, then
A)
B)
C)
D)
Correct Answer: D
Solution :
\[\int{\,(sin2x-cos\,2x)dx}\,\,=\,\,\frac{1}{\sqrt{2}}\,sin(2x-a)+b\] \[\Rightarrow \,\,-\frac{1}{2}(sin\,2x+cos2x)=\frac{1}{\sqrt{2}}\sin (2x-a)+b\] \[\Rightarrow \,\,-\,\left[ \frac{1}{\sqrt{2}}\,sin\,2x+\frac{1}{\sqrt{2}}cos2x) \right]=\,\,\sin (2x-a)+b\sqrt{2}\] \[\Rightarrow \,\,\,\sin \left( 2x+\frac{5\pi }{4} \right)=\sin (2x-a)+b\sqrt{2}\] \[\Rightarrow \,\,\,b\,\,is\,\,any\,\,cons\tan \operatorname{t}\,\,and\,\,a=\frac{-5\pi }{4}\]
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