∫≤ft((cos x-sin x)/(1+sin 2x))dx is equal to

The correct answer is Option (2) → \(\displaystyle\frac{-1}{\cos x+\sin x}+C\), where C is constant of integration

Since \(1 = \sin^2 x + \cos^2 x\) and \(\sin 2x = 2\sin x\cos x\), the denominator is a perfect square:

    \(1 + \sin 2x = \cos^2 x + \sin^2 x + 2\sin x\cos x = (\cos x + \sin x)^2\)

So the integral is

    \( \displaystyle \int \frac{\cos x - \sin x}{(\cos x + \sin x)^2}\,dx \)

The numerator is the derivative of \(\cos x + \sin x\). Substituting \(t = \cos x + \sin x\):

    \(dt = (\cos x - \sin x)\,dx\)

    \( \displaystyle \int \frac{dt}{t^2} = -\frac{1}{t} + C \)

    \( \displaystyle = \frac{-1}{\cos x + \sin x} + C \)

NCERT (2026-27 edition):
Integration Using Trigonometric Identities