LATIHAN 5.5 MTK
-
Upload
hendra-bangkit-pramana -
Category
Documents
-
view
230 -
download
0
description
Transcript of LATIHAN 5.5 MTK
Nama : Budi Setiawan
NIM : H1A114015
LATIHAN 5.5
1. ∫ sec2 (3 x+1 ) . dx
2. ∫ tan2 x sec2 x .dx3. ∫ tan5 x sec4 x .dx4. ∫ sec5 x tan3 x .dx5. ∫ tan5 xsec x .dx6. ∫ tan4 xsec x .dx
JAWAB
1. ∫ sec2 (3 x+1 ) . dx = ∫ sec2u.du3
Misalkan : u = 3x + 1 = 13∫ sec
2u.du
dudx
= 3 = 13
tan u + c
=13
tan (3x+1) + c
2. ∫ tan2 x sec2 x .dx = ∫ tan2 ¿¿
Misal : U = tan x
dU = sec2x.dx
=∫U 2 . dx
= 13U 2
+C
= 13tan2 x+C
3. ∫ tan5 x sec4 x .dx = ∫ tan4 x sec3 x tan x secx .dx= ∫¿¿¿.sec3x tan x secx .dx
Misal : U = sec x
du = tan x sec x.dx
dx = 1
tan x sec xdu
= ∫¿¿¿.U 3.du
= ∫(U ¿¿ 4−2U2−1¿)¿¿.U 3.du
= ∫U 7−2U 5−U 3.du
= 18
U 8 - 26
U 5- 14
U 4 + c
= 18
sec8x - 13
sec6- 14
sec4 + c
4. ∫ sec5 x tan3 x .dx = ∫ tan4 x sec x tan2 x tan x .dx= ∫ sec4 x tan2 x sec x tan x .dx= ∫ sec4 x(sec2 x−1¿ sec x tan x.dx
Misal : U = sec x
dU = tan x sec x.dx
= ∫U 4 x(U 2−1¿ . dU
= ∫U 6−U 4.dU
= 17U 7
- = 15U 5
+ C
= 17sec7- =
15sec5+ C
5. ∫ tan5 xsec x .dx = ∫ tan4 x tan x sec x .dx= ∫¿¿x-1¿6 tan x sec x.dx
Misal : u = sec x
du = tan sec x dx
= ∫¿¿¿.du
= ∫(U ¿¿ 4−2u2−1)¿.du
= 15
U 5 - 23U 3
- u + c
= 15
sec5 - 23sec3- sec x + c
6. ∫ tan4 xsec x .dx = ∫¿¿
= ∫¿¿
= ∫(sec 4 x−2 sec2+1)sec x .dx
= ∫ sec3 x .dx - ∫2 sec3 x .dx + ∫ sec x . dx= sec
3 x tan x4
+ 34
(sec3x.dx – 2 ∫ sec3 x.dx + tan x + c
= 14
sec3x.tan x - 54
(sec3 x.dx + tan x + c
= 14
sec3x.tan x - 54
¿ + 12
∫ sec x . dx) + tan x + c
= 14
sec3x.tan x - 58
sec x tan x58
tan x + tan x + c
= 14
sec3x.tan x - 58
sec x tan x + 38
tan x + c