\[\int_{{}}^{{}}{{{x}^{2}}\sin 2x}\ dx=\] [IIT 1974]

1.	\[\int_{{}}^{{}}{{{x}^{2}}\sin 2x}\ dx=\] [IIT 1974]
A.	\[\frac{1}{2}{{x}^{2}}\cos 2x+\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x+c\]
B.	\[-\frac{1}{2}{{x}^{2}}\cos 2x+\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x+c\]
C.	\[\frac{1}{2}{{x}^{2}}\cos 2x-\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x+c\]
D.	None of these
Answer» C. \[\frac{1}{2}{{x}^{2}}\cos 2x-\frac{1}{2}x\sin 2x+\frac{1}{4}\cos 2x+c\]

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\[\int_{{}}^{{}}{{{x}^{2}}\sin 2x}\ dx=\] [IIT 1974]

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