Question

$\int \frac{\cos (x-a)}{\cos (x-b)}dx=Ax+B\ln |\cos (x-b)|+C$,
then $B/A=$

• A. $\tan (a–b)$
• B. $\tan (a+b)$
• C. $\tan (b–a)$
• D. $–\tan (a+b)$

Answer 1

The correct option is C

$\tan (b–a)$

$\int \frac{\cos (x-b+b-a)}{\cos (x-b)}dx$

$=\int \left[\cos \right(b-a)-\sin (b-a\left)\tan \right(x-a\left)\right]dx$

$=x\cos (b-a)+\sin (b-a)\ln |\cos (x-a)|+C$

$\therefore \frac{B}{A}=\tan (b-a)$