Text Using the half-angle formula sin2x=1−cos2x2\sin^{2}x=\dfrac{1-\cos 2x}{2}sin2x=21−cos2x, compute ∫0πsin2x dx\displaystyle\int_{0}^{\pi}\sin^{2}x\,dx∫0πsin2xdx. Solution ∫0πsin2x dx=∫0π1−cos2x2 dx=[x2−sin2x4]0π=π2≈1.5708.\int_{0}^{\pi}\sin^{2}x\,dx=\int_{0}^{\pi}\frac{1-\cos 2x}{2}\,dx=\left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_{0}^{\pi}=\frac{\pi}{2}\approx1.5708.∫0πsin2xdx=∫0π21−cos2xdx=[2x−4sin2x]0π=2π≈1.5708. π2≈1.5708\boxed{\dfrac{\pi}{2}\approx1.5708}2π≈1.5708