What is the integral of sqrt (4x - x^2) dx?

Question

What is the integral of sqrt (4x - x^2) dx?

Answer 1

∫ √(4x - x²) dx =



first, complete the square inside the root by adding and subtracting 4:



∫ √(4 - 4 + 4x - x²) dx =



∫ √[4 - (x² - 4x + 4)] dx =



∫ √[4 - (x - 2)²] dx =



substitute:



(x - 2) = 2sinθ



sinθ = (x - 2)/2



θ = arcsin[(x - 2)/2]



x = 2sinθ + 2



dx = 2cosθ dθ



yielding:



∫ √[4 - (x - 2)²] dx = ∫ √[4 - (2sinθ)²] 2cosθ dθ =



∫ √(4 - 4sin²θ) 2cosθ dθ =



factor out 4:



∫ √[4(1 - sin²θ)] 2cosθ dθ =



∫ 2√(1 - sin²θ) 2cosθ dθ =



replace (1 - sin²θ) with cos²θ:



∫ 2√(cos²θ) 2cosθ dθ =



∫ 2cosθ 2cosθ dθ =



∫ 4 cos²θ dθ =



recall the power-reducing formula cos²θ = (1/2)[1 + cos(2θ)]:



∫ 4 (1/2)[1 + cos(2θ)] dθ =



∫ [2 + 2cos(2θ)] dθ =



break it up into:



∫ 2 dθ + ∫ 2cos(2θ) dθ =



2θ + sin(2θ) + C =



that is, according to double-angle identities:



2θ + 2sinθ cosθ + C



now recall that:



θ = arcsin[(x - 2)/2]



sinθ = (x - 2)/2



hence:



cosθ = √(1 - sin²θ) = √{1 - [(x - 2)/2]²} = √{1 - [(x - 2)²/4]} = √{[4 - (x² - 4x + 4)]/4} =

√[(4 - x² + 4x - 4)/4] = [√(4x - x²)]/2



thus, substituting back, you get:



2θ + 2sinθ cosθ + C = 2arcsin[(x - 2)/2] + 2[(x - 2)/2] {[√(4x - x²)]/2} + C =



in conclusion:



∫ √(4x - x²) dx = 2arcsin[(x - 2)/2] + (1/2)(x - 2)√(4x - x²) + C







I hope it helps..

Answer 2

I just threw this in wolframalpha.com



Possible intermediate steps:

integral sqrt(4 x-x^2) dx

For the integrand sqrt(4 x-x^2), complete the square:

= integral sqrt(4-(x-2)^2) dx

For the integrand sqrt(4-(x-2)^2), substitute u = x-2 and du = dx:

= integral sqrt(4-u^2) du

For the integrand, sqrt(4-u^2) substitute u = 2 sin(s) and du = 2 cos(s) ds. Then sqrt(4-u^2) = sqrt(4-4 sin^2(s)) = 2 cos(s) and s = sin^(-1)(u/2):

= 4 integral cos^2(s) ds

Write cos^2(s) as 1/2 cos(2 s)+1/2:

= 4 integral (1/2 cos(2 s)+1/2) ds

Integrate the sum term by term and factor out constants:

= 4 integral 1/2 ds+2 integral cos(2 s) ds

For the integrand cos(2 s), substitute p = 2 s and dp = 2 ds:

= integral cos(p) dp+4 integral 1/2 ds

The integral of 1/2 is s/2:

= integral cos(p) dp+2 s

The integral of cos(p) is sin(p):

= sin(p)+2 s+constant

Substitute back for p = 2 s:

= 2 s+sin(2 s)+constant

Substitute back for s = sin^(-1)(u/2):

= 1/2 sqrt(4-u^2) u+2 sin^(-1)(u/2)+constant

Substitute back for u = x-2:

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

Factor the answer a different way:

= 1/2 sqrt(4-x) x^(3/2)-sqrt(-(x-4) x)+2 sin^(-1)((x-2)/2)+constant

Which is equivalent for restricted x values to:

= (sqrt(-(x-4) x) (sqrt(x-4) (x-2) sqrt(x)-8 log(2 (sqrt(x-4)+sqrt(x)))))/(2 sqrt(x-4) sqrt(x))+constant

Answer 3

This function is defined where the stuff insides the square root is positive. So we can write:

4x - x^2 = 0 = x(4 - x) which gives x = 0 and x = 4

We must evaluate the integral between these limits



Substitution.

x = 4cos^2(A) ..... flip this to get A = ACOS{SQRT[(x/4)]}

dx = -8cos(A)sin(A)dA



4x - x^2 = 4[4cos^2(A)] - [4cos^2(A)]^2

4x - x^2 = 16cos^2(A)[1 - cos^2(A)] = 16cos^2(A)sin^2(A)

SQRT[4x - x^2] = 4cos(A)sin(A)



SQRT[4x - x^2]dx = [4cos(A)sin(A)][-8cos(A)sin(A)dA]

SQRT[4x - x^2]dx = -32cos^2(A)sin^2(A)dA



Trig identity: sin(2A)/2 = cos(A)sin(A)



SQRT[4x - x^2]dx = -32[sin(2A)/2]^2dA = -8sin^2(2A)dA



Trig identity: sin^2(2A) = [1 - cos(4A)]/2



SQRT[4x - x^2]dx = -8{[1 - cos(4A)]/2}dA =-4dA + 4cos(4A)dA



We can now integrate this to get: -4A + sin(4A)



Go back to the original substitution : A = ACOS{SQRT[(x)/4]}



-4{ACOS[SQRT(x/4)]} + sin{4*ACOS[SQRT(x/4)]}



Evaluate this between x = 0 and x = 4



x = 0 gives A = -4{ACOS[0]} + sin{4*ACOS[0]}

ACOS(0) = pi/2

sin[4*(pi/2)] = sin(2*pi) = 0

A = -4(pi/2) + 0 = -2*pi



x = 4 gives A = -4{ACOS[1]} + sin{4*ACOS[1]}

ACOS(1) = 0

sin(4*0) = sin(0) = 0

So this is just 0



A = [0] - [-2*pi] = 2*pi



The integral is 2*pi