A) \[x\cos a-\sin a.\log (x-a)+c\]
B) \[x\sin a+c\]
C) \[x\sin a+\sin a.\log \sin (x-a)+c\]
D) \[x\cos a+\sin a.\log \sin (x-a)+c\]
E) \[x\cos a+\cos a.\log \sin (x-a)+c\]
Correct Answer: D
Solution :
Let \[I=\int{\frac{\sin x}{\sin (x-a)}}dx\] Put \[x-a=t\] \[\Rightarrow \] \[dx=dt\] and \[x=t+a\] \[\therefore \] \[I=\int{\frac{\sin (t+a)}{\sin t}}dt\] \[=\int{\frac{\sin t\cos a+\cos t\sin a}{\sin t}}dt\] \[=\cos a\int{1}\,dt+\sin a\int{\frac{\cos t}{\sin t}}dt\] \[=\cos a.t+\sin a\log \sin t+c\] \[=(x-a)\cos a+\sin a\log \sin (x-a)+c\] \[=x\cos a+\sin a\log \sin (x-a)+c\]
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