2018

$\begin{aligned} I & = \int_0^{2018\pi} \sin^2 x\ dx & \small \color{#3D99F6} \text{Note that }\sin^2 x \text{ repeats every }\pi. \\ & = 2018 \int_0^\pi {\color{#3D99F6}\sin^2 x} \ dx & \small \color{#3D99F6} \text{and } \sin^2 \theta = \frac 12(1-\cos 2\theta) \\ & = \frac {2018}{\color{#3D99F6}2} \int_0^\pi {\color{#3D99F6}(1-\cos 2x)} \ dx \\ & = 1009\pi - 1009 \int_0^\pi \cos 2x \ dx \\ & = 1009\pi - \frac {1009}2 \sin 2x \ \bigg|_0^\pi \\ & = 1009\pi - 0 \\ & = 1009\pi \\ & \approx \boxed{3169.867} \end{aligned}$

• ${\sin}^{2} (x)=\frac{1-\cos(2x)}{2}$
• $\displaystyle{\int_{0}^{2018\pi}\frac{1-\cos(2x)}{2}}\,dx=[\frac{x}{2}-\frac{\sin(2x)}{4}]\bigg|_0^{2018\pi}$
• $(\frac{2018\pi}{2}-\frac{\sin(2(2018\pi)}{4})-(\frac{0}{2}-\frac{\sin(2(0))}{4})$
• $(1009\pi-0)-(0-0)$
• $1009\pi \approx \boxed{3169.866987}$