Question

Tolong no 1 aja, besok dikumpul, tapi harus benar ya

Answer 1

1) Int (x^2/√(x^3 - 5) dx
= Int x^2 (x^3 - 5)^(-1/2) dx
Misal
u = x^3 - 5 => du = 3x^2 dx => dx = du/(3x^2)
= Int x^2 (x^3 -5)^(-1/2) dx
= Int x^2 . u^(-1/2) du/(3x^2)
= Int (1/3)u^(-1/2) du
= (1/3)/(1/2) u^(1/2) + C
= (1/3).(2/1) √u + C
= (2/3) √(x^3 - 5) + C

2) int cos^4 2x . Sin 2x dx
Misal
u = cos 2x => du = -2 sin 2x dx => dx = du/(-2 sin 2x)
Int cos^4 2x sin 2x dx
= Int u^4 . sin 2x du/(-2 sin 2x)
= Int (-1/2)u^4 du
= (-1/2)/5 u^5 + C
= (-1/10) cos^5 2x + C

1) int (x^2 / √(x^2 - 5)) dx
misal u = x^2 - 5 => du = 2x dx => dx = du/2x
u = x^2 - 5 => x^2 = u + 5 => x = √(u + 5)

int (x^2 / √(x^2 - 5)) dx
= int (x^2 / √u) du/2x
= int x/(2√u) du
= int √(u + 5) / 2√u du
buntu :(

1) int (x^2 /√(x^2 - 5)) dx
misal x = √5 sec a => sec a = (x/√5) => a = arc sec (x/√5)
x = √5 sec a => dx = √5 tan a sec a da
x = √5 sec a => x^2 = 5 sec^2 a

int (x^2 / √(x^2 - 5)) dx
= int [(5 sec^2 a)/√(5 sec^2 a - 5)] (√5 tan a sec a da)
= int [(5 sec^2 a)/√(5(sec^2 a - 1))] (√5 tan a sec a da)
= int [(5 sec^2 a)/√(5 tan^2 a)] (√5 tan a sec a da)
= int [5 sec^2 a)/√5 tan a] (√5 tan a sec a da)
= int 5 sec^3 a da
= .... ?????