If int (sin 2x - cos 2x) ,dx = frac{1}{sqrt{2 | vedclass

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(D) Given $\int (\sin 2x - \cos 2x) \,dx = \frac{1}{\sqrt{2}} \sin(2x - a) + b$. Integrating the left side: $\int \sin 2x \,dx - \int \cos 2x \,dx = -

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

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(D) Given $\int (\sin 2x - \cos 2x) \,dx = \frac{1}{\sqrt{2}} \sin(2x - a) + b$. Integrating the left side: $\int \sin 2x \,dx - \int \cos 2x \,dx = -\frac{1}{2} \cos 2x - \frac{1}{2} \sin 2x + C$. So,$-\frac{1}{2} (\sin 2x + \cos 2x) + C = \frac{1}{\sqrt{2}} \sin(2x - a) + b$. Multiply both sides by $-\sqrt{2}$: $\frac{1}{\sqrt{2}} \sin 2x + \frac{1}{\sqrt{2}} \cos 2x - \sqrt{2}C = -\sin(2x - a) - \sqrt{2}b$. Using $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we have $\sin(2x + \frac{\pi}{4}) = \frac{1}{\sqrt{2}} \sin 2x + \frac{1}{\sqrt{2}} \cos 2x$. Thus,$\sin(2x + \frac{\pi}{4}) = -\sin(2x - a) - \sqrt{2}b + \sqrt{2}C$. Since $-\sin 	heta = \sin(	heta - \pi)$,we have $\sin(2x + \frac{\pi}{4}) = \sin(2x - a - \pi) - \sqrt{2}b + \sqrt{2}C$. Comparing the arguments: $2x + \frac{\pi}{4} = 2x - a - \pi \Rightarrow a = -\pi - \frac{\pi}{4} = -\frac{5\pi}{4}$. Also,$b$ is an arbitrary constant.

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

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