# Evaluate : `1int0 `√( x-x^2) dx

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You need to complete the square `x - x^2` , using the following formula, such that:

`(a - b)^2 = a^2 - 2ab - b^2`

Reasoning by analogy yields that you need to add and subtract the missing term `(1/2)^2` , such that:

`x - x^2 = -(x^2 - x + 1/4 - 1/4) => -(x - 1/2)^2 + 1/4`

Hence, you may substitute `1/4 - (x - 1/2)^2` for `x - x^2` , such that:

`int_0^1 sqrt(1/4 - (x - 1/2)^2)dx`

You need to factor out `1/4` , such that:

`int_0^1 sqrt((1/4)(1 - 4(x - 1/2)^2))dx`

You need to use the following trigonometric substitution, such that:

`2(x - 1/2) = sin u => 2dx = cos udu => dx = (cos u)/2 du`

You need to change first the limits of integration, such that:

`x = 0 => sin u = 0 => u = 0`

`x = 1 => sin u = 1 => u = pi/2`

Changing the variable, yields:

`(1/2)int_0^1 (sqrt(1 - 4(x - 1/2)^2))dx = (1/2) int_0^(pi/2) (sqrt(1 - sin^2 u))(cos u)/2 du`

Using the fundamental formula of trigonometry yields:

`1 - sin^2 u = cos^2 u`

`(1/4) int_0^(pi/2) (sqrt(cos^2 u))(cos u) du = (1/4) int_0^(pi/2) cos u*cos u du`

`(1/2) int_0^(pi/2) (sqrt(1 - sin^2 u))(cos u)/2 du = (1/4) int_0^(pi/2) cos^2 udu`

You need to use half angle formula such that:

`cos^2 u = (1 + cos 2u)/2`

`(1/4) int_0^(pi/2) cos^2 udu = (1/4) int_0^(pi/2) (1 + cos 2u)/2 du`

You need to use the property of linearity of integral, hence, you need to split the integral in two simpler integrals, such that:

`(1/4) int_0^(pi/2) (1 + cos 2u)/2 du = (1/8) int_0^(pi/2) du + (1/8) int_0^(pi/2) cos 2u du`

`(1/4) int_0^(pi/2) (1 + cos 2u)/2 du = (1/8)*u|_0^(pi/2) + (1/16)sin 2u|_0^(pi/2)`

Using the fundamental theorem of calculus, yields:

`(1/4) int_0^(pi/2) (1 + cos 2u)/2 du = (1/8)*(pi/2 - 0) + (1/16)(sin(2*pi/2) - sin 0)`

Since `sin pi = sin 0 = 0` yields:

`(1/4) int_0^(pi/2) (1 + cos 2u)/2 du = pi/16`

**Hence, evaluating the definite integral, using trigonometric substitution, yields **`int_0^1 sqrt(x - x^2)dx = pi/16.`

the one is on top and the 0 is on the bottom of the integration