A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{8}\]
C) \[\frac{\pi }{16}\]
D) \[\frac{\pi }{32}\]
Correct Answer: C
Solution :
\[I=\int\limits_{0}^{\pi /2}{\sin x\sin 2x\sin 3x\sin 4xdx}\]?(i) Replacing\[x\]by \[\frac{\pi }{2}-x,\]we get \[I=\int\limits_{0}^{\pi /2}{\cos x\sin 2x\cos 3x\sin 4xdx}\] ?(2) Adding (1) and (2), we get \[2I=\int\limits_{0}^{\pi /2}{\cos 2x\sin 2x\sin 4xdx=\frac{\pi }{8}}\] \[\therefore \] \[I=\frac{\pi }{16}\]
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