CUET UG 2025 — Mathematics Algebra

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

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

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

The numerator becomes:

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

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

$$= 2(\cos^2 x - \cos^2 \alpha)$$=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):

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

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The fraction becomes:

$$\frac{2(\cos x - \cos \alpha)(\cos x + \cos \alpha)}{\cos x - \cos \alpha}$$cosx−cosα2(cosx−cosα)(cosx+cosα)​

Canceling the common term $(\cos x - \cos \alpha)$(cosx−cosα):

$$= 2(\cos x + \cos \alpha)$$=2(cosx+cosα)

$$= 2\cos x + 2\cos \alpha$$=2cosx+2cosα

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The integral becomes:

$$\int (2\cos x + 2\cos \alpha) \, dx$$∫(2cosx+2cosα)dx

Since $\alpha$α is a constant:

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

$$\int 2\cos\alpha \, dx = 2x\cos\alpha$$∫2cosαdx=2xcosα

Therefore:

$$\int \left(\frac{\cos 2x - \cos 2\alpha}{\cos x - \cos \alpha}\right) dx = 2\sin x + 2x\cos\alpha + c$$∫(cosx−cosαcos2x−cos2α​)dx=2sinx+2xcosα+c