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

The upper curve of the unit circle is the set

${\left\{(x,y)\mid x^{2}+y^{2}=1,\,-1\leq x\leq 1,\,y\geq 0\right\}}.$

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

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

Applying substitution with

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

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

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

In particular, we get that

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

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

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