A) \[\frac{{{\pi }^{2}}}{32}\]
B) \[\frac{{{\pi }^{2}}}{16}\]
C) \[\frac{\pi }{32}\]
D) none of these
Correct Answer: A
Solution :
Let \[I=\int_{0}^{\pi /2}{x{{\sin }^{2}}x{{\cos }^{2}}x}dx\] ...(i) \[I=\int_{0}^{\pi /2}{\left( \frac{\pi }{2}-x \right)}{{\sin }^{2}}x{{\cos }^{2}}x\,dx\] ...(ii) Adding Eqs. (i) and (ii) we get \[2I=\frac{\pi }{2}\int_{0}^{\pi /2}{{{\sin }^{2}}x}{{\cos }^{2}}x\,dx\] \[=\frac{\pi }{8}\int_{0}^{\pi /2}{{{\sin }^{2}}2x}\,dx\] \[\Rightarrow \]\[2I=\frac{\pi }{8}\left[ x-\frac{\sin 4x}{4} \right]_{0}^{\pi /2}=\frac{\pi }{8}\left[ \frac{\pi }{2} \right]\]\[\Rightarrow \]\[I=\frac{{{\pi }^{2}}}{32}\]
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