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Question

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

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Solution

## Step 1: Separate the numeratorThe given expression is $\frac{x-\mathrm{sin}x}{1-\mathrm{cos}x}$.Let its integration be $I$.$\therefore$$I=\int \frac{x-\mathrm{sin}x}{1-\mathrm{cos}x}dx$ $=\int \frac{x}{1-\mathrm{cos}x}dx-\int \frac{\mathrm{sin}x}{1-\mathrm{cos}x}dx$Step 2: Substitute the trigonometric identitiesWe know that $1-\mathrm{cos}x=2{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)$ and $\mathrm{sin}x=2\mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)$$\therefore$$I=\int \frac{x}{2{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)}dx-\int \frac{2\mathrm{sin}\left(\frac{x}{2}\right)\mathrm{cos}\left(\frac{x}{2}\right)}{2{\mathrm{sin}}^{2}\left(\frac{x}{2}\right)}dx$ $=\frac{1}{2}\int x\mathrm{cos}e{c}^{2}\left(\frac{x}{2}\right)dx-\int cot\left(\frac{x}{2}\right)dx$Step 3: Apply Integration By partsUsing 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 $\frac{x-\mathrm{sin}x}{1-\mathrm{cos}x}$ is integrated we get$-xcot\left(\frac{x}{2}\right)+c$.

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