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