$\int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} I &=\int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}ight)} d x \ &=\int \frac{\sin \left(x-\frac{\pi}{4}+\frac{\pi}{4}ight)}{\sin \left(x-\frac{\pi}{4}ight)} d x=\int \frac{\sin \left(x-\frac{\pi}{4}ight) \cos \frac{\pi}{4}+\cos \left(x-\frac{\pi}{4}ight) \sin \frac{\pi}{4}}{\sin \left(x-\frac{\pi}{4}ight)} \ &=\int\left(\frac{1}{\sqrt{2}}+\cot \left(x-\frac{\pi}{4}ight) \cdot \frac{1}{\sqrt{2}}ight) d x=\int \frac{1}{\sqrt{2}}\left[1+\cot \left(x-\frac{\pi}{4}ight)ight] d x \ &=\frac{1}{\sqrt{2}}\left[\int 1 d x+\int \cot \left(x-\frac{\pi}{4}ight)ight] d x=\frac{1}{\sqrt{2}}\left[x+\log \left(\sin \left(x-\frac{\pi}{4}ight)ight]+cight.\end{aligned}$

$\int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x=$

Solution