CUET · MATHS · PYQ PAPER 2025
\(\int \frac{\cos x-\sin x}{1+\sin 2 x} d x\) is equal to
- A \(\frac{1}{\cos x+\sin x}+C\), where C is constant of integration
- B \(\frac{-1}{\cos x+\sin x}+C\), where C is constant of integration
- C \(\frac{1}{1+\sin 2 x}+C\), where C is constant of integration
- D \(\frac{1}{1-\sin 2 x}+C\), where C is constant of integration
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 \(u = \sin x+\cos x\) \(d u = (\cos x-\sin x) d x\) \(\int \frac{1}{u^2} d u = -\frac{1}{u}+C\) \(= -\frac{1}{\sin x+\cos x}+C\)
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