\[\int_{{}}^{{}}{x\cos {{x}^{2}}\ dx}\] is equal to [MP PET 1999; Pb. CET 2000]

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

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

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

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

\[\int_{{}}^{{}}{x\cos {{x}^{2}}\ dx}\] is equal to [MP PET 1999; Pb.

1.	\[\int_{{}}^{{}}{x\cos {{x}^{2}}\ dx}\] is equal to [MP PET 1999; Pb. CET 2000]
A.	\[-\frac{1}{2}{{\sin }^{2}}x+c\]
B.	\[\frac{1}{2}{{\sin }^{2}}x+c\]
C.	\[-\frac{1}{2}\sin {{x}^{2}}+c\]
D.	\[\frac{1}{2}\sin {{x}^{2}}+c\]
Answer» E.

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\[\int_{{}}^{{}}{x\cos {{x}^{2}}\ dx}\] is equal to [MP PET 1999; Pb. CET 2000]

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