Integral Calculus Question 11
Question: $ \int{\frac{dx}{\sin (x-a)\sin (x-b)}} $ is
[Kerala (Engg.) 2005]
Options:
A) $ \frac{1}{\sin (a-b)}\log | \frac{\sin (x-a)}{\sin (x-b)} |+c $
B) $ \frac{-1}{\sin (a-b)}\log | \frac{\sin (x-a)}{\sin (x-b)} |+c $
C) $ \log \sin (x-a)\sin (x-b)+c $
D) $ \log | \frac{\sin (x-a)}{\sin (x-b)} | $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ \int{\frac{dx}{\sin (x-a)\sin (x-b)}} $ $ =\frac{1}{\sin (a-b)}\int{\frac{\sin { (x-b)-(x-a) }}{\sin (x-a)\sin (x-b)}}\ dx $ $ =\frac{1}{\sin (a-b)}\int{\frac{\sin (x-b)\cos (x-a)-\cos (x-b)\sin (x-a)}{\sin (x-a)\sin (x-b)}dx} $ $ =\frac{1}{\sin (a-b)}[ \int{\cot (x-a)dx-\int{\cot (x-b)dx}} ] $ $ =\frac{1}{\sin (a-b)}\ [ \log \sin (x-a)-\log \sin (x-b) ]+c $ $ =\frac{1}{\sin (a-b)}\ \log | \frac{\sin (x-a)}{\sin (x-b)} |+c $ .
BETA
