Integral Calculus Question 88
Question: If $ \int \frac{\sin x}{\sin (x-\alpha )}dx=Ax+B\log \sin (x-\alpha )+c, $ then the value of (A, B) is
Options:
A) $ (sin\alpha ,,cos\alpha ) $
B) $ (cos\alpha ,,\sin \alpha ) $
C) $ (-\sin \alpha ,,\cos \alpha ) $
D) $ (-\cos \alpha ,,\sin \alpha ) $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ I=\int{\frac{\sin x}{\sin (x-\alpha )}dx} $ Let $ x-\alpha =t\Rightarrow dx=dt\therefore x=(t+\alpha ) $
$ \therefore I=\int{\frac{\sin (t+\alpha )}{\sin t}dt} $ $ =\int{\frac{\sin t\cos \alpha +\cos t\sin \alpha }{\sin t}dt} $ $ =\int{\cos \alpha ,dt+\int{\sin \alpha \frac{\cos t}{\sin t}dt}} $ $ =\cos \alpha \int{1dt+\sin \alpha \int{\frac{\cos t}{\sin t}dt}} $ $ =\cos \alpha (t)+sin\alpha ,log,\sin t+c $ $ =\cos \alpha (x-\alpha )+\sin \alpha \log ,\sin (x-\alpha )+c $ $ =x\cos \alpha +\sin \alpha \log \sin (x-\alpha )-\alpha cos\alpha +c $ $ =x\cos \alpha +sing,\alpha log,sin(x-\alpha )+c $ But $ \int{\frac{\sin x}{\sin (x-\alpha )}dx=Ax+B\log \sin (x-\alpha )+c} $
$ \therefore x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+c $ $ =Ax+B\log \sin (x-\alpha )+c $ On comparing, we get $ A=\cos \alpha $ $ B=\sin \alpha $
BETA
