If $\int(\sin 2 x-\cos 2 x) d x=\frac{1}{\sqrt{2}} \sin (2 x-a)+b$, then

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

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Application +4 | -1

\int(\sin 2 x-\cos 2 x) d x=\frac{1}{\sqrt{2}} \sin (2 x-a)+b

, then

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Correct option: D

a=-\frac{5 \pi}{4}, b=

any constant

Option B ([1,3/2]) is correct because it represents the interval where the expression inside the square root is non-negative.

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