2 h ) 2 h
lim
¿
1
h→ 0
2 h
1 sin 1 2 h lim cos x+
¿ ⋅lim
2 h
1
h→ 0 ( ) h → 0
2 h
1 sin ( h )
1
2 cos x + ( )
¿ ⋅lim
2
1
( ) h → 0
2 h
¿ cos x ⋅1
cos x ⋯ ( 1 )
¿
( x ) =
Misal: f cos x, maka
f ( x+ h ) − f ( x ) ' f ( x ) lim
=
h h → 0
( )
cos x +h − cos x
¿ lim h h→ 0
1 − 2 x +h+x sin x+h−x
( ) ( )
2
⋅sin 12 ¿ lim h h→ 0
1 − 2 2 x +h sin h
( ) ( )
2
⋅sin 12 ¿ lim h h→ 0
1
1
1
2 2 x+ sin ( h ) − ⋅sin 2 2 h
2
( ) ( )
lim
¿ h h→ 0
1
1 − 2 ⋅sin x + sin
( 2 h ) 2 h
lim
¿
h
h→ 01
1 2 x + sin − ⋅sin 2 h 2 h
( )
¿ lim ⋅1
h
h→ 01
1
1 − 2 ⋅sin x + sin
( 2 h ) 2 h
2 lim
¿ ⋅
h
1
h→ 0
2
1
1 − sin x + sin 2 h 2 h
( ) ¿ lim
1
h→ 0
2 h
1 sin 1 2 h
¿ − lim sin x + ⋅ lim ( 2 h )
1
h → 0 h→ 0
2 h
1 sin ( h )
1
2
( ) ¿ − sin x+ ⋅lim
(
2 )
1
h→ 0
2 h
¿ − sin x ⋅1
( ) ¿ − sin x ⋯
2 Rumus Turunan Fungsi Trigonometri
' ( x )
Jika f = ( x ) cos x sin x, maka f =
' ( x )
= ( ) Jika f cos x, maka f x =− sin x
'
2 ( ) x =
Jika f ( x ) = sec x tan x, maka f
'
2 ( x ) =
Jika f ( x ) =− csc x cot x, maka f
( x ) = ( x )
Contoh: Jika diketahui f ! sin 2 x, maka hitunglah f ' Jawab:
( ) f x =
sin 2 x, maka
' f ( x ) = 2∙ cos2 x
' f ( x ) = 2cos 2 x
Jadi,
C. Teorema L’Hopital
1
= lim
1 '
y
) ]
∆ x+x ) ⋅g ( x
g (
⋅
( ∆ x +x
∆ x→ 0 [ f ( ∆ x +x ) ⋅ g ( x ) − f ( x ) ⋅ g ( ∆ x +x ) ∆ x
= lim
1 '
y
∆ x ]
1
∆ x→ 0 [ f
) ⋅ g
( x ) − f ( x ) ⋅ g
lim
1
) ]
∆ x+ x ) ⋅g ( x
g (
1
∆ x →0 [
]
( x
) ∆ x
( ∆ x+ x
) ⋅ g
( x
− f
)
( ∆ x +x ) g ( ∆ x +x ) ⋅ g ( x ) ⋅
∆ x→ 0 [ f ( ∆ x +x ) ⋅ g
( x )
Misal: y
= lim
1 '
y
f ( x ) g ( x )
=
1
Bukti:
−
x →c f ' ( x ) g ' ( x )
= lim
x →c f ( x ) g ( x )
Jika g ≠ 0 dan x ≠ c, maka lim
Teorema
f ' ( x ) = 2 cos 2 x
∆ x→ 0 [ f ( ∆ x+x ) g ( ∆ x +x )
f ( x ) g ( x ) ∆ x
= lim
( ∆ x+x
1 '
y
]
) ∆ x
( x
) ⋅g
f ( x ) ⋅ g ( ∆ x+x ) g
]
−
( x )
∆ x→ 0 [ f ( ∆ x+ x ) ⋅g
= lim
1 '
y
⋅g
( ∆ x +x )- 0−f ( x )
∆ x→ 0 [ f ( ∆ x +x ) ⋅ g
- f
- f
)
{ g{ f (
∆ x +x )
− f
( x
) } g
( x
) ∆ x
−
f (
x
( ∆ x+ x
= lim
)
− g
( x
) }
∆ x ]
lim
∆ x →0 [
1
g ( ∆ x+x ) ⋅g ( x ) ] y
1 '
∆ x→ 0 [
1 '
[
( x )
} ∆ x
]
lim
∆ x→ 0 [
1
g ( ∆ x +x ) ⋅ g ( x )
] y
1 '
= lim
∆ x→ 0 [ f ( ∆ x +x ) ⋅ g
−
g ( ∆ x+x ) ⋅ g ( x ) ] y
f ( x ) ⋅ g ( x )
∆ x
−
f ( x ) ⋅ g ( ∆ x +x )
−
f ( x ) ⋅ g ( x )
∆ x ]
lim
∆ x →0 [
1
=
lim
−
∆ x →0
[
lim
∆ x →0 [ f ( ∆ x + x ) − f ( x ) ∆ x
]
lim
∆ x → 0
[ g
( x
)
]− lim
[ f
1 '
( x
)
] lim
∆ x→ 0 [ g ( ∆ x +x ) − g ( x ) ∆ x
] ]
lim
∆ x →0 [
1
g ( ∆ x+ x ) ⋅g ( x )
]
=
] y
∆ x →0 [
∆ x→ 0 [
f
{ f
( ∆ x +x
)
− f
( x
) } g
( x
) ∆ x
]
− lim
(
xg ( ∆ x+x ) ⋅g ( x )
) { g
( ∆ x +x
)
− g
( x
) } ∆ x
] ]
lim
∆ x →0 [
1
f ( x ) ⋅ g ( x )
( ∆ x + x )
= lim
= lim
) ∆ x
]
lim
∆ x → 0 [
1
g (
∆ x +x ) ⋅ g ( x
) ] y
1 '
∆ x→ 0 [ { f ( ∆ x +x )
) ⋅ g
⋅ g ( x ) − f ( x )
⋅ g ( x ) } + { f ( x )
⋅ g ( x ) − f ( x )
⋅ g ( ∆ x +x ) }
∆ x ]
lim
∆ x →0 [
1
g ( ∆ x+ x ) ⋅g ( x )
] y
( ∆ x +x
( x
= lim
) ⋅ g
1 '
∆ x ]
lim
∆ x → 0 [
y
g ( ∆ x +x ) ⋅ g ( x ) ] y
1 '
= lim
∆ x→ 0 [ f
( ∆ x +x
( x
− f
)
− f
( x
) ⋅ g
( x
)
( x
)
⋅ g( x
)
1 '
∆ x→ 0 [
{
f ( x )
⋅ gg (
f ( x ) ⋅ g ( x )
} − {
f ( x )
⋅ g( ∆ x +x )
−
f ( x ) ⋅g ( x )
} ∆ x
]
lim
∆ x→ 0 [
1
∆ x +x )
( x )
⋅ g ( x
) ] y
1 '
= lim
∆ x→ 0 [ { f ( ∆ x +x ) ⋅ g
( x )
−
f ( x ) ⋅ g ( x )
} ∆ x
−
−
∆ x→ 0 [ { f ( ∆ x +x ) ⋅ g
{ f (
( x
∆ x +x )
⋅ g ( x
)
− f
( x
) ⋅ g
( x
) } − { − f
( x
) ⋅ g
)
= lim
( x
) ⋅ g
) }
∆ x ]
lim
∆ x → 0 [
1
g ( ∆ x +x ) ⋅ g ( x )
] y
1 '
( ∆ x +x
- 0−f ( x )
- f
−
lim
∆ x→ 0 [
1
g ( ∆ x +x ) ⋅ g ( x )
] ] y
2 '
= lim
∆ x→ 0 [
lim
∆ x → 0 [
{ f ( ∆ x + x ) ⋅ g
( x )
f ( x ) ⋅g ( x )
{ f ( ∆ x +x ) ⋅ g ( x ) − f ( x ) ⋅g ( x ) } + { f ( x ) ⋅ g ( x ) − f ( x ) ⋅ g ( ∆ x +x ) } ∆ x
} − { − f ( x ) ⋅ g
( x )
( ∆ x+x ) }
∆ x ]
lim
∆ x→ 0 [
1
g ( ∆ x + x ) ⋅ g ( x ) ]
] y
2 '
= lim
∆ x→ 0 [
]
∆ x → 0 [
∆ x → 0 [ { f ( ∆ x +x ) ⋅ g
)
( ∆ x + x
) ⋅ g
( x
)
− f
( x
) ⋅g
( x
)
( x
)
⋅g
( x
− f
lim
( x
) ⋅ g
( ∆ x+x
) ∆ x
]
lim
∆ x →0 [
1
g ( ∆ x+x ) ⋅ g ( x )
] ] y
2 '
= lim
∆ x→ 0 [
lim
( x )
lim
1
= lim
∆ x→ 0 [
lim
∆ x → 0 [ f ( ∆ x +x ) ⋅ g
( x ) − f ( x ) ⋅g
( x ) ∆ x
−
f ( x ) ⋅ g ( ∆ x +x ) − f ( x )
⋅g ( x )
∆ x ]
lim
∆ x→ 0 [
g ( ∆ x +x ) ⋅ g ( x )
] ] y
] ] y
2 '
= lim
∆ x→ 0 [
lim
∆ x → 0 [ { f ( ∆ x + x ) − f ( x ) } g ( x )
∆ x
−
f ( x ) { g ( ∆ x +x ) − g ( x ) }
∆ x
]lim
∆ x→ 0 [
1
g ( ∆ x + x ) ⋅ g ( x )
2 '
g ( ∆ x+ x ) ⋅ g ( x )
−
2 '
f ( x ) ⋅g ( x )
} − {
f ( x )
⋅ g( ∆ x +x )
−
f ( x ) ⋅ g ( x )
} ∆ x
]
lim
∆ x →0 [
1
g ( ∆ x+x ) ⋅g ( x )
] ] y
= lim
1
∆ x→ 0 [
lim
∆ x → 0 [ { f ( ∆ x + x )
⋅ g ( x ) − f ( x )
⋅g ( x ) }
∆ x
−
{ f ( x )
⋅ g( ∆ x+x ) − f ( x ) ⋅ g
( x ) } ∆ x
]
lim
∆ x → 0 [
∆ x → 0 [ f
∆ x→ 0 [
] ]
= lim
f ' ( x ) g ( x )
−
f ( x ) g ' ( x )
[ g (
x ) ]
2
⋯ ( 1 ) Misal: y
2
=
f ' ( x ) g ' ( x )
y
2
∆ x →0 [ f ( ∆ x+ x ) g ( ∆ x+x )
1 '
−
f ( x ) g ( x ) ∆ x
] y
2 '
= lim
∆ x→ 0 [
lim
∆ x → 0 [ f ( ∆ x + x ) g ( ∆ x+ x )
−
f ( x ) g ( x ) ∆ x
] ] y
2 '
= lim
=
y
lim
g ( x ) ⋅ g ( x )
y
1 '
=[ f '
( x
) g
( x
) − f
( x
) g '
( x
) ]
1
y
2
1 '
=[ f ' (
x ) g
( x
)
− f
( x
) g '
( x
) ]
1 [ g
( x )
]
∆ x→ 0 [
∆ x → 0 [ f ( ∆ x +x ) ⋅ g ( x ) − f ( x ) ⋅ g ( ∆ x +x ) g ( ∆ x +x ) ⋅ g ( x )
= lim
y
∆ x → 0 [ f ( ∆ x + x ) ⋅ g
( x ) − f ( x ) ⋅g
( ∆ x + x )
∆ x]
lim
∆ x→ 0 [
1
g (
∆ x + x )
⋅ g ( x
) ]
]
2 '
∆ x→ 0 [
= lim
∆ x→ 0 [
lim
∆ x → 0 [ f ( ∆ x +x ) ⋅ g
( x )
⋅ g ( ∆ x+x )
∆ x ]
lim
∆ x →0 [
1
g ( ∆ x+x ) ⋅ g ( x )
] ] y
2 '
lim
= lim
∆ x ]
]
]
y
2 '
= lim
∆ x→ 0 [
lim
∆ x → 0 [ f ( ∆ x+ x ) ⋅ g ( x ) − f ( x ) ⋅ g ( ∆ x+ x ) g
( ∆ x+ x
) ⋅ g ( x
) ⋅
1
∆ x ]
y
2 '
2 '
= lim
∆ x→ 0 [
lim
∆ x → 0 [ f ( ∆ x+ x ) ⋅ g
( x )
−
f ( x ) ⋅ g
( ∆ x+ x )
∆ x ⋅
1
g ( ∆ x +x ) ⋅ g ( x )
] ]
y
- f ( x ) ⋅ g
f ∆ x+x f x g x f x g ∆ x+ x g x ( ) − ( ) ( ) ( ) ( ) − ( )
' { } { }
1
y = lim lim − lim lim
2 ∆ x→ 0 ∆ x→ 0 ∆ x ∆ x →0 ∆ x ∆ x→ 0 g ( ∆ x +x ) ( x )
[ ] [ ] ⋅ g [ ]
[ ]
[ ] f ∆ x+x f x g ∆ x+ x g x
( ) − ( ) ( ) − ( )
1
' y = lim lim lim ( x ) ( x ) lim
[ g ]− lim [ f ] lim
2 ∆ x
∆ x ( ) ( ) ∆ x→ 0 ∆ x→ 0 ∆ x→ 0 ∆ x→ 0 ∆ x →0 ∆ x→ 0 g ∆ x + x ⋅ g x
[ ] [ ] [ ] [ [
] ] '
1
y = lim ( x ) g ( x ) − f ( x ) g ' ( x )
[ f ' ]
2 ∆ x→ 0 g ( x ) ⋅g ( x )
[ ] '
1
( x ) g ( x ) − f ( x ) g ' ( x ) 2 =[ f ' ]
y
( ) ( ) g x ⋅ g x
'
1
( x ) g ( x ) − f ( x ) g ' ( x ) 2 =[ f ' ]
2
y
( x )
[ g ]
f ' ( x ) g ( x ) f ( x ) g ' ( x ) ' −
2
2
y
= ⋯ ( 2 )
x
[ g ( ) ]
(
1 ) ( 2 ) Dari dan
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ' f ' x g x − f x g ' x ' f ' x g x − f x g ' x y
= =
1 2 dan y
2 2 , maka x x
[ g ( ) ] [ g ( ) ]
f ( x ) f ' ( x ) ' ' y = lim
Jika y = , maka lim
1
2 x →c g ( x ) x →c g ' ( x )
( x ) ( x )
= = Contoh: Jika diketahui f sin 2 x dan g 3 x pada saat x=0, maka sin 2 x tentukan lim 3 x !
x →0 ( x ) sin 2 x
= Diketahui: f
g ( x ) = 3 x ( ) f x
Ditanya: ' ?
g x ( )
[ ]
Jawab:
( ) f x = sin 2 x f ' ( x ) 2 cos 2 x
=
g ( x ) 3 x
=
g ' ( x ) =
3
( ) ( ) f x f ' x
lim lim =
x →0 g ( x ) x→ 0 g ' ( x )
2 cos2 x
¿ lim
3
x→ 0
2cos 0
¿
3 2 ( 1 )
¿
3
2
¿
3 sin 2 x
2 Jadi, lim 3 x =
3
x →0
D. Integral Fungsi Trigonometri
Misal:
f ( x ) cos x
=
∫ f ( x ) dx=
⋅cos 12 (
x+h+ x
f ( x ) = sin x ∫ f ( x ) dx=−cos x +C
Misal:
h ¿ sin x+C
sin ( x+h ) − sin x
h→ 0
lim
¿
) h
( x +h−x
2
1
sin
)
2
Jika f
¿ lim h→ 0
) h
( h
2
1
sin
)
2 x +h
⋅cos 12 (
2
¿ lim h→ 0
( h )
h
2
1
Rumus Integral Fungsi Trigonometri
( x )
) )
2 x, maka
∫
∫ f ( x ) dx=
= 2 cos2 x, maka
f ( x )
Jawab:
) dx!
( x
∫ f
= 2 cos2 x, maka hitunglah
( x )
Contoh: Jika diketahui f
) dx=−cot x +C
( x
∫ f
Jika f ( x ) = csc
= cos x, maka
) dx=tan x +C
( x
∫ f
2 x, maka
Jika f ( x ) = sec
) dx=−cos x +C
( x
∫ f
= sin x, maka
( x )
Jika f
) dx=sin x+C
( x
∫ f
sin
1 2 h
∫
1 2 h
lim
¿
1 2 h 1 2 h
1 2 h ) sin
( x +
cos
¿ lim h→ 0
1 2 h 1 2 h
sin
)
⋅lim
h → 01 2 h
( x+
cos
¿ lim h→ 0
)
2
( h
2
1
sin
( ) ) dx ⋅ lim h →0
2
1
( x +
cos
∫
⋅1dx ¿
cos x
¿ ∫
cos x dx
h→ 0
⋅cos ( x +
(
x +
h ⋅1 ¿
2
(
2
1
2 ⋅cos
h ¿ lim h→ 0
1 2 h
sin
)
1 2 h
( x +
2 ⋅cos
h→ 0
lim
1 2 h
1 2 h
sin
)
1 2 h
⋅cos ( x +
2
¿ lim h→ 0
2
1
2
1
h ⋅
1 2 h
sin
)
2 cos2 x dx
¿
2
∫
cos2 x dx
¿
2
- C
[
1 2 sin 2 x
]
¿
sin 2 x+C Jadi,
∫ f
( x
) dx=sin 2 x +C