int_{-2}^{pi} frac{sin^2 x}{[frac{x}{pi}] + f | vedclass

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(B) Let $I = \int_{-2}^{\pi} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \,dx$. We split the integral based on the value of $[\frac{x}{\pi}]$. For

Vedclass · Accepted Answer

$\pi - 2 + \sin 2 \cos 2$

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(B) Let $I = \int_{-2}^{\pi} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \,dx$. We split the integral based on the value of $[\frac{x}{\pi}]$. For $x \in [-2, 0)$,$\frac{x}{\pi} \in [-\frac{2}{\pi}, 0)$,so $[\frac{x}{\pi}] = -1$. For $x \in [0, \pi)$,$\frac{x}{\pi} \in [0, 1)$,so $[\frac{x}{\pi}] = 0$. Thus,$I = \int_{-2}^{0} \frac{\sin^2 x}{-1 + 1/2} \,dx + \int_{0}^{\pi} \frac{\sin^2 x}{0 + 1/2} \,dx$. $I = -2 \int_{-2}^{0} \sin^2 x \,dx + 2 \int_{0}^{\pi} \sin^2 x \,dx$. Using $\sin^2 x = \frac{1 - \cos 2x}{2}$,we get: $I = -2 \int_{-2}^{0} \frac{1 - \cos 2x}{2} \,dx + 2 \int_{0}^{\pi} \frac{1 - \cos 2x}{2} \,dx$. $I = -[x - \frac{\sin 2x}{2}]_{-2}^{0} + [x - \frac{\sin 2x}{2}]_{0}^{\pi}$. $I = -[0 - (-2 - \frac{\sin(-4)}{2})] + [(\pi - \frac{\sin 2\pi}{2}) - 0]$. $I = -[2 - \frac{\sin 4}{2}] + \pi = \pi - 2 + \frac{\sin 4}{2}$. Since $\sin 4 = 2 \sin 2 \cos 2$,we have $I = \pi - 2 + \sin 2 \cos 2$.

$\int_{-2}^{\pi} \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}} \,dx$ is equal to (where $[\cdot]$ denotes the greatest integer function).

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