I was recently asked by Mike Grigsby to integrate:

∫ √x √(1-x) dx

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Make the substitution:

x = sin^2 θ

Note:

√x = sin θ ,

θ = asin x ,

dx = 2 sin θ cos θ dθ ,

and

cos θ = √(1 - sin^2 θ )

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Hence:

∫ √x √(1-x) dx (x to θ)

= ∫ sin θ * √(1 - sin^2 θ )* 2 *sin θ * cos θ dθ

= 2 ∫ sin^2 θ cos^2 θ dθ

= 2 ∫ (1 - cos^2 θ ) * cos^2 θ dθ

= 2 ∫ cos^2 θ - cos^4 θ dθ

= 2 ∫ 1/8 - 1/8 * cos (4θ) dθ (See Note A)

= 2 ( θ/8 - 1/8 * 1/4* sin (4θ))

= 2 ( θ/8 - 1/32 * sin (4θ))

= 2 ( θ / 8 - 1/32 * [8 sin θ cos^3 θ - 4 sin θ cos θ]) (See Note B)

(θ to x)

= 2 ( asin √x / 8 - 1/32 * [8 √x (1 - x)^3/2 - 4 √x √(1-x)] )

= 2 ( asin √x / 8 - (√x √(1-x) (2 - 2x - 1))/8 )

= asin √x / 4 - (√x √(1-x) (1 - 2x)) / 4

Final:

∫ √x √(1-x) dx = 1/4 * ( asin √x - (√x √(1-x) (1 - 2x)) + C

for some constant C

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Note A

cos 2a = 2 cos^2 a - 1

cos^2 a = 1/2 * (cos 2a + 1)

cos^4 a

= [1/2 * (cos 2a + 1)]^2

= 1/4 * (cos^2 2a + 2 cos 2a + 1)

= 1/4 * ( (cos 4a + 1)/2 + 2 cos 2a + 1)

= 1/8 * (cos 4a + 4 cos 2a + 3)

= 1/8 * cos 4a + 1/2 * cos 2a + 3/8

cos^2 a - cos^4 a

= (1/2 * cos 2a + 1/2) - (1/8 * cos 4a + 1/2 * cos 2a + 3/8)

= 1/8 - 1/8 * cos 4a

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Note B

sin 2a = 2 sin a cos a

sin 4a

= 2 sin 2a cos 2a

= 8 sin a cos^3 a - 4 sin a cos a

Using the triangle from above:

sin θ = √x

cos θ = √(1-x)

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This blog is property of Edward Shore. © 2012

You got it. I differentiated the antiderivative and got back the original function.

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