_ ^ x - x 1 - x dx = - Vidyadip

MCQ

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

A
$x\cot \frac{x}{2} + c$
✓
$ - x\cot \frac{x}{2} + c$
C
$\cot \frac{x}{2} + c$
D
None of these

✓

Answer

Correct option: B.

$ - x\cot \frac{x}{2} + c$

b
(b)$\int_{}^{} {\frac{{x - \sin x}}{{1 - \cos x}}\,dx} = \int_{}^{} {\frac{x}{{1 - \cos x}}\,dx} - \int_{}^{} {\frac{{\sin x}}{{1 - \cos x}}\,dx} $
$ = \frac{1}{2}\int_{}^{} {x\,{\rm{cose}}{{\rm{c}}^{\rm{2}}}\left( {\frac{x}{2}} \right)\,dx} - \int_{}^{} {\frac{{2\sin (x2)\cos (x2)}}{{2{{\sin }^2}(x2)}}\,dx} $
$ = \frac{1}{2}\int_{}^{} {x\,{\rm{cose}}{{\rm{c}}^{\rm{2}}}\left( {\frac{x}{2}} \right)\,dx} - \int_{}^{} {\cot \left( {\frac{x}{2}} \right)\,dx} $$ = - x\cot \left( {\frac{x}{2}} \right) + c$.

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