CUET UG 2025 — Mathematics Calculus

The integral is $\displaystyle \int \frac{\cos x - \sin x}{1 + \sin 2x} dx$∫1+sin2xcosx−sinx​dx

Using the identity $\sin 2x = 2\sin x \cos x$sin2x=2sinxcosx:

$$1 + \sin 2x = 1 + 2\sin x \cos x$$1+sin2x=1+2sinxcosx

Since $\sin^2 x + \cos^2 x = 1$sin2x+cos2x=1:

$$1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x \cos x$$1+sin2x=sin2x+cos2x+2sinxcosx

$$= (\sin x + \cos x)^2$$=(sinx+cosx)2

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The integral becomes:

$$\int \frac{\cos x - \sin x}{(\sin x + \cos x)^2} dx$$∫(sinx+cosx)2cosx−sinx​dx

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Let $t = \sin x + \cos x$t=sinx+cosx

Differentiating both sides:

$$dt = (\cos x - \sin x) dx$$dt=(cosx−sinx)dx

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Substituting into the integral:

$$\int \frac{1}{t^2} dt$$∫t21​dt

$$= \int t^{-2} dt$$=∫t−2dt

$$= \frac{t^{-1}}{-1} + C$$=−1t−1​+C

$$= -\frac{1}{t} + C$$=−t1​+C

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Substituting back $t = \sin x + \cos x$t=sinx+cosx:

$$-\frac{1}{\sin x + \cos x} + C$$−sinx+cosx1​+C

Therefore, the answer is $\frac{-1}{\cos x + \sin x} + C$cosx+sinx−1​+C