Question

Find : \( \int \frac{x - \sin x}{1 - \cos x} \, dx \)

Hint:

Splitting complex fractions involving \(1 \pm \cos x\) into simpler parts often leads to integrals that cancel out using integration by parts.
Look for the pattern \( \int [f(x) + xf'(x)] dx = xf(x) + C \). Here, if \( f(x) = -\cot(x/2) \), then \( f'(x) = \frac{1}{2}\text{cosec}^2(x/2) \).

Answer 1

Concept:
• Half-angle identities: \( 1 - \cos x = 2\sin^2(x/2) \) and \( \sin x = 2\sin(x/2)\cos(x/2) \).
• Integration by parts: \( \int u \, dv = uv - \int v \, du \).
• Basic integral: \( \int \text{cosec}^2(ax) \, dx = -\frac{1}{a}\cot(ax) \).

Step 1: Simplify the integrand using trigonometric identities
Let \( I = \int \frac{x - \sin x}{1 - \cos x} \, dx \). Substitute \( 1 - \cos x = 2\sin^2(x/2) \) and \( \sin x = 2\sin(x/2)\cos(x/2) \): \[ I = \int \left( \frac{x}{2\sin^2(x/2)} - \frac{2\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} \right) \, dx \] \[ I = \int \left( \frac{x}{2}\text{cosec}^2(x/2) - \cot(x/2) \right) \, dx \]

Step 2: Split the integral and apply integration by parts to the first term
\[ I = \int \frac{x}{2}\text{cosec}^2(x/2) \, dx - \int \cot(x/2) \, dx \] For the first integral \( I_1 = \int \frac{x}{2}\text{cosec}^2(x/2) \, dx \), let \( u = x \) and \( dv = \frac{1}{2}\text{cosec}^2(x/2) \, dx \). Then \( du = dx \) and \( v = \int \frac{1}{2}\text{cosec}^2(x/2) \, dx = -\cot(x/2) \). Applying integration by parts: \[ I_1 = x(-\cot(x/2)) - \int (-\cot(x/2)) \, dx \] \[ I_1 = -x\cot(x/2) + \int \cot(x/2) \, dx \]

Step 3: Combine the results
Substituting the value of \( I_1 \) back into the expression for \( I \): \[ I = \left( -x\cot(x/2) + \int \cot(x/2) \, dx \right) - \int \cot(x/2) \, dx \] The integral terms cancel out: \[ I = -x\cot(x/2) + C \]