Integral Calculus Question 199
Question: $ \int_{{}}^{{}}{\frac{\sin x}{\sin (x-\alpha )}dx=} $
[RPET 1999; Kerala (Engg.) 2002; AIEEE 2004]
Options:
A) $ x\cos \alpha -\sin \alpha \log \sin (x-\alpha )+c $
B) $ x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+c $
C) $ x\sin \alpha -\sin \alpha \log \sin (x-\alpha )+c $
D) None of these
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Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{\sin x}{\sin (x-\alpha )},dx=\int_{{}}^{{}}{\frac{\sin (x-\alpha +\alpha )}{\sin (x-\alpha )},dx}} $ $ =\int_{{}}^{{}}{\frac{{ (\sin (x-\alpha )\cos \alpha +\cos (x-\alpha )\sin \alpha }}{\sin (x-\alpha )},dx} $ $ =\int_{{}}^{{}}{\cos \alpha ,dx+\int_{{}}^{{}}{\sin \alpha ,.,\cot ,(x-\alpha ),dx}} $ $ =x\cos \alpha +\sin \alpha ,.,\log \sin (x-\alpha )+c $ .
BETA
