$\displaystyle\int_{-\pi/4}^{\pi/4} \sin^2 x\,dx$ equals

1. $\sin^2 x$ is even, so $\int_{-\pi/4}^{\pi/4}=2\int_0^{\pi/4}$. 2. Use $\sin^2 x=\dfrac{1-\cos 2x}{2}$, giving $\int_0^{\pi/4}(1-\cos 2x)\,dx$. 3. $=\left[x-\tfrac12\sin 2x\right]_0^{\pi/4}=\dfrac{\pi}{4}-\tfrac12\sin\tfrac{\pi}{2}$. 4. $=\dfrac{\pi}{4}-\dfrac{1}{2}$. Option D wrongly treats $\sin^2 x$ as odd. _Source: syllabus Mathematics Ch 7 "Integrals", p.344_