dx (x - a) (x - b) is - Vidyadip

MCQ

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

✓
$\frac{1}{{\sin (a - b)}}\log \left| {\frac{{\sin (x - a)}}{{\sin (x - b)}}} \right| + c$
B
$\frac{{ - 1}}{{\sin (a - b)}}\log \left| {\frac{{\sin (x - a)}}{{\sin (x - b)}}} \right| + c$
C
$\log \sin (x - a)\sin (x - b) + c$
D
$\log \left| {\frac{{\sin (x - a)}}{{\sin (x - b)}}} \right|$

✓

Answer

Correct option: A.

$\frac{1}{{\sin (a - b)}}\log \left| {\frac{{\sin (x - a)}}{{\sin (x - b)}}} \right| + c$

a
(a) Let $ I =$ $\int {\frac{{dx}}{{\sin (x - a)\sin (x - b)}}} $
$ = \frac{1}{{\sin (a - b)}}\int {\frac{{\sin \left\{ {(x - b) - (x - a)} \right\}}}{{\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)}}\left[ {\int {\cot (x - a)dx - \int {\cot (x - b)dx} } } \right]$
$ = \frac{1}{{\sin (a - b)}}\;\left[ {\log \sin (x - a) - \log \sin (x - b)} \right] + c$
$ I =$ $ = \frac{1}{{\sin (a - b)}}\;\log \left| {\frac{{\sin (x - a)}}{{\sin (x - b)}}} \right| + c$.

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