Q.$\int \dfrac{\cos 2x-\cos 2\theta}{\cos x-\cos\theta}\,dx$ is equal to
(A) $2(\sin x + x\cos\theta) + C$
(B) $2(\sin x - x\cos\theta) + C$
(C) $2(\sin x + 2x\cos\theta) + C$
(D) $2(\sin x - 2x\cos\theta) + C$

Concept: U Substitution — simplify the integrand using trigonometric identities before substituting.

Step 1: Simplify the numerator

Use $\cos 2x - \cos 2\theta = -2\sin(x+\theta)\sin(x-\theta)$ and $\cos x - \cos\theta = -2\sin\frac{x+\theta}{2}\sin\frac{x-\theta}{2}$.

Step 2: Cancel common factors

The ratio becomes

Using $\sin(x+\theta) = 2\sin\frac{x+\theta}{2}\cos\frac{x+\theta}{2}$ and $\sin(x-\theta) = 2\sin\frac{x-\theta}{2}\cos\frac{x-\theta}{2}$, we get

$\frac{4\sin\frac{x+\theta}{2}\cos\frac{x+\theta}{2}\sin\frac{x-\theta}{2}\cos\frac{x-\theta}{2}}{2\sin\frac{x+\theta}{2}\sin\frac{x-\theta}{2}} = 2\cos\frac{x+\theta}{2}\cos\frac{x-\theta}{2}.$ …

The key to this problem is noticing that the integrand looks messy, but the numerator and denominator are both differences of cosines. That structure is a strong hint: we can use the identity $\cos 2x = 2\cos^2 x - 1$ (and similarly for $\cos 2\theta$) to rewrite everything in terms of $\cos x$ and $\cos\theta$, then factor.

Let’s walk through it.

1. Rewrite the numerator using the double-angle identity. Recall: $\cos 2u = 2\cos^2 u - 1$. So

The $-1$ terms cancel neatly — that’s the first simplification.

2. Factor the difference of squares.

Therefore

3. Cancel the denominator. The integrand becomes

provided $\cos x \neq \cos\theta$ (which is fine for indefinite integration — we treat it as an algebraic identity).

4. Integrate term by term. …

Method: Simplify a cosine-difference quotient before integrating

A quotient like $\dfrac{\cos 2x-\cos 2\theta}{\cos x-\cos\theta}$ looks hard but collapses algebraically once you expand $\cos 2x$.

Steps

Step 1: Expand with the double-angle identity.

$\cos 2x=2\cos^2 x-1$, so the numerator $\cos 2x-\cos 2\theta=2(\cos^2 x-\cos^2\theta)$; the $-1$'s cancel.

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

Common Mistakes

Mistake 1: Integrating $\cos\theta$ as if it depended on $x$.

Why it's wrong: $\theta$ is a fixed parameter, so $\int\cos\theta\,dx=x\cos\theta$, not $\sin\theta$. Correct approach: pull the constant $\cos\theta$ outside the $x$-integral.

Mistake 2: Forcing product-to-sum on the sines. …