Integral Calculus Question 1
Question: $ \int_{{}}^{{}}{\frac{dx}{\cos (x-a)\cos (x-b)}=} $
Options:
A) $ cosec(a-b)\log \frac{\sin (x-a)}{\sin (x-b)}+c $
B) $ cosec(a-b)\log \frac{\cos (x-a)}{\cos (x-b)}+c $
C) $ cosec(a-b)\log \frac{\sin (x-b)}{\sin (x-a)}+c $
D) $ cosec(a-b)\log \frac{\cos (x-b)}{\cos (x-a)}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{dx}{\cos (x-a)\cos (x-b)}} $
$ =\frac{1}{\sin (a-b)}\int_{{}}^{{}}{\frac{\sin { (x-b)-(x-a) }}{\cos (x-a),.,\cos (x-b)},dx} $
$ =\frac{1}{\sin (a-b)}\int_{{}}^{{}}{{ \frac{\sin (x-b)}{\cos (x-b)}-\frac{\sin (x-a)}{\cos (x-a)} }dx} $
$ =cosec,(a-b)\log \frac{\cos (x-a)}{\cos (x-b)}+c $ .
BETA
