MCQ
$\int_{}^{} {\frac{{dx}}{{\cos (x - a)\cos (x - b)}} = } $
  • A
    ${\rm{cosec}}\,\,(a - b)\log \frac{{\sin (x - a)}}{{\sin (x - b)}} + c$
  • ${\rm{cosec}}(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + c$
  • C
    ${\rm{cosec}}(a - b)\log \frac{{\sin (x - b)}}{{\sin (x - a)}} + c$
  • D
    ${\rm{cosec}}(a - b)\log \frac{{\cos (x - b)}}{{\cos (x - a)}} + c$

Answer

Correct option: B.
${\rm{cosec}}(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + c$
b
(b)$\int_{}^{} {\frac{{dx}}{{\cos (x - a)\cos (x - b)}}} $
$ = \frac{1}{{\sin (a - b)}}\int_{}^{} {\frac{{\sin \left\{ {(x - b) - (x - a)} \right\}}}{{\cos (x - a)\,.\,\cos (x - b)}}\,dx} $
$ = \frac{1}{{\sin (a - b)}}\int_{}^{} {\left\{ {\frac{{\sin (x - b)}}{{\cos (x - b)}} - \frac{{\sin (x - a)}}{{\cos (x - a)}}} \right\}dx} $
$ = {\rm{cosec}}\,(a - b)\log \frac{{\cos (x - a)}}{{\cos (x - b)}} + c$.

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