Limit, Turunan, dan Integral Fungsi Trigonometri
A. Limit Fungsi Trigonometri
sinx=BCOB
BC=OB⋅sinx
BC=r⋅sinx⋯(1)
tanx=OAAD
AD=OA⋅tanx
AD=r⋅tanx⋯(2)
x
2π=
Ljuring O AD
π r2
2π⋅LjuringOAD=x⋅π r2
Ljuring OAD=x⋅π r
2 2π Ljuring OAD=x⋅r
2 2
Ljuring OAD=12⋅x⋅r2⋯(3)
L⊿OBC<LjuringOAD<L⊿OAD
1
2⋅OC⋅BC<12⋅x⋅r2<12⋅OA⋅AD 1
2⋅OC⋅r⋅sinx<12⋅x⋅r2<12⋅OA⋅r⋅tanx∨:21r2
OC⋅sinx
r <x<OA⋅rtanx
OC
r ⋅sinx<x<OAr ⋅tanx
cosx⋅sinx<x<rr⋅tanx cosx⋅sinx<x<tanx∨:sinx
cosx⋅sinx
cosx⋅sinx
sinx <sinxx<
sinx
cosx
sinx
cosx<sinxx<cos1x lim
x →0cosx<limx →0
x
sinx<limx→0 1 cosx
cos0<lim
x→0
x
sinx<cos01 1<lim
x→0
x
sinx<11 1<lim
x→0
x
sinx<1, maka
lim
x →0
x
sinx=1⋯(1)
cosx⋅sinx<x<tanx∨:tanx cosx⋅sinx
tanx <tanxx<tantanxx
cosx⋅sinx
sinx
cosx
<tanxx<1 cosx⋅sinx⋅cossinxx<tanxx<1 cosx⋅cos1 x<tanxx<1 cos2x<tanxx<1 lim
x →0cos 2
x<lim
x →0
x
tanx<limx →01 cos20<lim
x →0
x
tanx<1
1<lim
x→0
x
tanx<1, maka
lim
x →0
x
tanx=1⋯(2)
Rumus Limit Fungsi Trigonometri lim
x →0
x
sinx=1
lim
x →0
x
tanx=1
lim
x →0 sinx
x =1
lim
x →0 tanx
x =1
lim
x →0
ax
sinax=1
lim
x →0
ax
tanax=1
lim
x →0 sinax
ax =1
lim
x →0 tanax
ax =1
Contoh: Hitunglah lim
x →0 sin 2x
lim
x →0 sin 2x
3x =limx →0 sin 2x
3x ⋅1
¿lim
x→0 sin 2x
3x ⋅22xx
¿lim
x→0 sin 2x
2x ⋅23xx
¿1⋅23 ¿23 Jadi,
lim
x →0 sin 2x
3x =23
B. Turunan Fungsi Trigonometri
Misal: f(x)=sinx, maka
f'(x)=lim
h →0
f(x+h)−f(x)
h
¿lim
h→0
sin(x+h)−sinx
h
¿lim
h→0
2⋅cos12(x+h+x)sin12(x+h−x)
h
¿lim
h→0
2⋅cos12(2x+h)sin12(h)
h
¿lim
h→0
2⋅cos12
(
2(
x+12h)
)
sin12(h)h
¿lim
h→0
2⋅cos
(
x+12h)
sin12hh
¿lim
h→0
2⋅cos
(
x+12h)
sin12hh ⋅1
¿lim
h→0
2⋅cos
(
x+12h)
sin12hh ⋅
1 2 1 2
¿lim
h→0
cos
(
x+12h)
sin12h 12h ¿lim
h→0cos
(
x+ 12h
)
⋅limh →0sin12h
1 2h ¿cos
(
x+12(0))
⋅limh →0
sin12(h) 1 2h ¿cosx⋅1
Misal: f(x)=cosx, maka
f'(x)=lim
h →0
f(x+h)−f(x)
h
¿lim
h→0
cos(x+h)−cosx
h
¿lim
h→0
−2⋅sin12(x+h+x)sin12(x+h−x)
h
¿lim
h→0
−2⋅sin12(2x+h)sin12(h)
h
¿lim
h→0
−2⋅sin12
(
2(
x+12h)
)
sin12(h)h
¿lim
h→0
−2⋅sin
(
x+12h)
sin12hh
¿lim
h→0
−2⋅sin
(
x+12h)
sin12hh ⋅1
¿lim
h→0
−2⋅sin
(
x+12h)
sin12hh ⋅
1 2 1 2
¿lim
h→0
−sin
(
x+12h)
sin12h 12h ¿−lim
h →0sin
(
x+ 12h
)
⋅limh→0sin12h
1 2h ¿−sin
(
x+12(0))
⋅limh→0
sin12(h) 1 2h ¿−sinx⋅1
¿−sinx⋯(2)
Rumus Turunan Fungsi Trigonometri
Jika f(x)=sinx, maka f'(x)=cosx
Jika f(x)=cosx, maka f'(x)=−sinx
Jika f(x)=tanx, maka f'(x)=sec2x
Jika f(x)=cotx, maka f'(x)=−csc2x
Contoh: Jika diketahui f(x)=sin 2x, maka hitunglah f '(x)!
Jawab:
f(x)=sin 2x, maka
f'
(x)=2∙cos2x
f'
f '(x)=2 cos 2x
C. Teorema L’Hopital
Teorema
Jika g ≠0 dan x ≠ c, maka
lim
x →c
f (x)
g(x)=limx →c
f '(x)
g '(x)
Bukti:
Misal: y1=
f(x)
g(x)
y
1'=lim
∆ x→0
[
f(∆ x+x)
g(∆ x+x)−
f (x)
g(x)
∆ x
]
y
1'=lim
∆ x→0
[
f(∆ x+x)⋅g(x)−f(x)⋅g(∆ x+x)
g(∆ x+x)⋅g(x)
∆ x
]
y1
'
=lim
∆ x→0
[
f(∆ x+x)⋅g(x)−f(x)⋅g(∆ x+x)
g(∆ x+x)⋅g(x) ⋅
1
∆ x
]
y1
'
=lim
∆ x→0
[
f(∆ x+x)⋅g(x)−f(x)⋅g(∆ x+x)
∆ x ⋅g(∆ x+1x)⋅g(x)
]
y1'=lim
∆ x→0
[
f(∆ x+x)⋅g(x)−f(x)⋅g(∆ x+x)
∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
y1'=lim
∆ x→0
[
f(∆ x+x)⋅g(x)+0−f (x)⋅g(∆ x+x)
∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
y1
'
=lim
∆ x→0
[
f(∆ x+x)⋅g(x)−f(x)⋅g(x)+f(x)⋅g(x)−f(x)⋅g(∆ x+x)
∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
y1
'
=lim
∆ x→0
[
{f(∆ x+x)⋅g(x)−f(x)⋅g(x)}+{f (x)⋅g(x)−f(x)⋅g(∆ x+x)}
∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
y1'
=lim
∆ x→0
[
{f(∆ x+x)⋅g(x)−f(x)⋅g(x)}−{−f(x)⋅g(x)+f(x)⋅g(∆ x+x)}
∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
y1'=lim
∆ x→0
[
{f(∆ x+x)⋅g(x)−f(x)⋅g(x)}−{f(x)⋅g(∆ x+x)−f(x)⋅g(x)}
∆ x
]
∆ x→lim0[
1
g(∆ x+x)⋅g(x)
]
y1
'
=lim
∆ x→0
[
{f(∆ x+x)⋅g(x)−f(x)⋅g(x)}
∆ x −
{f(x)⋅g(∆ x+x)−f (x)⋅g(x)}
∆ x
]
∆ x→lim0[
1
g(∆ x+x)⋅g(x)
]
y1
'
=lim
∆ x→0
[
f(∆ x+x)⋅g(x)−f(x)⋅g(x)
∆ x −f(x)⋅g(∆ x+∆ xx)−f(x)⋅g(x)
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
y1'
= lim
∆ x→0
[
{f (∆ x+x)−f(x)}g(x)
∆ x −f
(x){g(∆ x+x)−g(x)} ∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
y1
'
=
[
lim∆ x →0
[
{f(∆ x+x)−f(x)}g(x)
∆ x
]
−∆ x→lim0[
f(x){g(∆ x+x)−g(x)}
∆ x
]
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
y1'=
[
lim∆ x →0
[
f (∆ x+x)−f(x)
∆ x
]
∆ x →lim0[g(x)]−∆ x →lim0[f(x)]∆ x→lim0[
g(∆ x+x)−g(x)
∆ x
]
]
∆ x →lim0[
1
y1'=[f '(x)g(x)−f(x)g '(x)] g 1 (x)⋅g(x) y1
'
=[f '(x)g(x)−f(x)g '(x)] 1 [g(x)]2 y1'=
f '(x)g(x)−f(x)g '(x)
[g(x)]2 ⋯(1)
Misal: y2=
f '(x)
g '(x)
y 2=lim
∆ x →0
[
f(∆ x+x)
g(∆ x+x)−
f(x)
g(x)
∆ x
]
y
2'=lim
∆ x→0
[
∆ x →lim0[
f(∆ x+x)
g(∆ x+x)−
f(x)
g(x)
∆ x
]
]
y 2
'
=lim
∆ x→0
[
∆ x →lim0[
f(∆ x+x)⋅g(x)−f(x)⋅g(∆ x+x)
g(∆ x+x)⋅g(x)
∆ x
]
]
y2
'
=lim
∆ x→0
[
∆ x →lim0[
f (∆ x+x)⋅g(x)−f(x)⋅g(∆ x+x)
g(∆ x+x)⋅g(x) ⋅
1
∆ x
]
]
y2'=lim
∆ x→0
[
∆ x →lim0[
f (∆ x+x)⋅g(x)−f(x)⋅g(∆ x+x)
∆ x ⋅g(∆ x+1x)⋅g(x)
]
]
y2
'
=lim
∆ x→0
[
∆ x →lim0[
f (∆ x+x)⋅g(x)−f(x)⋅g(∆ x+x)
∆ x
]
∆ x→lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2
'
=lim
∆ x→0
[
∆ x →lim0[
f (∆ x+x)⋅g(x)+0−f(x)⋅g(∆ x+x)
∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2
'
=lim
∆ x→0
[
∆ x →lim0[
f (∆ x+x)⋅g(x)−f(x)⋅g(x)+f(x)⋅g(x)−f(x)⋅g(∆ x+x)
∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2'=lim
∆ x→0
[
∆ x →lim0[
{f(∆ x+x)⋅g(x)−f(x)⋅g(x)}+{f(x)⋅g(x)−f (x)⋅g(∆ x+x)}
∆ x
]
∆ x→lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2'=lim
∆ x→0
[
∆ x →lim0[
{f(∆ x+x)⋅g(x)−f(x)⋅g(x)}−{−f(x)⋅g(x)+f(x)⋅g(∆ x+x)}
∆ x
]
∆ x→lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2'
=lim
∆ x→0
[
∆ x →lim0[
{f(∆ x+x)⋅g(x)−f(x)⋅g(x)}−{f(x)⋅g(∆ x+x)−f(x)⋅g(x)}
∆ x
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2'
=lim
∆ x→0
[
∆ x →lim0[
{f(∆ x+x)⋅g(x)−f(x)⋅g(x)}
∆ x −{f(x)⋅g(∆ x+∆ xx)−f(x)⋅g(x)}
]
∆ x →lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2'=lim
∆ x→0
[
∆ x →lim0[
f (∆ x+x)⋅g(x)−f(x)⋅g(x)
∆ x −f (x)⋅g(∆ x+∆ xx)−f(x)⋅g(x)
]
∆ x→lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2'
=lim
∆ x→0
[
∆ x →lim0[
{f(∆ x+x)−f(x)}g(x)
∆ x −f(x){g(∆ x∆ x+x)−g(x)}
]
∆ x→lim0[
1
y2
'
=lim
∆ x→0
[
[
∆ x→lim0[
{f(∆ x+x)−f (x)}g(x)
∆ x
]
−∆ x →lim0[
f(x){g(∆ x+x)−g(x)}
∆ x
]
]
∆ x→lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2'
=lim
∆ x→0
[
[
∆ x→lim0[
f(∆ x+x)−f(x)
∆ x
]
∆ x→lim0[g(x)]−∆ x→lim0[f(x)]∆ x →lim0[
g(∆ x+x)−g(x)
∆ x
]
]
∆ x→lim0[
1
g(∆ x+x)⋅g(x)
]
]
y2'=lim
∆ x→0
[
[f '(x)g(x)−f(x)g '(x)] 1g(x)⋅g(x)
]
y2
'
=[f '(x)g(x)−f(x)g '(x)] g 1 (x)⋅g(x) y2'=[f '(x)g(x)−f(x)g '(x)] 1
[g(x)]2 y2
'
=f '(x)g(x)−f(x)g '(x) [g(x)]2 ⋯(2)
Dari (1) dan (2)
y1'=
f '(x)g(x)−f(x)g '(x)
[g(x)]2 dan y2
'
=f '(x)g(x)−f(x)g '(x)
[g(x)]2 , maka
Jika y1
'
=y2
'
, maka lim
x →c
f(x)
g(x)=limx →c
f '(x)
g '(x)
Contoh: Jika diketahui f(x)=sin 2x dan g(x)=3x pada saat x=0, maka
tentukan lim
x →0 sin 2x
3x !
Diketahui: f(x)=sin 2x
g(x)=3x
Ditanya:
[
fg(x)(x)
]
' ? Jawab:f(x)=sin 2x
f '(x)=2 cos 2x
g(x)=3x
g '(x)=3
lim
x →0
f (x)
g(x)=limx→0
f '(x)
g '(x)
¿lim
x→0
2 cos2x
3 ¿2cos 03 ¿2(31) ¿23
Jadi, lim
x →0 sin 2x
3x =23 D. Integral Fungsi Trigonometri
Misal:
∫
f(x)dx=∫
cosx dx ¿∫
cosx⋅1dx¿
∫
cos(
x+12(0))
dx⋅limh →0
sin12(h) 1 2h ¿lim
h→0cos
(
x+ 12h
)
⋅limh →0sin12h
1 2h
¿lim
h→0
cos
(
x+12h)
sin12h 12h
¿lim
h→0
2⋅cos
(
x+12h)
sin12hh ⋅
1 2 1 2
¿lim
h→0
2⋅cos
(
x+12h)
sin12hh ⋅1
¿lim
h→0
2⋅cos
(
x+12h)
sin12hh
¿lim
h→0
2⋅cos12
(
2(
x+12h)
)
sin12(h)h
¿lim
h→0
2⋅cos12(2x+h)sin12(h)
h
¿lim
h→0
2⋅cos12(x+h+x)sin12(x+h−x)
h
¿lim
h→0
sin(x+h)−sinx
h
¿sinx+C Misal:
f(x)=sinx
∫
f(x)dx=−cosx+CRumus Integral Fungsi Trigonometri
Jika f(x)=cosx, maka
∫
f(x)dx=sinx+CJika f(x)=sinx, maka
∫
f(x)dx=−cosx+CJika f(x)=sec2x, maka
∫
f (x)dx=tanx+CJika f(x)=csc2x, maka
∫
f(x)dx=−cotx+CContoh: Jika diketahui f(x)=2 cos2x, maka hitunglah
∫
f(x)dx!Jawab:
f(x)=2 cos2x, maka
¿2
∫
cos2x dx ¿2[
12 sin 2x]
+C ¿sin 2x+C Jadi,