Explanation

(b)
$\int_{}^{}{(\sin{2x - \cos{2x)dx = \frac{1}{\sqrt{2}}\sin{(2x - a) + b}}}}$
$\Rightarrow \sqrt{2}\ \int_{}^{}{\left( \frac{1}{\sqrt{2}}\sin{2x - \frac{1}{\sqrt{2}}\cos{2x}} \right)dx = \frac{1}{\sqrt{2}}\sin{(2x - a) + b}}$
$\Rightarrow - \sqrt{2}\ \int_{}^{}\left( \frac{1}{2}\cos{2x - \frac{1}{\sqrt{2}}\sin{2x}} \right)dx = \frac{1}{\sqrt{2}}\sin{(2x - a) + b}$
$\Rightarrow \ - \sqrt{2}\ \int_{}^{}{\left( \ 2x + \frac{\pi}{4}\ \right)dx = \frac{1}{\sqrt{2}}\sin{(2x - a) + b}}$
$\Rightarrow \ - \frac{\sqrt{2}}{2}\sin{\left( \ 2x + \frac{\pi}{4}\ \right) + c = \frac{1}{\sqrt{2}}\sin{(2x - a) + b}}$
$\Rightarrow \ \frac{1}{\sqrt{2}}\sin{\left( \ 2x + \frac{5\pi}{4}\ \right) + c = \frac{1}{\sqrt{2}}\sin{(2x - a) + b}}$
$\therefore a = - \frac{5\pi}{4},\ b \in R$