MCQ
$\int \frac{x-\sin x}{1-\cos x} \cdot d x=$
  • A
    $x \cot \left(\frac{x}{2}\right)+ C$
  • $- x \cot \left(\frac{x}{2}\right)+ c$
  • C
    $\cot \left(\frac{x}{2}\right)+c$
  • D
    $x \tan \left(\frac{x}{2}\right)+ C$

Answer

Correct option: B.
$- x \cot \left(\frac{x}{2}\right)+ c$
$- x \cot \left(\frac{x}{2}\right)+ c$
Hint :
$\int \frac{x-\sin x}{1-\cos x} d x=\int \frac{x-2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{2 \sin ^2\left(\frac{x}{2}\right)} d x$
$= \frac{1}{2} \int x \operatorname{cosec}^2\left(\frac{x}{2}\right) d x-\int \cot \left(\frac{x}{2}\right) d x$
$ =\frac{1}{2}\left[x \int \operatorname{cosec}^2\left(\frac{x}{2}\right) d x-\int\left[\frac{d}{d x}(x) \int \operatorname{cosec}^2\left(\frac{x}{2}\right) d x\right] d x\right. \\ -\int \cot \left(\frac{x}{2}\right) d x$
$=\frac{1}{2}\left[x\left\{\frac{-\cot \left(\frac{x}{2}\right)}{\left(\frac{1}{2}\right)}\right\}-\int 1 \cdot \frac{-\cot \left(\frac{x}{2}\right)}{\left(\frac{1}{2}\right)} d x-\int \cot \left(\frac{x}{2}\right) d x\right.]$
$=-x \cot \left(\frac{x}{2}\right)+\int \cot \left(\frac{x}{2}\right) d x-\int \cot \left(\frac{x}{2}\right) d x$
$=-x \cot \left(\frac{x}{2}\right)+c.$

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