CUET UG 2025 — Mathematics Algebra

The integral to evaluate is $\displaystyle \int \frac{\cos 2x - \cos 2α}{\cos x - \cos α} dx$∫cosx−cosαcos2x−cos2α​dx, where α is a constant.

Using the double angle formula $\cos 2θ = 2\cos^2 θ - 1$cos2θ=2cos2θ−1:

$$\cos 2x = 2\cos^2 x - 1$$cos2x=2cos2x−1

$$\cos 2α = 2\cos^2 α - 1$$cos2α=2cos2α−1

The numerator becomes:

$$\cos 2x - \cos 2α = (2\cos^2 x - 1) - (2\cos^2 α - 1)$$cos2x−cos2α=(2cos2x−1)−(2cos2α−1)

$$= 2\cos^2 x - 1 - 2\cos^2 α + 1$$=2cos2x−1−2cos2α+1

$$= 2\cos^2 x - 2\cos^2 α$$=2cos2x−2cos2α

$$= 2(\cos^2 x - \cos^2 α)$$=2(cos2x−cos2α)

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Using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$a2−b2=(a−b)(a+b):

$$2(\cos^2 x - \cos^2 α) = 2(\cos x - \cos α)(\cos x + \cos α)$$2(cos2x−cos2α)=2(cosx−cosα)(cosx+cosα)

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The integral becomes:

$$\int \frac{2(\cos x - \cos α)(\cos x + \cos α)}{\cos x - \cos α} dx$$∫cosx−cosα2(cosx−cosα)(cosx+cosα)​dx

$$= \int 2(\cos x + \cos α) dx$$=∫2(cosx+cosα)dx

$$= \int 2\cos x \, dx + \int 2\cos α \, dx$$=∫2cosxdx+∫2cosαdx

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Evaluating each term:

$$\int 2\cos x \, dx = 2\sin x$$∫2cosxdx=2sinx

$\displaystyle \int 2\cos α \, dx = 2x\cos α$∫2cosαdx=2xcosα (since α is a constant)

Therefore:

$$\int \frac{\cos 2x - \cos 2α}{\cos x - \cos α} dx = 2\sin x + 2x\cos α + C$$∫cosx−cosαcos2x−cos2α​dx=2sinx+2xcosα+C