Adv.H Maths on integration?

50)
∫ sqrt( 25-x^2) dx

Let x= 5 sin θ
dx = 5 cos θ dθ
25-x^2 = 25-25 sin^2 θ = 25(1-sin^2 θ) = 25 cos^2 θ
sqrt(25-x^2) = 5 cos θ

∫ sqrt( 25-x^2) dx = ∫ (5 cos θ) ( 5 cos θ) dθ
= 25 ∫ cos^2 θ dθ
= 25 ∫ (1/2) ( 1+cos 2θ) dθ
= (25/2) ∫ ( 1+ cos 2θ) dθ
= (25/2) ∫ dθ + (25/2) ∫ cos 2θ dθ
= (25/2) θ + (25/2)(1/2) sin 2θ
= (25/2) θ + (25/2)(1/2) (2 sin θ cos θ)
= (25/2) θ + (25/2) sin θ cos θ

x = 5 sin θ
sin θ = x/5
θ = sin^-1(x/5)
cos θ = sqrt( 1-sin^2 θ) = sqrt( 1- x^2/25) = (1/5) sqrt(25-x^2)

(25/2) θ + (25/2) sin θ cos θ = (25/2) sin^-1(x/5) + (25/2) (x/5) (1/5) sqrt(25-x^2)
= (25/2) sin^-1(x/5) + (1/2) x sqrt( 25-x^2)

∫ sqrt( 25-x^2) dx = (25/2) sin^-1(x/5) + (1/2) x sqrt( 25-x^2)

Let F(x) = (25/2) sin^-1(x/5) + (1/2) x sqrt( 25-x^2)

substitute the upper limit 5/√2
F(5/√2) = 25π/8 + 25/4
substitute the lower limit 0
F(0) = 0

subtract:
F(3)-F(0) = 25π/8 + 25/4

51)

∫ sqrt(1-4x^2) dx

Let x= (1/2) sin θ
dx = (1/2) cos θ dt
1-4x^2 = 1- (4)(1/4) sin^2 θ = 1-sin^2 θ = cos^2 θ
sqrt(1-4x^2) = cos θ

∫ sqrt(1-4x^2) dx = (1/2) ∫ (cos θ) (cos θ) dt
= (1/2) ∫ cos^2 θ dθ


∫ cos^2 θ dθ = ∫ (1+cos 2θ) dθ / 2
= ∫ (1/2) dθ + ∫ (1/2) cos 2θ dθ
= θ / 2 + (1/2) ∫ cos 2θ dθ ------------------ (1)

let u = 2θ
du = 2 dθ
dθ = (1/2) du
(1/2) ∫ cos 2θ dθ = (1/2)(1/2) ∫ cos u du
(1/2) ∫ cos 2θ dθ = (1/4) ∫ cos u du
(1/2) ∫ cos 2θ dθ = (1/4) sin u
(1/2) ∫ cos 2θ dθ = (1/4) sin 2θ
= (1/4) ( 2 sin θ cos θ )
= (1/2) sin θ cos θ
substitute this into (1)

∫ cos^2 θ dθ = (1/2) θ + (1/2) sin θ cos θ
(1/2) ∫ cos^2 θ dθ = (1/4) θ + (1/4) sin θ cos θ


transform θ back to x
x = (1/2) sin θ
sin θ = 2x
θ = sin^-1(2x)
cos θ = sqrt( 1-sin^2 θ) = sqrt(1- 4x^2)

(1/2) θ + (1/2) sin θ cos θ = (1/4) sin^-1(2x) + (1/4) (2x) sqrt(1-4x^2)

∫ sqrt(1-4x^2) dx = (1/4) sin^-1(2x) + (1/2) x sqrt(1-4x^2)

Let F(x) = (1/4) sin^-1(2x) + (1/2) x sqrt(1-4x^2)

substitute the upper limit 1/4
F(1/4) = (1/4) sin^-1(1/2) + (1/2) (1/4) sqrt(1-4/16)
substitute the lower limit 0
F(0) = √3/16 + π/24

subtract:
F(1/4)-F(0) = √3/16+π/24

52)
6 ∫ dx/ sqrt(2x-x^2)

2x-x^2 = -(x^2-2x) = -( (x^2-2x+1) -1) = - ( (x-1)^2 -1) = 1-(x-1)^2
6 ∫ dx/ sqrt(2x-x^2) = 6 ∫ dx/sqrt(1- (x-1)^2 )

Let u= x-1
du = dx
6 ∫ dx/sqrt(1- (x-1)^2 ) = 6 ∫ du/dqrt(1-u^2) = 6 sin^-1(u)
= 6 sin^-1(x-1)

Let F(x) = 6 sin^-1(x-1)

substitute the upper limit 1
F(1) = 6 sin^-1(0) = 0
substitute the lower limit 1/2
F(1/2) = 6 sin^-1(1/2-1) = 6sin^-1(-1/2) = 6(-pi/6) = -pi

subtract:
F(1)-F(1/2) = 0-(-pi) = pi

The final answer should be