Explanation

(d)
Let $I = \int_{}^{}\frac{dx}{\sin{x - \cos{x + \sqrt{2}}}}$
$= \int_{}^{}\frac{dx}{\sin{x\frac{\sqrt{2}}{\sqrt{2}} - \cos{x\frac{\sqrt{2}}{\sqrt{2}} + \sqrt{2}}}}$
$= \int_{}^{}\frac{dx}{\sqrt{2}\left( \sin{x\sin\frac{\pi}{4} - \cos{x\cos\frac{\pi}{4} + 1}} \right)}$
$= \frac{1}{\sqrt{2}}\int_{}^{}\frac{dx}{1 - \cos\left( x + \frac{\pi}{4} \right)}$
$= \frac{1}{\sqrt{2}}\int_{}^{}\frac{dx}{1 - \cos{2\left( \frac{x}{2} + \frac{\pi}{8} \right)}}$
$= \frac{1}{\sqrt{2}}\ \int_{}^{}\frac{dx}{2\sin^{2}\left( \frac{x}{2} + \frac{\pi}{8} \right)}$
$= \frac{1}{2\sqrt{2}}\ \int_{}^{}{cosec^{2}\left( \frac{x}{2} + \frac{\pi}{8} \right)}dx$
$= \frac{1}{2\sqrt{2}}\ .\ \frac{- \cot\left( \frac{x}{2} + \frac{\pi}{8} \right)}{\frac{1}{2}} + c$
$= - \frac{1}{\sqrt{2}}\cot{\left( \frac{x}{2} + \frac{\pi}{8} \right) + c}$