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