Question

how do you integrate y* sqrt(6+4y-4y^2)

Answer 1

first, 6+4y-4y^2 = -(4y^2 - 4y -6) = -((2y - 1)^2 - 7))

u = 2y - 1 ------> y = (u+1)/2
du = 2 dy
du/2 = dy

see the integral above becomes :
int ( (u+1)/2) * sqrt(-(u^2 - 7)) 1/2 du
= 1/4 int (u+1) * sqrt(7-u^2) du

looks we have to use trigono subt

u= sqrt(7)sinθ --------> u^2 = 7sin^2 θ, then
sqrt(7-u^2) = sqrt(7-7sin^2 θ) = sqrt(7(1-sin^2 θ)) = sqrt(7cos^2 θ) = sqrt(7)cosθ
du = sqrt(7)cosθ dθ

1/4 int (sqrt(7)sinθ + 1) * sqrt(7)cosθ * sqrt(7)cosθ dθ
= 1/4 int (sqrt(7)sinθ + 1) * 7 cos^2 θ dθ
= 7/4 int (sqrt(7)sinθcos^2 θ + cos^2 θ) dθ
split the integral above be 2 parts :

7/4 int (sqrt(7)sinθcos^2 θ) dθ + 7/4 intcos^2 θ dθ

Answer 2

the 1st part is using u-sub, and the 2nd part is change the equation by using identity trigono :
cos^2 θ = (1 + cos2θ)/2 = 1/2 + 1/2 cos2θ