[SCRA 1980, 91]
A) $ \pi $
B) 0
C) 1
D) $ {{\pi }^{2}} $
Correct Answer: A
$ I=\int_0^{\pi }{x\sin xdx=\int_0^{\pi }{(\pi -x)\sin xdx}} $
therefore $ 2I=\pi \int_0^{\pi }{\sin xdx=\pi [-\cos x]_0^{\pi }\Rightarrow I=\pi } $ .
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