Q.Integrate the following function: $\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}$

The key idea is to use the cosine double-angle identity to simplify the numerator, then factor and cancel the denominator.

First, rewrite $\cos 2x$ and $\cos 2\alpha$ using $\cos 2\theta = 2\cos^2\theta - 1$:

Factor the difference of squares:

The integrand becomes: …

We start with the integral

The presence of $\cos 2x$ and $\cos 2\alpha$ suggests using the double-angle identity:

This identity is often more useful here than $\cos 2\theta = 1 - 2\sin^2 \theta$ because the denominator involves $\cos x$ and $\cos \alpha$, so expressing everything in terms of cosines will let us factor cleanly.

1. Rewrite the numerator using the double-angle identity

The $-1$ and $+1$ cancel, leaving a simple difference of squares.

2. Factor the numerator

3. Cancel the common factor with the denominator The denominator is $\cos x - \cos \alpha$. Provided $\cos x \neq \cos \alpha$ (the integrand is undefined at those points, but we integrate over intervals where it is defined), we cancel: $\frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} = 2(\cos x + \cos \alpha).$ …

Method: Factor a difference of cosines via the double-angle identity, then cancel

When both numerator and denominator are differences of cosines, expand the double angles so the numerator factors and the denominator cancels, leaving something elementary.

Steps

Step 1: Expand the double-angle terms.

so the numerator $\cos 2x-\cos 2\alpha=2(\cos^2 x-\cos^2\alpha)$.

Step 2: Factor as a difference of squares and cancel.

Dividing by $(\cos x-\cos\alpha)$ leaves $2(\cos x+\cos\alpha)$. …

Common Mistakes

Mistake 1: Trying to cancel $\cos 2x$ against $\cos x$ directly.

Why it's wrong: $\cos 2x$ and $\cos x$ are different functions; the denominator only cancels after $\cos 2x$ is rewritten via the double-angle identity. Correct approach: use $\cos 2x=2\cos^2 x-1$ so the numerator factors as $2(\cos x-\cos\alpha)(\cos x+\cos\alpha)$.

Mistake 2: Treating $\cos\alpha$ as a variable and mis-integrating $2\cos\alpha$. …