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