Question

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

• A. $x\cos (b-a)-\sin (b-a)\log \left|\sec (x-b)\right|+c$
• B. $x\cos (b+a)-\sin (b-a)\log \left|\sec (x+b)\right|+c$
• C. $x\cos (b-a)-\sin (b+a)\log \left|\sec (x-b)\right|+c$
• D. $x\cos (b-a)-\sin (b+a)\log \left|\sec (x+b)\right|+c$

Answer 1

The correct option is A $x\cos (b-a)-\sin (b-a)\log \left|\sec (x-b)\right|+c$

$\int \frac{\cos (x-a)}{\cos (x-b)}dx=\int \frac{\cos \left[\right(x-b)+(b-a\left)\right]}{\cos (x-b)}dx$

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

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

$=\cos (b-a)\int 1dx-\sin (b-a)\int \tan (x-b)dx$

$=x\cos (b-a)+\sin (b-a)\log |\cos (x-b)|+c$ (Ans)