## University Calculus: Early Transcendentals (3rd Edition)

$$\int^{\pi}_{0}\sin^25rdr=\frac{\pi}{2}$$
$$A=\int^{\pi}_{0}\sin^25rdr$$ Recall the identity: $$\sin^2a=\frac{1-\cos2a}{2}$$ Apply the identity here, we have $$A=\int^{\pi}_{0}\frac{1-\cos10r}{2}dr$$ $$A=\int^{\pi}_{0}\frac{1}{2}dr-\frac{1}{2}\int^{\pi}_{0}\cos10rdr$$ $$A=\frac{r}{2}\Big]^{\pi}_{0}-\Big(\frac{1}{2}\times\frac{1}{10}\sin10r\Big)\Big]^{\pi}_{0}$$ $$A=\frac{\pi}{2}-\Big(\frac{\sin10r}{20}\Big)\Big]^{\pi}_{0}$$ $$A=\frac{\pi}{2}-\frac{1}{20}(\sin10\pi-\sin0)$$ $$A=\frac{\pi}{2}-\frac{1}{20}(0-0)$$ $$A=\frac{\pi}{2}$$