Question

Bantuin yaa..........

Answer 1

4). int (sin x)^2 cos x dx = ...
misal
u = sin x
du = cos x dx
int (sin x)^2 cos x dx = int u^2 du = 1/3 u^3 + C = 1/3 cos^3 x + C

3) m = y' = 2x - 1/x^2 = 2x - x^-2
y = int (2x - x^-2) dx
=> y = x^2 + x^(-1) + C
=> y = x^2 + 1/x + C ......... melalui (1,4)
=> 4 = 1^2 + 1/1 + C
=> 2 = C
y = x^2 + 1/x + 2

pers garis singgung di x = -1
y = (-1)^2 + 1/(-1) + 2 = 2 => (-1,2)
m = 2x - 1/x^2 = 2(-1) - 1/(-1)^2 = -2 - 1 = -3
y - y1 = m(x - x1)
=> y - 2 = -3(x - (-1))
=> y - 2 = -3x - 3
=> 3x + y + 1 = 0