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

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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...Show more

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