Q.Evaluate: $\int \dfrac{\cos x-\cos 2x}{1-\cos x}\,dx$

Concept: Power Rule Integration — but first, simplify the integrand using trigonometric identities.

Step 1: Use the identity $\cos 2x = 2\cos^2 x - 1$ in the numerator:

Step 2: Factor the denominator using $1 - \cos x = 2\sin^2\frac{x}{2}$, but here it's simpler to rewrite the whole fraction:

A cleaner approach: directly simplify by splitting terms.

Step 3: Write …

1. Rewrite $\cos 2x$. Using $\cos 2x = 2\cos^2 x - 1$,

2. Factor the numerator. Since $2\cos^2 x - \cos x - 1 = (2\cos x + 1)(\cos x - 1)$,

3. Cancel with the denominator. For $\cos x \neq 1$, …

Method: Simplify a trig quotient by factoring before integrating

Use this when a trig fraction like $\dfrac{\cos x-\cos 2x}{1-\cos x}$ appears — expand the double angle, factor, and cancel so an elementary integrand remains.

Steps

Step 1: Expand any multiple angles.

so the numerator becomes $\cos x-(2\cos^2x-1)=-2\cos^2x+\cos x+1$.

Step 2: Factor the numerator.

Step 3: Cancel the common factor with the denominator. …

Common Mistakes

Mistake 1: Integrating the quotient directly.

Why it's wrong: $\dfrac{\cos x-\cos 2x}{1-\cos x}$ hides a cancellable factor; integrating as-is is intractable. Correct approach: expand $\cos 2x$, factor, and cancel first.

Mistake 2: Factoring the numerator incorrectly.

Why it's wrong: the correct factorisation is $(2\cos x+1)(1-\cos x)$; a wrong factoring won't cancel the $(1-\cos x)$. Correct approach: verify by expanding the product back. …