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
Correct Answer: B
Solution :
Let \[I=\int_{{}}^{{}}{{{x}^{2}}\sin 2x\,dx}=\frac{-{{x}^{2}}\cos 2x}{2}+\int_{{}}^{{}}{\frac{2x\cos 2x}{2}\,dx}+c\] \[=-\frac{{{x}^{2}}\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}+c.\]
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