\[\int_{{}}^{{}}{x{{\sin }^{2}}x\ dx=}\] [BIT Ranchi 1977; IIT 1972]

\[\frac{{{x}^{2}}}{4}+\frac{x}{4}\sin 2x+\frac{1}{8}\cos 2x+c\]

\[\frac{{{x}^{2}}}{4}-\frac{x}{4}\sin 2x+\frac{1}{8}\cos 2x+c\]

\[\frac{{{x}^{2}}}{4}+\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c\]

\[\frac{{{x}^{2}}}{4}-\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c\]

\[\int_{{}}^{{}}{x{{\sin }^{2}}x\ dx=}\] [BIT Ranchi 1977; IIT 19

1.	\[\int_{{}}^{{}}{x{{\sin }^{2}}x\ dx=}\] [BIT Ranchi 1977; IIT 1972]
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.	\[\frac{{{x}^{2}}}{4}-\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c\]
Answer» E.

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\[\int_{{}}^{{}}{x{{\sin }^{2}}x\ dx=}\] [BIT Ranchi 1977; IIT 1972]

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