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Question: Answered & Verified by Expert
$\int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x=$
MathematicsIndefinite IntegrationMHT CETMHT CET 2020 (14 Oct Shift 2)
Options:
  • A $\frac{1}{\sqrt{2}}\left[x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|\right]+c$
  • B $x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$
  • C $x-\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$
  • D $\frac{1}{\sqrt{2}}\left[x-\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|\right]+c$
Solution:
1784 Upvotes Verified Answer
The correct answer is: $\frac{1}{\sqrt{2}}\left[x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|\right]+c$
$\begin{aligned} I &=\int \frac{\sin x}{\sin \left(x-\frac{\pi}{4}\right)} d x \\ &=\int \frac{\sin \left(x-\frac{\pi}{4}+\frac{\pi}{4}\right)}{\sin \left(x-\frac{\pi}{4}\right)} d x=\int \frac{\sin \left(x-\frac{\pi}{4}\right) \cos \frac{\pi}{4}+\cos \left(x-\frac{\pi}{4}\right) \sin \frac{\pi}{4}}{\sin \left(x-\frac{\pi}{4}\right)} \\ &=\int\left(\frac{1}{\sqrt{2}}+\cot \left(x-\frac{\pi}{4}\right) \cdot \frac{1}{\sqrt{2}}\right) d x=\int \frac{1}{\sqrt{2}}\left[1+\cot \left(x-\frac{\pi}{4}\right)\right] d x \\ &=\frac{1}{\sqrt{2}}\left[\int 1 d x+\int \cot \left(x-\frac{\pi}{4}\right)\right] d x=\frac{1}{\sqrt{2}}\left[x+\log \left(\sin \left(x-\frac{\pi}{4}\right)\right]+c\right.\end{aligned}$

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