int_{0}^{^{n}C_{r}} { sin^{2}{x} } dx is equa | vedclass

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(C) The integral is $I = \int_{0}^{^{n}C_{r}} \{ \sin^{2}\{x\} \} dx$. Since $\{x\} \in [0, 1)$,$\sin^{2}\{x\} \in [0, \sin^{2} 1)$. Since $\sin^{2} 1

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$\frac{1}{2} ^{n}C_{r}(1 - \sin 1 \cos 1)$

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(C) The integral is $I = \int_{0}^{^{n}C_{r}} \{ \sin^{2}\{x\} \} dx$. Since $\{x\} \in [0, 1)$,$\sin^{2}\{x\} \in [0, \sin^{2} 1)$. Since $\sin^{2} 1 Thus,$I = \int_{0}^{^{n}C_{r}} \sin^{2}\{x\} dx$. Since $\sin^{2}\{x\}$ is a periodic function with period $1$,we have $I = ^{n}C_{r} \int_{0}^{1} \sin^{2} x dx$. Using the identity $\sin^{2} x = \frac{1 - \cos 2x}{2}$,we get: $I = ^{n}C_{r} \int_{0}^{1} \frac{1 - \cos 2x}{2} dx = \frac{^{n}C_{r}}{2} [x - \frac{\sin 2x}{2}]_{0}^{1}$. $I = \frac{^{n}C_{r}}{2} (1 - \frac{\sin 2}{2}) = \frac{^{n}C_{r}}{2} (1 - \sin 1 \cos 1)$.

$\int_{0}^{^{n}C_{r}} \{ \sin^{2}\{x\} \} dx$ is equal to (where $\{.\}$ denotes the fractional part function and $n, r \in N$)

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