CUET · MATHS · PYQ PAPER 2023
\(\int \frac{\cos x - \sin x}{1 + \sin 2x} dx\) is equal to:
- A \(\frac{1}{\sin x + \cos x} + C\), where C is a constant.
- B \(\frac{-1}{\sin x + \cos x} + C\), where C is a constant.
- C \(\frac{1}{\cos x - \sin x} + C\), where C is a constant.
- D \(\frac{2}{\sin x + \cos x} + C\), where C is a constant.
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sin x + \cos x} + C\), where C is a constant.
Step-by-step Solution
Detailed explanation
Let \(u = \sin x + \cos x\).Then \(du = (\cos x - \sin x) dx\). The integral becomes \(\int \frac{\cos x - \sin x}{( \sin x + \cos x)^2} dx = \int \frac{1}{u^2} du\). \(= -\frac{1}{u} + C\)\(= -\frac{1}{\sin x + \cos x} + C\)
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