JEE Advanced 2021 Paper 2 · Q12 · Definite Integrals

Correct answer:1.5

Solution

$S_2 = \int_{\pi/8}^{3\pi/8} \sin^2 x \cdot|4x-\pi|\,dx$. Substitute $x \to \pi/2 - x$ on the second half: this gives $S_2 = \int_{\pi/8}^{3\pi/8}\cos^2 x\cdot|4x-\pi|\,dx$, so $2S_2 = \int_{\pi/8}^{3\pi/8}|4x-\pi|\,dx$. Splitting at $x=\pi/4$: $2S_2 = 2\int_{\pi/8}^{\pi/4}(\pi-4x)\,dx = 2\left[\pi x - 2x^2\right]_{\pi/8}^{\pi/4} = 2\left[\left(\dfrac{\pi^2}{4}-\dfrac{\pi^2}{8}\right) - \left(\dfrac{\pi^2}{8}-\dfrac{\pi^2}{32}\right)\right] = \dfrac{\pi^2}{16}$. So $S_2 = \dfrac{\pi^2}{32}$, and $\dfrac{48 S_2}{\pi^2} = \dfrac{48}{32} = 1.5$.

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