Unit circle/Integral of square root of 1-x^2/Example

The upper curve of the unit circle is the set

$\displaystyle { \left\{ (x,y) \mid x^2+y^2 [[Category:Wikiversity soft redirects|Unit circle/Integral of square root of 1-x^2/Example]] __NOINDEX__ 1 , \, -1 \leq x \leq 1 , \, y \geq 0 \right\} } .$

For a given ${\displaystyle {}x}$, ${\displaystyle {}-1\leq x\leq 1}$, there exists exactly one ${\displaystyle {}y}$ fulfilling this condition, namely ${\displaystyle {}y={\sqrt {1-x^{2}}}}$. Hence, the area of the upper half of the unit circle is the area beneath the graph of the function ${\displaystyle {}x\mapsto {\sqrt {1-x^{2}}}}$, above the interval ${\displaystyle {}[-1,1]}$, that is

${\displaystyle \int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx.}$

Applying substitution with

${\displaystyle x=\cos t{\text{ and }}t=\arccos x}$

(where ${\displaystyle {}\cos :[0,\pi ]\rightarrow [-1,1]}$ is bijective, due to), we obtain, using example, the identities

{\displaystyle {}{\begin{aligned}\int _{a}^{b}{\sqrt {1-x^{2}}}\,dx&=\int _{\arccos a}^{\arccos b}{\sqrt {1-\cos ^{2}t}}(-\sin t)\,dt\\&=-\int _{\arccos a}^{\arccos b}\sin ^{2}t\,dt\\&={\frac {1}{2}}(\sin t\cos t-t)|_{\arccos a}^{\arccos b}.\end{aligned}}}

In particular, we get that

${\displaystyle {}{\frac {1}{2}}{\left(x\cdot \sin {\left(\arccos x\right)}-\arccos x\right)}={\frac {1}{2}}{\left(x\cdot {\sqrt {1-x^{2}}}-\arccos x\right)}\,}$

is a primitive function for ${\displaystyle {}{\sqrt {1-x^{2}}}}$. Therefore,

{\displaystyle {}{\begin{aligned}\int _{-1}^{1}{\sqrt {1-x^{2}}}\,dx&={\frac {1}{2}}{\left(x\cdot {\sqrt {1-x^{2}}}-\arccos x\right)}|_{-1}^{1}\\&={\frac {1}{2}}(-\arccos 1+\arccos(-1))\\&=\pi /2.\end{aligned}}}