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