Question

The value of $\int \frac{x-\sin x}{1-\cos x}dx$ is equal to
(where $C$ is integration constant)

• A. $x\cot \frac{x}{2}+C$
• B. $-x\cot \frac{x}{2}+C$
• C. $2\cot \frac{x}{2}+C$
• D. $\cot \frac{x}{2}+C$

Answer 1

The correct option is B $-x\cot \frac{x}{2}+C$

$\int \frac{x-\sin x}{1-\cos x}dx=\int \frac{x}{1-\cos x}dx-\int \frac{\sin x}{1-\cos x}dx=\frac{1}{2}\int x{\text{cosec}}^2\left(\frac{x}{2}\right)dx-\int \frac{2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)}{2{\sin }^2\left(\frac{x}{2}\right)}dx\left[\begin{array}{c}\because 1-\cos x=2{\sin }^2\left(\frac{x}{2}\right)\\ \sin x=2\sin \left(\frac{x}{2}\right)\cdot \cos \left(\frac{x}{2}\right)\end{array}\right]=\int [\frac{x}{2}\cdot {\text{cosec}}^2\left(\frac{x}{2}\right)+(-\text{cot}\left(\frac{x}{2}\right))]dx$

This is of the form $\int \left[f\right(x)+x{f}^{\prime }(x\left)\right]dx$
where
$f\left(x\right)=-\cot \left(\frac{x}{2}\right);f^{\prime }\left(x\right)=\frac{1}{2}{\text{cosec}}^2\left(\frac{x}{2}\right)$
We know that,
$\int \left[f\right(x)+x{f}^{\prime }(x\left)\right]dx=xf\left(x\right)+C$
Thus,
$\int \frac{x-\sin x}{1-\cos x}dx=-x\cot \left(\frac{x}{2}\right)+C$