dx 1 - x - x = - Vidyadip

MCQ

$\int {\frac{{dx}}{{1 - \cos x - \sin x}}} = $

A
$\log |1 + \cot x/2| + c$
B
$\log |1 - \tan x/2| + c$
✓
$\log |1 - \cot x/2| + c$
D
$\log |1 + \tan x/2| + c$

✓

Answer

Correct option: C.

$\log |1 - \cot x/2| + c$

c
(c)$I = \int {\frac{{dx}}{{1 - \cos x - \sin x}}} $
$ = \int {\frac{{dx}}{{1 - \frac{{[(1 - {{\tan }^2}(x/2)]}}{{[(1 + {{\tan }^2}(x/2)]}} - \frac{{2\tan (x/2)}}{{1 + {{\tan }^2}(x/2)}}}}} $
$ = \int {\frac{{{{\sec }^2}(x/2).dx}}{{1 + {{\tan }^2}(x/2) - 1 + {{\tan }^2}(x/2) - 2\tan (x/2)}}} $
$ = \int {\frac{{{{\sec }^2}(x/2)\,dx}}{{2{{\tan }^2}(x/2) - 2\tan (x/2)}}} $ $ = \int {\frac{{\frac{1}{2}.{{\sec }^2}\left( {\frac{x}{2}} \right)\,dx}}{{{{\tan }^2}\left( {\frac{x}{2}} \right) - \tan \left( {\frac{x}{2}} \right)}}} $
Put $\tan (x/2) = t$==> $\frac{1}{2}{\sec ^2}(x/2)\,dx = dt$
$I = \int {\frac{{dt}}{{{t^2} - t}}} $ $ = \int {\frac{{dt}}{{t\,(t - 1)}}} $$ = \int {\left[ {\frac{1}{{t - 1}} - \frac{1}{t}} \right]\,dt} $ ( Put tan $x = t,$ $\therefore {\sec ^2}x{\rm{ }}dx = dt$)
$ = \int {\frac{{dt}}{{t - 1}} - \int {\frac{{dt}}{t}} } = \log (t - 1) - \log t + c = \log \left| {\frac{{t - 1}}{t}} \right| + c$
$ = \log \left| {\,\frac{{\tan (x/2) - 1}}{{\tan (x/2)}}\,} \right| + c$$ = \log \left| {\,1 - \cot \,\frac{x}{2}\,} \right| + c$.

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