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

$$\int^{\pi/2}_0\sin^2xdx=\frac{\pi}{4}$$
$$A=\int^{\pi/2}_0\sin^2xdx$$ This is Case 3, so we will need to rewrite $$\sin^2x=\frac{1-\cos2x}{2}$$ Therefore, $$A=\int^{\pi/2}_0\frac{1-\cos2x}{2}dx$$ $$A=\int^{\pi/2}_0\frac{1}{2}dx-\frac{1}{2}\int^{\pi/2}_0\cos2xdx$$ $$A=\frac{x}{2}\Big]^{\pi/2}_0-\frac{1}{2}\Big(\frac{1}{2}\sin2x\Big)\Big]^{\pi/2}_0$$ $$A=\frac{\pi}{4}-\frac{1}{4}(\sin\pi-\sin0)$$ $$A=\frac{\pi}{4}-\frac{1}{4}(0-0)$$ $$A=\frac{\pi}{4}$$