The value of $\sqrt{2} \int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x$, is
Answer & explanation
Correct answer: option 1
We have,
$\sqrt{2} \int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x$
$=\sqrt{2} \int \frac{\sin \left\{\left(x-\frac{\pi}{4}\right)+\frac{\pi}{4}\right\}}{\sin \left(x-\frac{\pi}{4}\right)} d x$
$=\sqrt{2} \int \frac{\sin \left(x-\frac{\pi}{4}\right) \cos \frac{\pi}{4}+\cos \left(x-\frac{\pi}{4}\right) \sin \frac{\pi}{4}}{\sin \left(x-\frac{\pi}{4}\right)} d x$
$=\int 1 . d x+\int \cot \left(x-\frac{\pi}{4}\right) d x=x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+C$