Question

Integrate $\frac{x-\sin x}{1-\cos x}$.

Answer 1

Step 1: Separate the numerator

The given expression is $\frac{x-\sin x}{1-\cos x}$.

Let its integration be $I$.

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

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

Step 2: Substitute the trigonometric identities

We know that $1-\cos x=2{\sin }^2\left(\frac{x}{2}\right)$ and $\sin x=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)$

$\therefore$$I=\int \frac{x}{2{\sin }^2\left({\displaystyle \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$

$=\frac{1}{2}\int x\cos e{c}^2\left(\frac{x}{2}\right)dx-\int cot\left(\frac{x}{2}\right)dx$

Step 3: Apply Integration By parts

Using integration by parts where $\int u.vdx=u\int vdx-\int u\text{'}\left(\int vdx\right)dx$ we get,

$I=\frac{1}{2}\left[x.2\left\{-cot\left(\frac{x}{2}\right)\right\}-\int 1.\left(-2\right)cot\left(\frac{x}{2}\right)dx\right]-\int cot\left(\frac{x}{2}\right)dx$

$=-xcot\left(\frac{x}{2}\right)+\int cot\left(\frac{x}{2}\right)dx-\int cot\left(\frac{x}{2}\right)dx$

$=-xcot\left(\frac{x}{2}\right)+c$

where $c$ is the constant of integration.

Hence, when $\frac{x-\sin x}{1-\cos x}$ is integrated we get$-xcot\left(\frac{x}{2}\right)+c$.