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) \[\frac{{{x}^{4}}}{4}+\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+c\]
Correct Answer: B
Solution :
\[\int_{{}}^{{}}{x{{\cos }^{2}}x\,dx}=\frac{1}{2}\int_{{}}^{{}}{x(1+\cos 2x)\,dx}\] \[=\frac{{{x}^{2}}}{4}+\frac{1}{2}\left[ \frac{x\sin 2x}{2}-\int_{{}}^{{}}{\frac{\sin 2x}{2}\,dx} \right]+c\] \[=\frac{{{x}^{2}}}{4}+\frac{x\sin 2x}{4}+\frac{\cos 2x}{8}+c.\]
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