$\displaystyle\int_{-\pi/4}^{\pi/4} \sin^2 x\,dx$ equals
A$\dfrac{\pi}{4}-\dfrac{1}{2}$
B$\dfrac{\pi}{4}+\dfrac{1}{2}$
C$\dfrac{\pi}{2}-1$
D$0$
Answer & Solution
Correct answer: A. $\dfrac{\pi}{4}-\dfrac{1}{2}$
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: NCERT Class 12 Mathematics Ch 7 "Integrals", p.344_
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