\[\int_{\,0}^{\,\pi /2}{\{x-[\sin x]\}\,dx}\] is equal to [AMU 1999]

1.	\[\int_{\,0}^{\,\pi /2}{\{x-[\sin x]\}\,dx}\] is equal to [AMU 1999]
A.	\[\frac{{{\pi }^{2}}}{8}\]
B.	\[\frac{{{\pi }^{2}}}{8}-1\]
C.	\[\frac{{{\pi }^{2}}}{8}-2\]
D.	None of these
Answer» B. \[\frac{{{\pi }^{2}}}{8}-1\]

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\[\int_{\,0}^{\,\pi /2}{\{x-[\sin x]\}\,dx}\] is equal to [AMU 1999]

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