A) \[\frac{1}{\sin (a-b)}\log \left| \frac{\sin (x-a)}{\sin (x-b)} \right|+c\]
B) \[\frac{-1}{\sin (a-b)}\log \left| \frac{\sin (x-a)}{\sin (x-b)} \right|+c\]
C) \[\log \sin (x-a)\sin (x-b)+c\]
D) \[\log \left| \frac{\sin (x-a)}{\sin (x-b)} \right|\]
E) \[\frac{1}{\sin (x-a)}\log \sin (x-a)\sin (x-b)+c\]
Correct Answer: A
Solution :
Let \[I=\int{\frac{dx}{\sin (x-a)\sin (x-b)}}\] \[=\frac{1}{\sin (b-a)}\int{\frac{\sin (b-a)}{\sin (x-a)\sin (x-b)}}dx\] \[=\frac{1}{\sin (b-a)}\int{\frac{\sin \{(x-a)-(x-b)\}}{\sin (x-a)\sin (x-b)}}dx\] \[=\frac{1}{\sin (b-a)}\] \[\int{\frac{\begin{align} & \{\sin (x-a)\cos (x-b)-\cos (x-a) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sin (x-b)\} \\ \end{align}}{\sin (x-a)\sin (x-b)}}dx\] \[=\frac{1}{\sin (b-a)}\] \[[\log \sin (x-b)-\log \sin (x-a)]+c\] \[=\frac{-1}{\sin (a-b)}\left[ \log \frac{\sin (x-b)}{\sin (x-a)} \right]+c\]
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