Question

Find the indefinite integral:
\[ \int \frac{x - \sin x}{1 - \cos x} \, dx \]

Hint:

Keep an eye out for integrals structured like $\int [f(x) + f'(x)] dx$. In many cases, performing integration by parts on one term creates a second term that cancels out the other half of the expression, saving you from having to compute both integrals!

Answer 1

Concept: To evaluate this trigonometric integral, we split the given fraction into two manageable parts and then apply half-angle trigonometric identities:
• $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$
• $\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$ After simplification, we can use the technique of Integration by Parts, which is defined as: \[ \int u \, dv = u v - \int v \, du \]

Step 1: Splitting and simplifying the integrand
Let $I = \int \frac{x - \sin x}{1 - \cos x} \, dx$. Splitting the numerator gives: \[ I = \int \frac{x}{1 - \cos x} \, dx - \int \frac{\sin x}{1 - \cos x} \, dx \] Now, apply the half-angle identities to both terms separately: \[ I = \int \frac{x}{2\sin^2\left(\frac{x}{2}\right)} \, dx - \int \frac{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}{2\sin^2\left(\frac{x}{2}\right)} \, dx \] Using the trigonometric definitions $\frac{1}{\sin^2\theta} = \csc^2\theta$ and $\frac{\cos\theta}{\sin\theta} = \cot\theta$: \[ I = \frac{1}{2}\int x \csc^2\left(\frac{x}{2}\right) \, dx - \int \cot\left(\frac{x}{2}\right) \, dx \quad \cdots (1) \]

Step 2: Integrating the first term using Integration by Parts
Let us focus on the first integral: $I_1 = \int x \csc^2\left(\frac{x}{2}\right) \, dx$. Using the ILATE rule, we choose:
• First function (algebraic): $u = x \implies du = dx$
• Second function (trigonometric): $dv = \csc^2\left(\frac{x}{2}\right) \, dx \implies v = \frac{-\cot\left(\frac{x}{2}\right)}{1/2} = -2\cot\left(\frac{x}{2}\right)$ Applying the integration by parts formula: \[ I_1 = x \left[-2\cot\left(\frac{x}{2}\right)\right] - \int \left[-2\cot\left(\frac{x}{2}\right)\right] \, dx \] \[ I_1 = -2x\cot\left(\frac{x}{2}\right) + 2\int \cot\left(\frac{x}{2}\right) \, dx \]

Step 3: Combining the terms back into the main equation
Now substitute this result for $I_1$ back into Equation (1): \[ I = \frac{1}{2} \left[ -2x\cot\left(\frac{x}{2}\right) + 2\int \cot\left(\frac{x}{2}\right) \, dx \right] - \int \cot\left(\frac{x}{2}\right) \, dx \] Distributing the factor of $\frac{1}{2}$ across the brackets: \[ I = -x\cot\left(\frac{x}{2}\right) + \int \cot\left(\frac{x}{2}\right) \, dx - \int \cot\left(\frac{x}{2}\right) \, dx \] Notice that the two remaining integrals cancel each other out completely: \[ I = -x\cot\left(\frac{x}{2}\right) + C \]