If ∫(sin⁡2x−cos⁡2x)dx=12sin⁡(2x−a)+b, Then

∫(sin⁡2x−cos⁡2x)dx=12sin⁡(2x−a)+b ⇒ −12(sin⁡2x+cos⁡2x)=12sin⁡(2x−a)+b⇒ −12sin⁡2x+12cos⁡2x∣=sin⁡(2x−a)+b2 ⇒ sin⁡2x+5π4=sin⁡(2x−a)+b2⇒ b is any constant and a=−5π4.

If ∫(sin⁡2x−cos⁡2x)dx=12sin⁡(2x−a)+b, Then

∫(sin⁡2x−cos⁡2x)dx=12sin⁡(2x−a)+b ⇒ −12(sin⁡2x+cos⁡2x)=12sin⁡(2x−a)+b⇒ −12sin⁡2x+12cos⁡2x∣=sin⁡(2x−a)+b2 ⇒ sin⁡2x+5π4=sin⁡(2x−a)+b2⇒ b is any constant and a=−5π4.

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

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$a = - \frac{π}{4}, b = 0$

$a = \frac{5 π}{4}, b = any constant$

$a = - \frac{5 π}{4}, b = any constant$

$a = \frac{π}{4}, b = 0$

answer is D.

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Detailed Solution

$\int (\sin 2 x - \cos 2 x) d x = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \begin{array}{l} \Rightarrow - \frac{1}{2} (\sin 2 x + \cos 2 x) = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow - [\frac{1}{\sqrt{2}} \sin 2 x + \frac{1}{\sqrt{2}} \cos 2 x ∣] = \sin (2 x - a) + b \sqrt{2} \end{array} \begin{array}{l} \Rightarrow \sin (2 x + \frac{5 π}{4}) = \sin (2 x - a) + b \sqrt{2} \\ \Rightarrow b is any constant and a = \frac{- 5 π}{4} . \end{array}$