# Prove that the area of half of the unit circle is pi/2. Hint: if you cannot evaluate the first integral you get, try a clever substitution based on your knowledge of the unit circle...

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### 1 Answer

You should use the general equation of the circle whose center is at origin (0,0) such that:

`x^2 + y^2 = r^2`

Since the radius of unit circle is r = 1 yields:

`x^2 + y^2 = 1 => y^2 = 1 - x^2 => y = +-sqrt(1-x^2)`

You should integrate the given function to evaluate the area such thatr:

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

You should use the following substitution such that:

`x = sint => dx = cos t dt`

`int sqrt(1 - x^2) dx = int sqrt(1 - sin^2 t) cost dt`

Using the fundamental formula of trigonometry yields:

`1 - sin^2 t = cos^2 t => sqrt(1 - sin^2 t) = sqrt(cos^2 t) = |cos t|`

`int sqrt(1 - sin^2 t) cos t dt = int cos^2 t dt`

Using the half angle identity yields:

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

`int cos^2 t dt = (1/2) (int dt + int cos 2t dt)`

`int cos^2 t dt = t/2 + (sin 2t)/4`

Substituting back `t = sin^(-1) x` yields:

`2int_(0)^(pi/2) sqrt(1 - x^2) dx = 2((sin^(-1) x)/2 + (sin 2sin^(-1) x)/4)_0^pi`

`2int_(0)^(pi/2) sqrt(1 - x^2) dx = 2((sin^(-1) pi/2)/2 + (sin 2sin^(-1) pi/2)/4)`

**Hence, evaluating the area of half of unit circle yields `2int_(0)^(pi/2) sqrt(1 - x^2) dx = 2((sin^(-1) pi/2)/2 + (sin 2sin^(-1) pi/2)/4).` **