Using trig substitution, what is the integral of (1-y)sqrt(9-16y^2)?

Modified:

∫ (1-y) sqrt(9-16y^2) dy

Let y = (3/4) sin t

dy = (3/4) cos t dt

9-16y^2 = 9 - 16(3/4)^2 sin^2 t = 9 - 9 sin^2 t = 9 cos^2 t

sqrt(9-y^2) = 3 cos t

∫ (1-y) sqrt(9-16y^2) dy = ∫ (1- (3/4)sin t ) 3 cos t (3/4) cos t dt

= (9/4) ∫ (1 -(3/4) sin t) cos^2 t dt

= (9/4) ∫ cos^2 t dt - (27/16) ∫ cos^2 t sin t dt ------(*)

Let us evaluate ∫ cos^2 t dt

∫ (1+cos 2t) dt / 2

∫ (1/2) dt + ∫ (1/2) cos 2t dt

= t / 2 + (1/2) ∫ cos 2t dt ------ (1)

let s = 2t

ds = 2 dt

dt = (1/2) ds

(1/2) ∫ cos 2t dt = (1/2)(1/2) ∫ cos s ds

(1/2) ∫ cos 2t dt = (1/4) ∫ cos s ds

(1/2) ∫ cos 2t dt = (1/4) sin s

(1/2) ∫ cos 2t dt = (1/4) sin 2t

= (1/4) ( 2 sin t cos t )

= (1/2) sin t cos t

substitute this into (1)

∫ cos^2 t dt = (1/2) t + (1/2) sin t cos t

(9/4) ∫ cos^2 t dt = (9/8) t + (9/8) sin t cos t

∫ cos^2 t sin t dt

Let u = cos t

du = -sin t dt

sint dt = -du

∫ cos^2 t sin t dt = - ∫ u^2 du = (-1/3) u^3 = (-1/3) cos^3 t

(27/16) ∫ cos^2 t sin t dt = (-9/16) cos^3 t

(*) becomes:

(9/8) t + (9/8) sin t cos t + (9/16) cos^3 t ------- (**)

our substitution was y = (3/4) sin t

t = sin^-1( 4y/3)

sin t = 4y/3

cos t = sqrt(1-sin^2 t) = sqrt ( 1- 16y^2/9) = (1/3) sqrt(9-16y^2)

cos^3 t = (1-sin^2 t)^(3/2) = (1/27) (9-16y^2)^(3/2)

(**) becomes:

(9/8) sin^-1(4y/3) +(9/8) (4y/3) (1/3) sqrt(9-16y^2)+(9/16)(1/27) (9-16y^2)^(3/2)

(9/8) sin^-1(4y/3) +(1/2) y sqrt(9-16y^2) + (1/48) (9-16y^2)^(3/2) + C

I believe the answer is "banana" but I could be wrong.