A) \[\frac{\pi }{3}\]
B) \[-\frac{\pi }{3}\]
C) \[\frac{\pi }{4}\]
D) \[-\frac{\pi }{4}\]
Correct Answer: C
Solution :
Given that \[\int_{{}}^{{}}{(\cos x-\sin x)\,dx}=\sqrt{2}\sin (x+\alpha )+c\] \[\Rightarrow \sin x+\cos x+c=\sqrt{2}\sin (x+\alpha )+c\] \[\Rightarrow \sqrt{2}\left( \frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}} \right)+c=\sqrt{2}\sin (x+\alpha )+c\] \[\Rightarrow \sqrt{2}\sin \left( x+\frac{\pi }{4} \right)+c=\sqrt{2}\sin (x+\alpha )+c\] \[\Rightarrow \sin \left( x+\frac{\pi }{4} \right)=\sin (x+\alpha )\Rightarrow \alpha =\frac{\pi }{4}.\]
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