Question

Integral dari (96x) (6x^2+8) sin^2(3x^2+4)

Answer 1

Int (96x)(6x²+8)sin²(3x²+4)  dx
= int (96x)(2)(3x²+4)sin²(3x²+4)  dx
= int (192x)(3x²+4)sin²(3x²+4)  dx
misal t = 3x²+4
dt = 6x dx
1/6 dt = x dx
maka, integral di atas dapat menjadi :
= int (192/6)(t)sin²(t) dt
= int 32t sin²(t)  dt
= int 32t (1 - cos(2t))/2  dt
= int 16t (1  -  cos(2t))  dt
= int 16t dt - int 16t cos(2t) dt
= 8t² - int 16t cos(2t) dt

gnakan integral parsial, untuk mnentukan hasil int 16tcos(2t)dt
formula :  int udv = uv - int u dv
u = 16t -----> du = 16 dt
dv =  cos(2t) dt
v = int cos(2t) dt = 1/2 sin(2t)
jadi,
int 16t cos(2t) dt = 16t (1/2 sin(2t) - int 1/2 sin(2t) (16) dt
= 8t sin(2t) - int 8 sin(2t) dt
= 8t sin(2t) - (-8/2) cos(2t) + C
= 8t sin(2t) + 4 cos(2t)

jadi,
= 8t² - int16t cos(2t) dt
= 8t² - [8t sin(2t) + 4 cos(2t)] + C
= 8t² - 8t sin(2t) - 4 cos(2t) + C
substitusi nilai t = 3x² + 4, maka
= 8(3x² + 4)² - 8(3x²+4)sin(2(3x²+4)) - 4 cos(2(3x²+4)) + C
= 8(3x²+4)² - 8(3x²+4)sin(6x²+8) - 4cos(6x²+8) + C

i hope i dont make mistake!