First slide

Introduction to integration

Question

$Evaluate \int \frac{x - \sin x}{1 - \cos x} d x$

Moderate

A

$- 2 x \cot \frac{x}{2} + C$
B

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

$- \frac{x}{2} \cot \frac{x}{2} + C$
D

None of these

Solution

$\begin{array}{l} Solution. \int \frac{x - \sin x}{1 - \cos x} d x = \int \frac{x}{1 - \cos x} - \int \frac{\sin x}{1 - \cos x} d x \\ = \int \frac{x}{2} {cosec}^{2} \frac{x}{2} d x - \int \frac{2 \sin x / 2 \cos x / 2}{2 \sin^{2} x / 2} d x \\ = \frac{1}{2} \int_{} x {cosec}_{}^{2} \frac{x}{2} - \int \cot \frac{x}{2} d x \\ \begin{aligned} = \frac{1}{2} \{x (- 2 \cot \frac{x}{2}) - \int 1 (- 2 \cot \frac{x}{2}) d x\} \end{aligned} - \int \cot \frac{x}{2} d x + C \\ = - x \cot \frac{x}{2} + \int \cot \frac{x}{2} d x - \int \cot \frac{x}{2} d x + C \\ = - x \cot \frac{x}{2} + C \end{array}$

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