## Calculus 8th Edition

$$\displaystyle{\int_{0}^{1}\sqrt{x-x^2}dx=\frac{\pi }{8}}$$
$\displaystyle{I=\int_{0}^{1}\sqrt{x-x^2}dx}\\ \displaystyle{I=\int_{0}^{1}\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}dx}\\$ $\displaystyle \left[\begin{array}{ll} x-\frac{1}{2}=\frac{1}{2}\sin\theta & \left({x-\frac{1}{2}}\right)^2=\frac{1}{4}\sin^2\theta \\ & \\ \frac{dx}{d\theta}=\frac{1}{2}\cos\theta & dx=\frac{1}{2}\cos\theta\ d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{\frac{1}{4}-\frac{1}{4}\sin^2\theta}\times\frac{1}{2}\cos\theta\ d\theta}\\ \displaystyle{I=\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{\frac{1}{4}\left(1-\sin^2\theta\right)}\times\frac{1}{2}\cos\theta\ d\theta}\\ \displaystyle{I=\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{2}\cos\theta\times\frac{1}{2}\cos\theta\ d\theta}\\ \displaystyle{I=\frac{1}{4}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos^2\theta\ d\theta}\\ \displaystyle{I=\frac{1}{8}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos2\theta+1\ d\theta}\\ \displaystyle{I=\frac{1}{8}\left[\frac{1}{2}\sin2\theta+\theta\right]_{-\frac{\pi }{2}}^{\frac{\pi }{2}}}\\ \displaystyle{I=\frac{1}{8}\left(\left(\frac{1}{2}\sin2\frac{\pi }{2}+\frac{\pi }{2}\right)-\left(\frac{1}{2}\sin\left(-2\frac{\pi }{2}\right)-\frac{\pi }{2}\right)\right)}\\ \displaystyle{I=\frac{\pi }{16}+\frac{\pi }{16}}\\ \displaystyle{I=\frac{\pi }{8}}\\$