A) \[0\]
B) \[\pi -\frac{{{\pi }^{3}}}{3}\]
C) \[2\pi -{{\pi }^{3}}\]
D) \[\frac{7}{2}-2{{\pi }^{3}}\]
Correct Answer: A
Solution :
Key Idea: If\[f(-x)=-f(x)\], then \[\int_{-a}^{a}{f(x)}\,dx=0\] Let \[I=\int_{-\pi }^{\pi }{(1-{{x}^{2}})}\sin x{{\cos }^{2}}x\,\,dx\] Let \[f(x)={{(1-x)}^{2}})(\sin (-x)){{\cos }^{2}}(-x)\] \[\Rightarrow \] \[=-(1-{{x}^{2}})\sin x{{\cos }^{2}}x\] \[\Rightarrow \] \[f(-x)=-f(x)\] \[\Rightarrow \]\[f(x)\]is an odd function \[\therefore \] \[\int_{-\pi }^{\pi }{(1-{{x}^{2}})}\sin x{{\cos }^{2}}x\,\,dx=0\]
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