Sunday, May 6, 2012

The Integral of ∫ √x * √(1-x) dx


I was recently asked by Mike Grigsby to integrate:

∫ √x √(1-x) dx

-----
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 )
-----

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


-----
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
-----

-----
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)
-----


This blog is property of Edward Shore. © 2012


1 comment:

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

    ReplyDelete

Next Week... and Plans for October 2017

I'm so excited, can't want for next week's HHC 2017 calculator conference in Nashville!  It is my annual calculator conference ...