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CUET · MATHS · PYQ PAPER 2025

\(\int \frac{\cos x-\sin x}{1+\sin 2 x} d x\) is equal to

  1. A \(\frac{1}{\cos x+\sin x}+C\), where C is constant of integration
  2. B \(\frac{-1}{\cos x+\sin x}+C\), where C is constant of integration
  3. C \(\frac{1}{1+\sin 2 x}+C\), where C is constant of integration
  4. D \(\frac{1}{1-\sin 2 x}+C\), where C is constant of integration
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{-1}{\cos x+\sin x}+C\), where C is constant of integration

Step-by-step Solution

Detailed explanation

\(\int \frac{\cos x-\sin x}{(\sin x+\cos x)^2} d x\) Let \(u = \sin x+\cos x\) \(d u = (\cos x-\sin x) d x\) \(\int \frac{1}{u^2} d u = -\frac{1}{u}+C\) \(= -\frac{1}{\sin x+\cos x}+C\)