置換積分
∫ 0 1 1 − x 2 d x = ∫ 0 π 2 1 − sin 2  ( u ) cos  ( u ) d u = ∫ 0 π 2 | cos  ( u ) | cos  ( u ) d u = ∫ 0 π 2 cos 2  ( u ) d u = ( u 2 + sin  ( 2 u ) 4 ) | 0 π 2 = π 4 + 0 = π 4 {\displaystyle {\begin{aligned}\int _{0}^{1}{\sqrt {1-x^{2}}}\,dx&=\int _{0}^{\frac {\pi }{2}}{\sqrt {1-\sin ^{2}(u)}}\cos(u)\,du\\&=\int _{0}^{\frac {\pi }{2}}|\cos(u)|\cos(u)\,du=\int _{0}^{\frac {\pi }{2}}\cos ^{2}(u)\,du\\&={\Bigg (}{{\frac {u}{2}}+{\frac {\sin(2u)}{4}}{\Bigg )}{\Bigg \vert }\,}_{0}^{\frac {\pi }{2}}={\frac {\pi }{4}}+0={\frac {\pi }{4}}\end{aligned}}}
と計算できる。