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

∫(sin⁡2x−cos⁡2x)dx=12sin⁡(2x−a)+b⇒ 2∫12sin⁡2x−12cos⁡2xdx=12sin⁡(2x−a)+b⇒−2∫12cos⁡2x−12sin⁡2xdx=12sin⁡(2x−a)+b⇒ −2∫cos⁡2x+π4dx=12sin⁡(2x−a)+b⇒ −22sin⁡2x+π4+C=12sin⁡(2x−a)+b⇒ 12sin⁡2x+5π4+C=12sin⁡(2x−a)+b⇒ a=−5π4,b∈R

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

∫(sin⁡2x−cos⁡2x)dx=12sin⁡(2x−a)+b⇒ 2∫12sin⁡2x−12cos⁡2xdx=12sin⁡(2x−a)+b⇒−2∫12cos⁡2x−12sin⁡2xdx=12sin⁡(2x−a)+b⇒ −2∫cos⁡2x+π4dx=12sin⁡(2x−a)+b⇒ −22sin⁡2x+π4+C=12sin⁡(2x−a)+b⇒ 12sin⁡2x+5π4+C=12sin⁡(2x−a)+b⇒ a=−5π4,b∈R

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

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None of these

$a = \frac{π}{4}, b \in R$

$a = - \frac{5 π}{4}, b \in R$

$a = \frac{5 π}{4}, b \in R$

answer is B.

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Detailed Solution

$\begin{aligned} \int (\sin 2 x - \cos 2 x) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow \sqrt{2} \int (\frac{1}{\sqrt{2}} \sin 2 x - \frac{1}{\sqrt{2}} \cos 2 x) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow - \sqrt{2} \int (\frac{1}{\sqrt{2}} \cos 2 x - \frac{1}{\sqrt{2}} \sin 2 x) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow - \sqrt{2} \int \cos (2 x + \frac{π}{4}) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow - \frac{\sqrt{2}}{2} \sin (2 x + \frac{π}{4}) + C = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow \frac{1}{\sqrt{2}} \sin (2 x + \frac{5 π}{4}) + C = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow a = - \frac{5 π}{4}, b \in R \end{aligned}$