MHTCET 22 Apr 2025 Shift 1 Paper

Concept:Use the double-angle identity $\cos 2\theta = 2\cos^2\theta - 1$ to rewrite the numerator as a difference of squares, then cancel the common factor.Explanation:Apply the double-angle formula to $\cos 2x$ and $\cos 2\alpha$.$\cos 2x - \cos 2\alpha = (2\cos^2 x - 1) - (2\cos^2 \alpha - 1) = 2(\cos^2 x - \cos^2 \alpha)$.Now factor the difference of squares.$\cos^2 x - \cos^2 \alpha = (\cos x - \cos \alpha)(\cos x + \cos \alpha)$.Substitute this factorisation into the integral.$\int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha} dx = \int \frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha} dx$.The term $(\cos x - \cos \alpha)$ cancels from numerator and denominator.$= 2\int (\cos x + \cos \alpha) dx$.Integrate term by term. Since $\cos \alpha$ is a constant, its integral is $x\cos \alpha$.$= 2\sin x + 2x\cos \alpha + c$.Here $c$ is the constant of integration.Answer:$2\sin x + 2x\cos \alpha + c$.This corresponds to option C.