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