The upper curve of the unit circle is the set
-
For a given
,
,
there exists exactly one
fulfilling this condition, namely
.
Hence, the area of the upper half of the unit circle is the area beneath the graph of the function
, above the interval
, that is
-
Applying
substitution
with
-
(where
is bijective, due to
fact),
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bd0a492fb2d174bb6d19446c92662fa45a11eee)
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)}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be5c609880df663d6b39fa84b82f75b63b238d03)
is a
primitive function
for
. 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a2a36a7911b2bafb3695a78907a82d4da07f9c9)