\[\int_{{}}^{{}}{x{{\cos }^{2}}}xdx=\] [IIT 1972]

1.	\[\int_{{}}^{{}}{x{{\cos }^{2}}}xdx=\] [IIT 1972]
A.	\[\frac{{{x}^{4}}}{4}-\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+c\]
B.	\[\frac{{{x}^{2}}}{4}+\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+c\]
C.	\[\frac{{{x}^{4}}}{4}-\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+c\]
D.	\[\frac{{{x}^{4}}}{4}+\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+c\]
Answer» C. \[\frac{{{x}^{4}}}{4}-\frac{1}{4}x\sin 2x+\frac{1}{8}\cos 2x+c\]

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\[\int_{{}}^{{}}{x{{\cos }^{2}}}xdx=\] [IIT 1972]

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