Q.$\dfrac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}$

Step 1. Factor the numerator. $\cos2x-\cos2\alpha=-2\sin(x+\alpha)\sin(x-\alpha)$.

Step 2. Factor the denominator. $\cos x-\cos\alpha=-2\sin\dfrac{x+\alpha}2\sin\dfrac{x-\alpha}2$.

Step 3. Write $\sin(x\pm\alpha)$ via half-angles. $\sin(x+\alpha)=2\sin\dfrac{x+\alpha}2\cos\dfrac{x+\alpha}2$ and $\sin(x-\alpha)=2\sin\dfrac{x-\alpha}2\cos\dfrac{x-\alpha}2$, so the numerator becomes $-8\sin\dfrac{x+\alpha}2\cos\dfrac{x+\alpha}2\sin\dfrac{x-\alpha}2\cos\dfrac{x-\alpha}2$.

Step 4. Divide by the denominator. The $\sin\dfrac{x+\alpha}2\sin\dfrac{x-\alpha}2$ factors cancel, leaving $4\cos\dfrac{x+\alpha}2\cos\dfrac{x-\alpha}2$. …

• Forgetting $\alpha$ is a constant, so $2\cos\alpha$ integrates to $2x\cos\alpha$, not $2\sin\alpha$ …