JEE Advanced 2021 Paper 2 · Q11 · Definite Integrals

Correct answer:2

Solution

$S_1 = \int_{\pi/8}^{3\pi/8} \sin^2 x\,dx$. Using $\sin^2(\pi/2 - x) = \cos^2 x$ and substitution $x \to \pi/2 - x$ on the second half: $S_1 = \int_{\pi/8}^{3\pi/8}\cos^2 x\,dx$ as well. Hence $2S_1 = \int_{\pi/8}^{3\pi/8} 1\,dx = \dfrac{3\pi}{8} - \dfrac{\pi}{8} = \dfrac{\pi}{4}$, so $S_1 = \dfrac{\pi}{8}$. Therefore $\dfrac{16 S_1}{\pi} = \dfrac{16}{\pi}\cdot\dfrac{\pi}{8} = 2$.

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