A) \[a=\frac{\pi }{4},\ b=0\]
B) \[a=-\frac{\pi }{4},\ b=0\]
C) \[a=\frac{5\pi }{4},\ b=\]any constant
D) \[a=-\frac{5\pi }{4},\ b=\]any constant
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{(\sin 2x-\cos 2x)\,dx=\frac{1}{\sqrt{2}}\sin (2x-a)+b}\] \[\Rightarrow -\frac{1}{2}(\sin 2x+\cos 2x)=\frac{1}{\sqrt{2}}\sin (2x-a)+b\] \[\Rightarrow -\left[ \frac{1}{\sqrt{2}}\sin 2x+\frac{1}{\sqrt{2}}\cos 2x \right]=\sin (2x-a)+b\sqrt{2}\] \[\Rightarrow \sin \left( 2x+\frac{5\pi }{4} \right)=\sin (2x-a)+b\sqrt{2}\] \[\Rightarrow b\] is any constant and \[a=\frac{-5\pi }{4}\].
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