Question

Intergal 3x cos 2x sin 2x dx
Hitunglah luas daerah yang dibatasi oleh kurva y=x-1 dan x=3-y2
Hitunglah Intergal 1 sampai 2 3x akar 3x2 -1 dx

Answer 1

1) 
∫ 3x cos 2x sin 2x dx =
∫ (3/2) (x). 2 sin 2x cos 2x dx
∫ (3/2) x . sin 4x dx
= (3/2) (x)(-1/4)(cos 4x) + (3/2)(1/16) sin 4x + c
= (-3/8) x cos 4x + (3/32) sin 4x + c
= 3/32 ( -4x cos 4x + sin 4x)  + c 
.
2)
y = x-1 --> x1 = y+1
x2= 3- y^2
x1=x2 
y +1 = 3-y^2
y^2 + y - 2 = 0
y = -2 atau y=1
L = ∫ x1 -x2 dy --> L =∫ y+1 -3+y^2 dy |1..-2|
L = 1/2 y^2 + 4y + 1/3 y^3 | 1..-2|
L = 1/2(1 -4) + 4(1+2) + 1/3 (1+8)
L = 13,5
.
3)
∫ 3x √ (3x^2 -1) dx |1...2|
u = 3x^2 - 1  
du = 6x dx
3xdx = 1/2 . du
..
u = 3x^2 -1
jika x = 1 ---> u = 2
jjika x = 2 --> u = 11
.
∫ 1/2 u^1/2 du |11...2|
= 1/2 (2/3) u^(3/2) | 11..2|
= 1/3 u√u |11...2|
= 1/3 (11√11 - 2√2)